Rd Sharma Xi 2020 2021 _volume 2 for Class 11 Science Maths Chapter 28 - Introduction To 3 D Coordinate Geometry

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Rd Sharma Xi 2020 2021 _volume 2 Solutions for Class 11 Science Maths Chapter 28 Introduction To 3 D Coordinate Geometry are provided here with simple step-by-step explanations. These solutions for Introduction To 3 D Coordinate Geometry are extremely popular among Class 11 Science students for Maths Introduction To 3 D Coordinate Geometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 2 Book of Class 11 Science Maths Chapter 28 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 2 Solutions. All Rd Sharma Xi 2020 2021 _volume 2 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.

Page No 28.10:

Question 9:

Show that the points (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of an isosceles right-angled triangle.

Answer:

Let A(0, 7, 10), B( $-$ 1, 6, 6), C( $-$ 4, 9, 6) be the coordinates of △ABC
Then, we have:
$AB = \sqrt{{(0 + 1)}^{2} + {(7 - 6)}^{2} + {(10 - 6)}^{2}} = \sqrt{18} = 3 \sqrt{2}$
$BC = \sqrt{{(- 1 + 4)}^{2} + {(6 - 9)}^{2} + {(6 - 6)}^{2}} = \sqrt{18} = 3 \sqrt{2}$
$AC = \sqrt{{(0 + 4)}^{2} + {(7 - 9)}^{2} + {(10 - 6)}^{2}} = \sqrt{36} = 6 Clearly, {AB}^{2} + {BC}^{2} = {AC}^{2} \Rightarrow ∠ ABC = 90 ° & AB = BC$
Hence △ABC is a right-angled isosceles triangle.

Page No 28.10:

Question 10:

Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.

Answer:

Let A(3,3,3) , B(0,6,3) , C( 1,7,7) and D (4,4,7) are the vertices of quadrilateral $□ A B C D$

We have :

AB = $\sqrt{{(0 - 3)}^{2} + {(6 - 3)}^{2} + {(3 - 3)}^{2}}$
      $= \sqrt{{(- 3)}^{2} + {(3)}^{2} + {(0)}^{2}} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2}$

BC = $\sqrt{{(1 - 0)}^{2} + {(7 - 6)}^{2} + {(7 - 3)}^{2}}$
      $= \sqrt{{(1)}^{2} + {(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

CD = $\sqrt{{(4 - 1)}^{2} + {(4 - 7)}^{2} + {(7 - 7)}^{2}}$
      $= \sqrt{{(3)}^{2} + {(- 3)}^{2} + {(0)}^{2}} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3 \sqrt{2}$

DA = $\sqrt{{(4 - 3)}^{2} + {(4 - 3)}^{2} + {(7 - 3)}^{2}}$
     $= \sqrt{{(1)}^{2} + {(1)}^{2} + {(4)}^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

AB = BC = CD = DA

AC = $\sqrt{{(1 - 3)}^{2} + {(7 - 3)}^{2} + {(7 - 3)}^{2}}$
      $= \sqrt{{(- 2)}^{2} + {(4)}^{2} + {(4)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$

BD = $\sqrt{{(4 - 0)}^{2} + {(4 - 6)}^{2} + {(7 - 3)}^{2}}$
     $= \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(4)}^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
$∴$ AC = BD
Since, all sides and diagonals of quadrilateral $□ A B C D$ are equal
Therefore, the points are the vertices of a square.

Page No 28.10:

Question 11:

Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.

Answer:

Let A(1,3,0),B( $-$ 5,5,2), C( $-$ 9, $-$ 1,2) and D( $-$ 3, $-$ 3,0) be the coordinates of quadrilateral $□ A B C D$

$A B = \sqrt{{(- 5 - 1)}^{2} + {(5 - 3)}^{2} + {(2 - 0)}^{2}} = \sqrt{36 + 4 + 4} = \sqrt{44} B C = \sqrt{{(- 9 + 5)}^{2} + {(- 1 - 5)}^{2} + {(2 - 2)}^{2}} = \sqrt{16 + 36 + 0} = \sqrt{52} C D = \sqrt{{(- 3 + 9)}^{2} + {(- 3 + 1)}^{2} + {(0 - 2)}^{2}} = \sqrt{36 + 4 + 4} = \sqrt{44} D A = \sqrt{{(1 + 3)}^{2} + {(3 + 3)}^{2} + {(0 - 0)}^{2}} = \sqrt{16 + 36 + 0} = \sqrt{52} Here, we see that A B = C D & B C = D A$
Since, opposite pair of sides are equal .

Therefore, $□ A B C D$ is a parallelogram .

$A C = \sqrt{{(- 9 - 1)}^{2} + {(- 1 - 3)}^{2} + {(2 - 0)}^{2}} = \sqrt{{(- 10)}^{2} + {(- 4)}^{2} + {(2)}^{2}} = \sqrt{100 + 16 + 4} = \sqrt{120}$

$B D = \sqrt{{(- 3 + 5)}^{2} + {(- 3 - 5)}^{2} + {(0 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(- 8)}^{2} + {(- 2)}^{2}} = \sqrt{4 + 64 + 4} = \sqrt{72}$

$∴ A C \neq B D$
∴ ABCD is not a rectangle.

Page No 28.10:

Question 12:

Show that the points A(1, 3, 4), B(–1, 6, 10), C(–7, 4, 7) and D(–5, 1, 1) are the vertices of a rhombus.

Answer:

Let A(1,3,4) , B( $-$ 1,6,10) , C( $-$ 7,4,7) and D ( $-$ 5,1,1) be the vertices of quadrilateral $□ A B C D$
$A B = \sqrt{{(- 1 - 1)}^{2} + {(6 - 3)}^{2} + {(10 - 4)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 B C = \sqrt{{(- 7 + 1)}^{2} + {(4 - 6)}^{2} + {(7 - 10)}^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 C D = \sqrt{{(- 5 + 7)}^{2} + {(1 - 4)}^{2} + {(1 - 7)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 D A = \sqrt{{(1 + 5)}^{2} + {(3 - 1)}^{2} + {(4 - 1)}^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 ∴ A B = B C = C D = D A$

Hence, ABCD is a rhombus.

Page No 28.10:

Question 13:

Prove that the tetrahedron with vertices at the points O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0) is a regular one.

Answer:

The faces of a regular tetrahedron are equilateral triangles.

Let us consider $△$ OAB
$O A = \sqrt{{(0 - 0)}^{2} + {(0 - 1)}^{2} + {(0 - 1)}^{2}} = \sqrt{2}$
$O B = \sqrt{{(1 - 0)}^{2} + {(0 - 0)}^{2} + {(1 - 0)}^{2}} = \sqrt{2}$
$A B = \sqrt{{(1 - 0)}^{2} + {(0 - 1)}^{2} + {(1 - 1)}^{2}} = \sqrt{2}$
Hence, this face is an equilateral triangle.
Similarly, $△$ OBC, $△$ OAC, $△$ ABC all are equilateral triangles.
Hence, the tetrahedron is a regular one.

Page No 28.10:

Question 14:

Show that the points (3, 2, 2), (–1, 4, 2), (0, 5, 6), (2, 1, 2) lie on a sphere whose centre is (1, 3, 4). Find also its radius.

Answer:

Let the points be A (3, 2, 2), B ( $-$ 1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere
whose centre be P (1, 3, 4).
Since, AP, BP, CP and DP are radii.
Hence, AP = BP = CP = DP

$A P = \sqrt{{(1 - 3)}^{2} + {(3 - 2)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(- 2)}^{2} + {(1)}^{2} + {(2)}^{2}} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 B P = \sqrt{{(1 + 1)}^{2} + {(3 - 4)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(2)}^{2}} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 C P = \sqrt{{(1 - 0)}^{2} + {(3 - 5)}^{2} + {(4 - 6)}^{2}} = \sqrt{{(1)}^{2} + {(- 2)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 D P = \sqrt{{(1 - 2)}^{2} + {(3 - 1)}^{2} + {(4 - 2)}^{2}} = \sqrt{{(- 1)}^{2} + {(2)}^{2} + {(2)}^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Here, we see that AP = BP = CP = DP
Hence, A (3, 2, 2), B ( $-$ 1, 4, 2), C (0, 5, 6) and D (2, 1, 2) lie on the sphere whose radius is 3 .

Page No 28.10:

Question 15:

Find the coordinates of the point which is equidistant from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).

Answer:

Let P (x, y, z) be the required point which is equidistant from the points O(0,0,0), A(2,0,0)
B(0,3,0) and C(0,0,8)

Then,
OP = AP
$\Rightarrow O P^{2} = A P^{2}$
$∴ x^{2} + y^{2} + z^{2} = {(x - 2)}^{2} + y^{2} + z^{2} \Rightarrow x^{2} = {(x - 2)}^{2} \Rightarrow x^{2} = x^{2} - 4 x + 4 \Rightarrow 4 x = 4 \Rightarrow x = \frac{4}{4} ∴ x = 1$

Similarly, we have:
OP = BP
$\Rightarrow O P^{2} = B P^{2}$
$∴ x^{2} + y^{2} + z^{2} = x^{2} + {(y - 3)}^{2} + z^{2} \Rightarrow y^{2} = y^{2} - 6 y + 9 \Rightarrow 6 y = 9 \Rightarrow y = \frac{9}{6} ∴ y = \frac{3}{2}$

Similarly, we also have:

OP = CP
$\Rightarrow O P^{2} = C P^{2}$
$\Rightarrow x^{2} + y^{2} + z^{2} = x^{2} + y^{2} + {(z - 8)}^{2} \Rightarrow z^{2} = z^{2} - 16 z + 64 \Rightarrow 16 z = 64 \Rightarrow z = \frac{64}{16} ∴ z = 4$

Thus, the required point is P $(1, \frac{3}{2}, 4)$ .

Page No 28.10:

Question 16:

If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.

Answer:

Let coordinates of point P be (x, y, z).
Given:
3PA = 2PB

$\Rightarrow 3 (\sqrt{{(x + 2)}^{2} + {(y - 2)}^{2} + {(z - 3)}^{2}}) = 2 (\sqrt{{(x - 13)}^{2} + {(y + 3)}^{2} + {(z - 13)}^{2}}) \Rightarrow 3 (\sqrt{x^{2} + 4 x + 4 + y^{2} - 4 y + 4 + z^{2} - 6 z + 9}) = 2 (\sqrt{x^{2} - 26 x + 169 + y^{2} + 6 y + 9 + z^{2} - 26 z + 169}) Squaring both sides, \Rightarrow 9 (x^{2} + y^{2} + z^{2} + 4 x - 4 y - 6 z + 17) = 4 (x^{2} + y^{2} + z^{2} - 26 x + 6 y - 26 z + 347) \Rightarrow 9 x^{2} + 9 y^{2} + 9 z^{2} + 36 x - 36 y - 54 z + 153 = 4 x^{2} + 4 y^{2} + 4 z^{2} - 104 x + 24 y - 104 z + 1388 \Rightarrow 5 x^{2} + 5 y^{2} + 5 z^{2} + 140 x - 60 y + 50 z - 1235 = 0 \Rightarrow 5 (x^{2} + y^{2} + z^{2}) + 140 x - 60 y + 50 z - 1235 = 0 ∴ 5 (x^{2} + y^{2} + z^{2}) + 140 x - 60 y + 50 z - 1235 = 0 is the locus of the point P .$

Page No 28.10:

Question 17:

Find the locus of P if PA² + PB²= 2k², where A and B are the points (3, 4, 5) and (–1, 3, –7).

Answer:

Let P (x, y, z) be the point if $P A^{2} + P B^{2} = 2 k^{2}$

$\Rightarrow {(\sqrt{{(x - 3)}^{2} + {(y - 4)}^{2} + {(z - 5)}^{2}})}^{2} + {(\sqrt{{(x + 1)}^{2} + {(y - 3)}^{2} + {(z + 7)}^{2}})}^{2} = 2 k^{2} \Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 + z^{2} - 10 z + 25 + x^{2} + 2 x + 1 + y^{2} - 6 y + 9 + z^{2} + 14 z + 49 = 2 k^{2} \Rightarrow 2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - 2 k^{2} = 0$

Hence, $2 x^{2} + 2 y^{2} + 2 z^{2} - 4 x - 14 y + 4 z + 109 - 2 k^{2} = 0$ is the locus of point P.

Page No 28.10:

Question 18:

Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.

Answer:

Let A(a,b,c) , B(b,c,a) and C(c,a,b) be the vertices of $△ A B C$ . Then,

AB = $\sqrt{{(b - a)}^{2} + {(c - b)}^{2} + {(a - c)}^{2}}$
     $= \sqrt{b^{2} - 2 a b + a^{2} + c^{2} - 2 b c + b^{2} + a^{2} - 2 c a + c^{2}} = \sqrt{2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a} = \sqrt{2 (a^{2} + b^{2} + c^{2} - a b - b c - c a)}$

BC = $\sqrt{{(c - b)}^{2} + {(a - c)}^{2} + {(b - a)}^{2}}$
     $= \sqrt{c^{2} - 2 b c + b^{2} + a^{2} - 2 c a + c^{2} + b^{2} - 2 a b + a^{2}} = \sqrt{2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a} = \sqrt{2 (a^{2} + b^{2} + c^{2} - a b - b c - c a)}$

CA = $\sqrt{{(a - c)}^{2} + {(b - a)}^{2} + {(c - b)}^{2}}$
     $= \sqrt{a^{2} - 2 c a + c^{2} + b^{2} - 2 a b + a^{2} + c^{2} - 2 b c + b^{2}} = \sqrt{2 a^{2} + 2 b^{2} + 2 c^{2} - 2 a b - 2 b c - 2 c a} = \sqrt{2 (a^{2} + b^{2} + c^{2} - a b - b c - c a)}$
$∴$ AB = BC = CA
Therefore, $△ A B C$ is an equilateral triangle.

Page No 28.10:

Question 19:

Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?

Answer:

Let A(3,6,9), B(10,20,30) and C( 25, $-$ 41,5) are vertices of $△ A B C$

AB = $\sqrt{{(10 - 3)}^{2} + {(20 - 6)}^{2} + {(30 - 9)}^{2}}$
     $= \sqrt{{(7)}^{2} + {(14)}^{2} + {(21)}^{2}} = \sqrt{49 + 196 + 441} = \sqrt{686} = 7 \sqrt{14}$

BC = $\sqrt{{(25 - 10)}^{2} + {(- 41 - 20)}^{2} + {(5 - 30)}^{2}}$
     $= \sqrt{{(15)}^{2} + {(- 61)}^{2} + {(- 25)}^{2}} = \sqrt{225 + 3721 + 625} = \sqrt{4571}$

CA= $\sqrt{{(3 - 25)}^{2} + {(6 + 41)}^{2} + {(9 - 5)}^{2}}$
     $= \sqrt{{(- 22)}^{2} + {(47)}^{2} + {(- 4)}^{2}} = \sqrt{484 + 2209 + 16} = \sqrt{2709} = 3 \sqrt[]{301}$

$A B^{2} + B C^{2} = {(7 \sqrt{14})}^{2} + {(\sqrt{4571})}^{2} = 686 + 4571 = 5257 C A^{2} = 2709 ∴ A B^{2} + B C^{2} \neq C A^{2}$

A triangle $△ A B C$ is right-angled at B if $C A^{2} = A B^{2} + B C^{2}$ .
But, $C A^{2}$ ≠ $A B^{2} + B C^{2}$
Hence, the points are not vertices of a right-angled triangle.

Page No 28.10:

Question 20:

Verify the following:
(i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
(ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
(iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
(iv) (5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.

Answer:

(i) Let A(0, 7, $-$ 10) , B(1, 6, $-$ 6) , C(4, 9, $-$ 6) be the vertices of $△ A B C$ .Then,
AB = $\sqrt{{(1 - 0)}^{2} + {(6 - 7)}^{2} + {(- 6 + 10)}^{2}}$
       $= \sqrt{1^{2} + {(- 1)}^{2} + 4^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

BC = $\sqrt{{(4 - 1)}^{2} + {(9 - 6)}^{2} + {(- 6 + 6)}^{2}}$
        $= \sqrt{3^{2} + 3^{2} + 0} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$

CA= $\sqrt{{(0 - 4)}^{2} + {(7 - 9)}^{2} + {(- 10 + 6)}^{2}}$
     $= \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Clearly, AB = BC
Thus, the given points are the vertices of an isosceles triangle.

(ii) Let A(0,7,10) , B( $-$ 1,6,6) and C( $-$ 4,9,6) be the vertices of $△ A B C$ . Then ,

AB = $\sqrt{{(- 1 - 0)}^{2} + {(6 - 7)}^{2} + {(6 - 10)}^{2}}$

      $= \sqrt{{(- 1)}^{2} + {(- 1)}^{2} + {(- 4)}^{2}} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3 \sqrt{2}$

BC = $\sqrt{{(- 4 + 1)}^{2} + {(9 - 6)}^{2} + {(6 - 6)}^{2}}$
      $= \sqrt{{(- 3)}^{2} + {(3)}^{2} + 0} = \sqrt{9 + 9} = \sqrt{18} = 3 \sqrt{2}$

AC = $\sqrt{{(- 4 - 0)}^{2} + {(9 - 7)}^{2} + {(6 - 10)}^{2}}$
      $= \sqrt{{(- 4)}^{2} + {(2)}^{2} + {(- 4)}^{2}} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
$A C^{2}$ $= A B^{2} + B C^{2}$
Thus, the given points are the vertices of a right-angled triangle.

(iii) Let A( $-$ 1, 2, 1) , B(1, $-$ 2, 5) , C(4, $-$ 7, 8), D(2, $-$ 3, 4) be the vertices of quadrilateral $□ A B C D$

    $A B = \sqrt{{(1 + 1)}^{2} + {(- 2 - 2)}^{2} + {(5 - 1)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 B C = \sqrt{{(4 - 1)}^{2} + {(- 7 + 2)}^{2} + {(8 - 5)}^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43} C D = \sqrt{{(2 - 4)}^{2} + {(- 3 + 7)}^{2} + {(4 - 8)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 D A = \sqrt{{(- 1 - 2)}^{2} + {(2 + 3)}^{2} + {(1 - 4)}^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43} ∴ A B = C D and B C = D A$
Since, each pair of opposite sides are equal.
Thus, quadrilateral $□ A B C D$ is a parallelogram.

(iv) Let A(5, $-$ 1, 1) , B(7, $-$ 4, 7) , C(1, $-$ 6, 10), D( $-$ 1, $-$ 3, 4) be the vertices of quadrilateral $□ A B C D$

    $AB = \sqrt{{(5 - 7)}^{2} + {(- 1 + 4)}^{2} + {(1 - 7)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 BC = \sqrt{{(7 - 1)}^{2} + {(- 4 + 6)}^{2} + {(7 - 10)}^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 CD = \sqrt{{(1 + 1)}^{2} + {(- 6 + 3)}^{2} + {(10 - 4)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 DA = \sqrt{{(- 1 - 5)}^{2} + {(- 3 + 1)}^{2} + {(4 - 1)}^{2}} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7 ∴ AB = BC = CD = DA$
Since, all the sides are equal.
Thus, quadrilateral ABCS is a rhombus.

Page No 28.10:

Question 21:

Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).

Answer:

Let P (x, y, z) be any point that is equidistant from the points A (1, 2, 3) and B (3, 2, −1).
Then, we have:
PA = PB
$\Rightarrow \sqrt{{(x - 1)}^{2} + {(y - 2)}^{2} + {(z - 3)}^{2}} = \sqrt{{(x - 3)}^{2} + {(y - 2)}^{2} + {(z + 1)}^{2}} \Rightarrow x^{2} - 2 x - 1 + y^{2} - 4 y + 4 + z^{2} - 6 z + 9 = x^{2} - 6 x + 9 + y^{2} - 4 y + 4 + z^{2} + 2 z + 1 \Rightarrow - 2 x - 4 y - 6 z + 14 = - 6 x - 4 y + 2 z + 14 \Rightarrow - 2 x + 6 x - 4 y + 4 y - 6 z - 2 z = 14 - 14 \Rightarrow 4 x - 8 z = 0 \Rightarrow x - 2 z = 0$
Hence, the locus is x $-$ 2z = 0

Page No 28.10:

Question 22:

Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.

Answer:

Let P (x, y, z) be any point , the sum of whose distance from the points A(4,0,0) and B( $-$ 4,0,0)
is equal to 10. Then,
PA + PB = 10

$\Rightarrow \sqrt{{(x - 4)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2}} + \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2}} = 10 \Rightarrow \sqrt{{(x + 4)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2}} = 10 - \sqrt{{(x - 4)}^{2} + {(y - 0)}^{2} + {(z - 0)}^{2}} \Rightarrow \sqrt{x^{2} + 8 x + 16 + y^{2} + z^{2}} = 10 - \sqrt{x^{2} - 8 x + 16 + y^{2} + z^{2}}$
$\Rightarrow x^{2} + 8 x + 16 + y^{2} + z^{2} = 100 + x^{2} - 8 x + 16 + y^{2} + z^{2} - 20 \sqrt{x^{2} - 8 x + 16 + y^{2} + z^{2}} \Rightarrow 16 x - 100 = - 20 \sqrt{x^{2} - 8 x + 16 + y^{2} + z^{2}} \Rightarrow 4 x - 25 = - 5 \sqrt{x^{2} - 8 x + 16 + y^{2} + z^{2}} \Rightarrow 16 x^{2} - 200 x + 625 = 25 (x^{2} - 8 x + 16 + y^{2} + z^{2}) \Rightarrow 16 x^{2} - 200 x + 625 = 25 x^{2} - 200 x + 400 + 25 y^{2} + 25 z^{2} \Rightarrow 9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0$

$∴ 9 x^{2} + 25 y^{2} + 25 z^{2} - 225 = 0 is the required locus .$

Page No 28.10:

Question 23:

Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.

Answer:

To show that ABCD is a parallelogram, we need to show that its two opposite sides are equal.
$A B = \sqrt{{(- 1 - 1)}^{2} + {(- 2 - 2)}^{2} + {(- 1 - 3)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 B C = \sqrt{{(2 + 1)}^{2} + {(3 + 2)}^{2} + {(2 + 1)}^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43} C D = \sqrt{{(4 - 2)}^{2} + {(7 - 3)}^{2} + {(6 - 2)}^{2}} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6 D A = \sqrt{{(1 - 4)}^{2} + {(2 - 7)}^{2} + {(3 - 6)}^{2}} = \sqrt{9 + 25 + 9} = \sqrt{43} A B = C D and B C = D A Since, opposite pairs of sides are equal . ∴ ABCD is a parallelogram A C = \sqrt{{(2 - 1)}^{2} + {(3 - 2)}^{2} + {(2 - 3)}^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3} B D = \sqrt{{(4 + 1)}^{2} + {(7 + 2)}^{2} + {(6 + 1)}^{2}} = \sqrt{25 + 81 + 49} = \sqrt{155} Since, A C \neq B D$

Thus, ABCD is not a rectangle.

Page No 28.10:

Question 24:

Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.

Answer:

Let P (x, y, z) be any point which is equidistant from A (3,4,5) and B ( $-$ 2,1,4) .Then,
PA = PB
$\Rightarrow \sqrt{{(x - 3)}^{2} + {(y - 4)}^{2} + {(z + 5)}^{2}} = \sqrt{{(x + 2)}^{2} + {(y - 1)}^{2} + {(z - 4)}^{2}} \Rightarrow \sqrt{x^{2} - 6 x + 9 + y^{2} - 8 y + 16 + z^{2} + 10 z + 25} = \sqrt{x^{2} + 4 x + 4 + y^{2} - 2 y + 1 + z^{2} - 8 z + 16} \Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 + z^{2} + 10 z + 25 = x^{2} + 4 x + 4 + y^{2} - 2 y + 1 + z^{2} - 8 z + 16 \Rightarrow - 10 x - 6 y + 18 z + 29 = 0 ∴ 10 x + 6 y - 18 z - 29 = 0$

Hence, 10x + 6y $-$ 18z $-$ 29 = 0 is the required equation.

Page No 28.19:

Question 1:

The vertices of the triangle are A(5, 4, 6), B(1, –1, 3) and C(4, 3, 2). The internal bisector of angle A meets BC at D. Find the coordinates of D and the length AD.

Answer:

AB = $\sqrt{4^{2} + 5^{2} + 3^{2}} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5 \sqrt{2}$
AC = $\sqrt{1^{2} + 1^{2} + 4^{2}} = \sqrt{18} = 3 \sqrt{2}$
AD is the internal bisector of $∠ B A C$ .
$∴ \frac{B D}{D C} = \frac{A B}{A C} = \frac{5}{3}$
Thus, D divides BC internally in the ratio 5:3.
$∴ D \equiv (\frac{5 \times 4 + 3 \times 1}{5 + 3}, \frac{5 \times 3 + 3 (- 1)}{5 + 3}, \frac{5 \times 2 + 3 \times 3}{5 + 3}) \Rightarrow D \equiv (\frac{23}{8}, \frac{12}{8}, \frac{19}{8}) \Rightarrow D \equiv (\frac{23}{8}, \frac{3}{2}, \frac{19}{8}) ∴ A D = \sqrt{{(5 - \frac{23}{8})}^{2} + {(4 - \frac{12}{8})}^{2} + {(6 - \frac{19}{8})}^{2}} = \sqrt{\frac{17^{2} + 20^{2} + 29^{2}}{8^{2}}} = \sqrt{\frac{289 + 400 + 841}{8^{2}}} = \frac{\sqrt{1530}}{8}$

Page No 28.19:

Question 2:

A point C with z-coordinate 8 lies on the line segment joining the points A(2, –3, 4) and B(8, 0, 10). Find its coordinates.

Answer:

Suppose C divides AB in the ratio $λ : 1$ .
Then, the coordinates of C are $(\frac{8 λ + 2}{λ + 1}, \frac{- 3}{λ + 1}, \frac{10 λ + 4}{λ + 1})$ .
The z-coordinate of C is 8.
$∴ \frac{10 λ + 4}{λ + 1} = 8 \Rightarrow 10 λ + 4 = 8 λ + 8 \Rightarrow 2 λ = 4 ∴ λ = 2$
Hence, the coordinates of C are as follows:
$(\frac{8 λ + 2}{λ + 1}, \frac{- 3}{λ + 1}, \frac{10 λ + 4}{λ + 1}) \Rightarrow (\frac{8 \times 2 + 2}{2 + 1}, \frac{- 3}{2 + 1}, \frac{10 \times 2 + 4}{2 + 1}) \Rightarrow (\frac{18}{3}, \frac{- 3}{3}, \frac{24}{3}) ∴ (6, - 1, 8)$
Hence, the coordinates of C are (6, −1, 8).

Page No 28.20:

Question 3:

Show that the three points A(2, 3, 4), B(–1, 2 – 3) and C(–4, 1, –10) are collinear and find the ratio in which C divides AB.

Answer:

Suppose C divides AB in the ratio $λ : 1$ .
Then, the coordinates of C are $(\frac{- λ + 2}{λ + 1}, \frac{2 λ + 3}{λ + 1}, \frac{- 3 λ + 4}{λ + 1})$ .
But, the coordinates of C are ( $-$ 4, 1, $-$ 10).
$∴ \frac{- λ + 2}{λ + 1} = - 4, \frac{2 λ + 3}{λ + 1} = 1, \frac{- 3 λ + 4}{λ + 1} = - 10 \Rightarrow - λ + 2 = - 4 λ - 4, 2 λ + 3 = λ + 1, - 3 λ + 4 = - 10 λ - 10 \Rightarrow 3 λ = - 6, λ = - 2, 7 λ = - 14 ∴ λ = - 2, λ = - 2, λ = - 2$

From these three equations, we have:
$λ = - 2$
So, C divides AB in the ratio 2:1 (externally).

Page No 28.20:

Question 4:

Find the ratio in which the line joining (2, 4, 5) and (3, 5, 4) is divided by the yz-plane.

Answer:

Let A $\equiv$ (2, 4, 5) and B $\equiv$ (3, 5, 4)
Let the line joining A and B be divided by the yz-plane at point P in the ratio $λ : 1$ .
Then, we have:
P $\equiv (\frac{3 λ + 2}{λ + 1}, \frac{5 λ + 4}{λ + 1}, \frac{4 λ + 5}{λ + 1})$
Since P lies on the yz-plane, the x-coordinate of P will be zero.
$∴ \frac{3 λ + 2}{λ + 1} = 0 \Rightarrow 3 λ + 2 = 0 ∴ λ = \frac{- 2}{3}$
Hence, the yz-plane divides AB in the ratio 2:3 (externally).

Page No 28.20:

Question 5:

Find the ratio in which the line segment joining the points (2, –1, 3) and (–1, 2, 1) is divided by the plane x + y + z = 5.

Answer:

Suppose the plane x + y + z = 5 divides the line joining the points A (2, −1, 3) and B ( $-$ 1, 2, 1) at a point C in the ratio $λ : 1$ .
Then, coordinates of C are as follows:
$(\frac{- λ + 2}{λ + 1}, \frac{2 λ - 1}{λ + 1}, \frac{λ + 3}{λ + 1})$
Now, the point C lies on the plane x + y + z = 5.
Therefore, the coordinates of C must satisfy the equation of the plane.
$\frac{- λ + 2}{λ + 1} + \frac{2 λ - 1}{λ + 1} + \frac{λ + 3}{λ + 1} = 5 \Rightarrow - λ + 2 + 2 λ - 1 + λ + 3 = 5 λ + 5 \Rightarrow 2 λ + 4 = 5 λ + 5 \Rightarrow - 3 λ = 1 ∴ λ = \frac{- 1}{3}$
So, the required ratio is 1:3 (externally).

Page No 28.20:

Question 6:

If the points A(3, 2, –4), B(9, 8, –10) and C(5, 4, –6) are collinear, find the ratio in which C divides AB.

Answer:

Suppose C divides AB in the ratio $λ : 1$ .
Then, coordinates of C are as follows:
$(\frac{9 λ + 3}{λ + 1}, \frac{8 λ + 2}{λ + 1}, \frac{- 10 λ - 4}{λ + 1})$
But, the coordinates of C are (5, 4, $-$ 6).
$∴ \frac{9 λ + 3}{λ + 1} = 5, \frac{8 λ + 2}{λ + 1} = 4, \frac{- 10 λ - 4}{λ + 1} = - 6$
These three equations give $λ = \frac{1}{2}$ .
So, C divides AB in the ratio 1:2.

Page No 28.20:

Question 7:

The mid-points of the sides of a triangle ABC are given by (–2, 3, 5), (4, –1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Answer:

Let $D (x_{1}, y_{1}, z_{1}), E (x_{2}, y_{2}, z_{2}) and F (x_{3}, y_{3}, z_{3})$ be the vertices of the given triangle.
And, let $A (- 2, 3, 5), B (4, - 1, 7) and C (6, 5, 3)$ be the mid-points of the sides EF, FA and DE, respectively.
Now, A is the mid-point of EF.
$∴ \frac{x_{2} + x_{3}}{2} = - 2, \frac{y_{2} + y_{3}}{2} = 3, \frac{z_{2} + z_{3}}{2} = 5 \Rightarrow x_{2} + x_{3} = - 4, y_{2} + y_{3} = 6, z_{2} + z_{3} = 10 (i)$
B is the mid-point of FD.
$∴ \frac{x_{1} + x_{3}}{2} = 4, \frac{y_{1} + y_{3}}{2} = - 1, \frac{z_{1} + z_{3}}{2} = 7 \Rightarrow x_{1} + x_{3} = 8, y_{1} + y_{3} = - 2, z_{1} + z_{3} = 14 (i i)$

C is the mid-point of DE.
$∴ \frac{x_{1} + x_{2}}{2} = 6, \frac{y_{1} + y_{2}}{2} = 5, \frac{z_{1} + z_{2}}{2} = 3 \Rightarrow x_{1} + x_{2} = 12, y_{1} + y_{2} = 10, z_{1} + z_{2} = 6 (i i i)$
Adding the first three equations in (i), (ii) and (iii):
$2 (x_{1} + x_{2} + x_{3}) = - 4 + 8 + 12 \Rightarrow x_{1} + x_{2} + x_{3} = 8$
Solving the first three equations in (i), (ii) and (iii) with $x_{1} + x_{2} + x_{3} = 8$ :
$x_{1} = 12, x_{2} = 0, x_{3} = - 4$
Adding the next three equations in (i), (ii) and (iii):
$2 (y_{1} + y_{2} + y_{3}) = 6 - 2 + 10 \Rightarrow y_{1} + y_{2} + y_{3} = 7$
Solving the next three equations in (i), (ii) and (iii) with $y_{1} + y_{2} + y_{3} = 7$ :
$y_{1} = 1, y_{2} = 9, y_{3} = - 3$
Adding the last three equations in (i), (ii) and (iii):
$2 (z_{1} + z_{2} + z_{3}) = 10 + 14 + 6 \Rightarrow z_{1} + z_{2} + z_{3} = 15$
Solving the last three equations in (i), (ii) and (iii) with $z_{1} + z_{2} + z_{3} = 15$ :
$z_{1} = 5, z_{2} = 1, z_{3} = 9$
Thus, the vertices of the triangle are (12, 1, 5), (0, 9, 1), (−4, −3, 9).

Page No 28.20:

Question 8:

A(1, 2, 3), B(0, 4, 1), C(–1, –1, –3) are the vertices of a triangle ABC. Find the point in which the bisector of the angle ∠BAC meets BC.

Answer:

AB = $\sqrt{{(- 1)}^{2} + 2^{2} + {(- 2)}^{2}} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$
AC = $\sqrt{{(- 2)}^{2} + {(- 3)}^{2} + {(- 6)}^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

AD is the internal bisector of $∠ B A C$ .
$∴ \frac{B D}{D C} = \frac{A B}{A C} = \frac{3}{7}$
Thus, D divides BC internally in the ratio 3:7.
$∴ D \equiv (\frac{3 (- 1) + 3 \times 0}{3 + 7}, \frac{3 (- 1) + 7 (4)}{3 + 7}, \frac{3 (- 3) + 7 \times 1}{3 + 7}) \Rightarrow D \equiv (\frac{- 3}{10}, \frac{25}{10}, \frac{- 2}{10}) \Rightarrow D \equiv (\frac{- 3}{10}, \frac{5}{2}, \frac{- 1}{5})$

Page No 28.20:

Question 9:

Find the ratio in which the sphere x² + y²+ z² = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).

Answer:

Let the sphere meet the line joining the given points at the point (x₁, y₁, z₁).

Then, we have:

Suppose the point (x₁, y₁, z₁) divides the line joining the points (12, – 4, 8) and (27, – 9, 18) in the ratio λ:1.

∴

Substitute these values in equation (1):

Thus, the sphere divides the line joining the given points in the ratio 2:3 and 2:– 3.

Hence, the given sphere divides the line joining the points (12, – 4, 8) and (27, – 9, 18) internally in the ratio 2:3 and externally in the ratio −2:3.

Page No 28.20:

Question 10:

Show that the plane ax + by + cz + d = 0 divides the line joining the points (x₁, y₁, z₁) and (x₂, y₂, z₂) in the ratio $- \frac{a x_{1} + b y_{1} + c z_{1} + d}{a x_{2} + b y_{2} + c z_{2} + d}$ .

Answer:

Let:
A = (x₁, y₁, z₁)
B = (x₂, y₂, z₂)

Now, let the line joining A and B be divided by the plane ax + by + cz + d = 0 at point P in the ratio $λ : 1$ .

∴ P = $(\frac{λ x_{2} + x_{1}}{λ + 1}, \frac{λ y_{2} + y_{1}}{λ + 1}, \frac{λ z_{2} + z_{1}}{λ + 1})$
Since P lies on the given plane,

ax + by + cz + d = 0

Thus,

$a \frac{λ x_{2} + x_{1}}{λ + 1} + b \frac{λ y_{2} + y_{1}}{λ + 1} + c \frac{λ z_{2} + z_{1}}{λ + 1} + d = 0 \Rightarrow a (λ x_{2} + x_{1}) + b (λ y_{2} + y_{1}) + c (λ z_{2} + z_{1}) + d (λ + 1) = 0 \Rightarrow λ (a x_{2} + b y_{2} + c z_{2} + d) + (a x_{1} + b y_{1} + c z_{1} + d) = 0 \Rightarrow λ (a x_{2} + b y_{2} + c z_{2} + d) = - (a x_{1} + b y_{1} + c z_{1} + d) \Rightarrow λ = \frac{- (a x_{1} + b y_{1} + c z_{1} + d)}{(a x_{2} + b y_{2} + c z_{2} + d)} \Rightarrow λ = - \frac{a x_{1} + b y_{1} + c z_{1} + d}{a x_{2} + b y_{2} + c z_{2} + d} Thus, the given plane divides the line joining (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) in the ratio - \frac{a x_{1} + b y_{1} + c z_{1} + d}{a x_{2} + b y_{2} + c z_{2} + d} .$

Page No 28.20:

Question 11:

Find the centroid of a triangle, mid-points of whose sides are (1, 2, –3), (3, 0, 1) and (–1, 1, –4).

Answer:

$Let A (x_{1}, y_{1}, z_{1}), B (x_{2}, y_{2}, z_{2}) and C (x_{3}, y_{3}, z_{3}) be the vertices of the given triangle, and let D (1, 2, - 3), E (3, 0, 1) and F (- 1, 1, - 4) be the mid points of the sides B C, C A and A B respectively . D is the mid point of B C ∴ \frac{x_{2} + x_{3}}{2} = 1, \frac{y_{2} + y_{3}}{2} = 2 and \frac{z_{2} + z_{3}}{2} = - 3 \Rightarrow x_{2} + x_{3} = 2, y_{2} + y_{3} = 4 and z_{2} + z_{3} = - 6 . . . (1) E is the mid point of C A ∴ \frac{x_{1} + x_{3}}{2} = 3, \frac{y_{1} + y_{3}}{2} = 0 and \frac{z_{1} + z_{3}}{2} = 1 \Rightarrow x_{1} + x_{3} = 6, y_{1} + y_{3} = 0 and z_{1} + z_{3} = 2 . . . (2) F is the mid point of A B ∴ \frac{x_{1} + x_{2}}{2} = - 1, \frac{y_{1} + y_{2}}{2} = 1 and \frac{z_{1} + z_{2}}{2} = - 4 \Rightarrow x_{1} + x_{2} = - 2, y_{1} + y_{2} = 2 and z_{1} + z_{2} = - 8 . . . (2) Adding first three equations we get, 2 (x_{1} + x_{2} + x_{3}) = 6, 2 (y_{1} + y_{2} + y_{3}) = 6 and 2 (z_{1} + z_{2} + z_{3}) = - 12 \Rightarrow x_{1} + x_{2} + x_{3} = 3, y_{1} + y_{2} + y_{3} = 3 and z_{1} + z_{2} + z_{3} = - 6 The coordinate of the centroid is given by (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3}) \Rightarrow (1, 1, - 2)$

Page No 28.20:

Question 12:

The centroid of a triangle ABC is at the point (1, 1, 1). If the coordinates of A and B are (3, –5, 7) and (–1, 7, –6) respectively, find the coordinates of the point C.

Answer:

Let G be the centroid of $∆$ ABC.
Given: G $\equiv (1, 1, 1)$
Let C $\equiv (x, y, z)$
$Then, 1 = \frac{3 - 1 + x}{3} \Rightarrow 3 = 3 - 1 + x \Rightarrow 3 = 2 + x \Rightarrow x = 1 and 1 = \frac{- 5 + 7 + y}{3} \Rightarrow 3 = 2 + y \Rightarrow y = 1 and 1 = \frac{7 - 6 + z}{3} \Rightarrow 3 = 1 + z \Rightarrow z = 2$

$∴ C \equiv (1, 1, 2)$

Page No 28.20:

Question 13:

Find the coordinates of the points which tisect the line segment joining the points P(4, 2, –6) and Q(10, –16, 6).

Answer:

Let P $\equiv$ (4, 2, $-$ 6) and Q $\equiv$ (10, $-$ 16, 6)
Let A and B be the point of trisection.
Then, we have:
PA = AB = BQ
∴ PA:AQ = 1:2
Thus, A divides PQ internally in the ratio 1:2.
∴ A $\equiv$ $(\frac{1 \times 10 + 2 \times 4}{1 + 2}, \frac{1 \times (- 16) + 2 \times 2}{1 + 2}, \frac{1 \times 6 + 2 \times (- 6)}{1 + 2})$
                     $\equiv (\frac{10 + 8}{3}, \frac{- 16 + 4}{3}, \frac{6 - 12}{3}) \equiv (\frac{18}{3}, \frac{- 12}{3}, \frac{- 6}{3}) \equiv (6, - 4, - 2)$
$\Rightarrow$ A $\equiv$ (6, $-$ 4, $-$ 2)
AB = BQ


Therefore, B is the mid-point of AQ.
∴ B $\equiv$ $(\frac{6 + 10}{2}, \frac{- 4 - 16}{2}, \frac{- 2 + 6}{2})$
                   $\equiv (\frac{16}{2}, \frac{- 20}{2}, \frac{4}{2}) \equiv (8, - 10, 2)$
$\Rightarrow$ B $\equiv$ (8, $-$ 10, 2)

Page No 28.20:

Question 14:

Using section formula, show that he points A(2, –3, 4), B(–1, 2, 1) and C(0, 1/3, 2) are collinear.

Answer:

Let, C divides AB in the ratio $λ$ :1.
Then, coordinates of C are $(\frac{- λ + 2}{λ + 1}, \frac{2 λ - 3}{λ + 1}, \frac{λ + 4}{λ + 1})$ .
But, the coordinates of C are $(0, \frac{1}{3}, 2)$ .
∴ $\frac{- λ + 2}{λ + 1}$ =0 , $\frac{2 λ - 3}{λ + 1}$ = $\frac{1}{3}$ , $\frac{λ + 4}{λ + 1}$ = 2
From each of these equations, we get $λ = 2$ .
Since each of these equations give the same value of $λ$ .
Therefore, the given points are collinear.

Page No 28.20:

Question 15:

Given that P(3, 2, –4), Q(5, 4, –6) and R(9, 8, –10) are collinear. Find the ratio in which Q divides PR.

Answer:

Let Q divide PR in the ratio $λ : 1$
Thus, the coordinates of Q are as follows:
$(\frac{9 λ + 3}{λ + 1}, \frac{8 λ + 2}{λ + 1}, \frac{- 10 λ - 4}{λ + 1})$
But, the coordinates of Q are (5, 4, −6) .
$∴ \frac{9 λ + 3}{λ + 1} = 5, \frac{8 λ + 2}{λ + 1} = 4, \frac{- 10 λ - 4}{λ + 1} = - 6$
These three equation gives $λ = \frac{1}{2}$ .
So, Q divides PR in the ratio $\frac{1}{2} : 1$ or 1:2

Page No 28.20:

Question 16:

Find the ratio in which the line segment joining the points (4, 8, 10) and (6, 10, –8) is divided by the yz-plane.

Answer:

Let A = (4, 8, 10) , B = (6, 10, $-$ 8)
Let the line joining A and B be divided by the YZ-plane at point P in the ratio $λ : 1$ .
∴ P= $(\frac{6 λ + 4}{λ + 1}, \frac{10 λ + 8}{λ + 1}, \frac{- 8 λ + 10}{λ + 1})$
Since P lies on the YZ-plane, the x-coordinate of P will be zero.
$∴ \frac{6 λ + 4}{λ + 1} = 0 \Rightarrow 6 λ + 4 = 0 \Rightarrow 6 λ = - 4 ∴ λ = \frac{- 2}{3}$
Hence, the YZ-plane divides AB in the ratio 2:3 (externally)

Page No 28.22:

Question 1:

The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
(a) 2 : 3
(b) 3 : 2
(c) –2 : 3
(d) 4 : –3

Answer:

(c) $-$ 2 : 3

Let A $\equiv$ (2, 4, 5) and B $\equiv$ (3, 5, $-$ 9)
Let the line joining A and B be divided by the yz-plane at point P in the ratio $λ : 1$ .
Then, we have,

P $\equiv (\frac{3 λ + 2}{λ + 1}, \frac{5 λ + 4}{λ + 1}, \frac{- 9 λ + 5}{λ + 1})$

Since P lies on the yz-plane, the x-coordinate of P will be zero.

$∴ \frac{3 λ + 2}{λ + 1} = 0 \Rightarrow 3 λ + 2 = 0 \Rightarrow λ = \frac{- 2}{3}$
Hence, the yz-plane divides AB in the ratio $-$ 2 : 3

Page No 28.22:

Question 2:

The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
(a) a : b
(b) b : c
(c) c : a
(d) c : b

Answer:

(d) c : b

Let A $\equiv$ (a, b, c) and B $\equiv$ ( $-$ a, $-$ c, $-$ b)
Let the line joining A and B be divided by the xy-plane at point P in the ratio $λ : 1$ .
Then, we have,

P $\equiv (\frac{- a λ + a}{λ + 1}, \frac{- c λ + b}{λ + 1}, \frac{- b λ + c}{λ + 1})$

Since P lies on the xy-plane, the z-coordinate of P will be zero.

$∴ \frac{- b λ + c}{λ + 1} = 0 \Rightarrow - b λ + c = 0 \Rightarrow λ = \frac{c}{b}$
Hence, the xz-plane divides AB in the ratio c : b

Page No 28.22:

Question 3:

If P(0, 1, 2), Q(4, –2, 1) and O(0, 0, 0) are three points, then ∠POQ =

(a) $\frac{π}{6}$

(b) $\frac{π}{4}$

(c) $\frac{π}{3}$

(d) $\frac{π}{2}$

Answer:

(d) $\frac{π}{2}$

$P Q^{2} = {(4 - 0)}^{2} + {(- 2 - 1)}^{2} + {(1 - 2)}^{2} = 16 + 9 + 1 = 26 O P^{2} = {(0 - 0)}^{2} + {(1 - 0)}^{2} + {(2 - 0)}^{2} = 0 + 1 + 4 = 5 Q O^{2} = {(0 - 4)}^{2} + {(0 + 2)}^{2} + {(0 - 1)}^{2} = 16 + 1 + 4 = 21 Since, P Q^{2} = O P^{2} + Q O^{2} Hence, ∠ P O Q = \frac{π}{2}$

Page No 28.22:

Question 4:

If the extremities of the diagonal of a square are (1, –2, 3 and (2, –3, 5), then the length of the side is

(a) $\sqrt{6}$

(b) $\sqrt{3}$

(c) $\sqrt{5}$

(d) $\sqrt{7}$

Answer:

(b) $\sqrt{3}$

Length of the diagonal = $\sqrt{{(2 - 1)}^{2} + {(- 3 + 2)}^{2} + {(5 - 3)}^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6}$

∴ Length of the side = $\frac{Length of diagonal}{\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3}$

Page No 28.23:

Question 5:

The points (5, –4, 2), (4, –3, 1), (7, 6, 4) and (8, –7, 5) are the vertices of
(a) a rectangle
(b) a square
(c) a parallelogram
(d) none of these

Answer:

(d) None of these

Suppose:
A(5, –4, 2)
B(4, –3, 1)
C(7, 6, 4)
D(8, –7, 5)

$A B = \sqrt{{(4 - 5)}^{2} + {(- 3 + 4)}^{2} + {(1 - 2)}^{2}} = \sqrt{{(- 1)}^{2} + {(1)}^{2} + {(- 1)}^{2}} = \sqrt{1 + 1 + 1} = \sqrt{3} B C = \sqrt{{(7 - 4)}^{2} + {(6 + 3)}^{2} + {(4 - 1)}^{2}} = \sqrt{{(3)}^{2} + {(9)}^{2} + {(3)}^{2}} = \sqrt{9 + 81 + 9} = \sqrt{99} = 3 \sqrt{11} C D = \sqrt{{(8 - 7)}^{2} + {(- 7 - 6)}^{2} + {(5 - 4)}^{2}} = \sqrt{{(1)}^{2} + {(- 13)}^{2} + {(1)}^{2}} = \sqrt{1 + 169 + 1} = \sqrt{171} D A = \sqrt{{(8 - 5)}^{2} + {(- 7 + 4)}^{2} + {(5 - 2)}^{2}} = \sqrt{{(3)}^{2} + {(- 3)}^{2} + {(3)}^{2}} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3 \sqrt{3}$

We see that none of the sides are equal.

Page No 28.23:

Question 6:

In a three dimensional space the equation x² – 5x + 6 = 0 represents
(a) points
(b) planes
(c) curves
(d) pair of straight lines

Answer:

(c) curves

Since, there is only one variable in the given equation. Also, it is quadratic equation.
Hence, It represents curves in yz plane.

Page No 28.23:

Question 7:

Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
(a) 2
(b) 3
(c) 6
(d) 7

Answer:

(b) 3

Suppose d is the diameter of the sphere. Then

$d^{2} = {(- 1 - 3)}^{2} + {(2 - 4)}^{2} + {(3 + 1)}^{2} \Rightarrow d^{2} = {(- 4)}^{2} + {(- 2)}^{2} + {(4)}^{2} \Rightarrow d^{2} = 16 + 4 + 16 \Rightarrow d^{2} = 36 \Rightarrow d = 6$

Hence, radius of the sphere is 3 units.

Page No 28.23:

Question 8:

XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
(a) 3 : 7
(b) 2 : 7
(c) –3 : 7
(d) –2 : 7

Answer:

(c) −3:7

Let A $\equiv$ (2, 3, 1) and B $\equiv$ (6, 7, 1)
Let the line joining A and B be divided by the xz-plane at point P in the ratio $λ : 1$ .
Then, we have,

P $\equiv (\frac{6 λ + 2}{λ + 1}, \frac{7 λ + 3}{λ + 1}, \frac{λ + 1}{λ + 1})$

Since P lies on the xz-plane, the y-coordinate of P will be zero.

$∴ \frac{7 λ + 3}{λ + 1} = 0 \Rightarrow 7 λ + 3 = 0 \Rightarrow λ = \frac{- 3}{7}$
Hence, the xz-plane divides AB in the ratio $-$ 3 : 7

Page No 28.23:

Question 9:

What is the locus of a point for which y = 0, z = 0?
(a) x - axis
(b) y - axis
(c) z - axis
(d) yz - plane

Answer:

We know that on x - axis both y = 0, z = 0.
Hence, the correct answer is option (a)

Page No 28.23:

Question 10:

The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are
(a) (3, 4, 0)
(b) (0, 4, 5)
(c) (3, 0, 5)
(d) (3, 0, 0)

Answer:

We know that the x - coordinate on yz - plane is 0.
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz - plane are (0, 4, 5).
Hence, the correct answer is option (b).

Page No 28.23:

Question 11:

The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are

(a) (6, 0, 0)
(b) (0, 7, 0)
(c) (0, 0, 8)
(d) (0, 7, 8)

Answer:

We know that the y and z coordinates on x - axis are 0
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are (6, 0, 0)
Hence, the correct answer is option (a).

Page No 28.23:

Question 12:

The perpendicular distance of the point P (6, 7, 8) from xy - plane is

(a) 8
(b) 7
(c) 6
(d) 10

Answer:

The distance of the point P (6, 7, 8) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 8.
Hence, the correct answer is option (a).

Page No 28.23:

Question 13:

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
(a) 10
(b) $\sqrt{34}$
(c) $\sqrt{113}$
(d) 512

Answer:

The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is given by

$\sqrt{3^{2} + 5^{2}} = \sqrt{34}$
Hence, the correct answer is option (b)

Page No 28.23:

Question 14:

The perpendicular distance of the point P(3, 3,4) from the x-axis is

(a) $3 \sqrt{2}$
(b) 5
(c) 3
(d) 4

Answer:

The perpendicular distance of the point P(3, 3,4) from the x-axis is given by

$\sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$
Hence, the correct answer is option (b)

Page No 28.23:

Question 15:

The length of the perpendicular drawn from the point P(a, b, c) from z-axis is

(a) $\sqrt{a^{2} + b^{2}}$

(b) $\sqrt{b^{2} + c^{2}}$

(c) $\sqrt{a^{2} + c^{2}}$

(d) $\sqrt{a^{2} + b^{2} + c^{2}}$

Answer:

The length of the perpendicular drawn from the point P(x, y, z) from z-axis is given by $\sqrt{y^{2} + x^{2}}$
Thus, the length of the perpendicular drawn from the point P(a, b, c) from z-axis is $\sqrt{a^{2} + b^{2}}$
Hence, the correct answer is option (a)

Page No 28.23:

Question 16:

L is the foot of the perpendicular drawn from a point (3, 4, 5) on x-axis. The coordinates of L are
(a) (3, 0, 0)
(b) (0, 4, 0)
(c) (0, 0, 5)
(d) none of these

Answer:

Let L be the foot of the perpendicular drawn from (3, 4, 5) an x-axis.
Since on x-axis, coordinates of y and z are zero.
$∴$ coordinates of L are (3, 0, 0)

Hence, the correct answer is option A.

Page No 28.23:

Question 17:

The coordinates of the foot of the perpendicular drawn from a point (3, 4, 5) on the xy-plane are
(a) (3, 0, 0)
(b) (0, 4, 5)
(c) (3, 0, 5)
(d) none of these

Answer:

The co-ordinate of the foot of perpendicular from a point (3, 4, 5) an xy plane is (3, 4, 0) [i.e z-coordinate is zero].

Hence, the correct answer is option D.

Page No 28.23:

Question 18:

If a parallelopiped is formed by planes drawn through the points (5, 8, 10) and (3, 6, 8) parallel to the coordinate planes, then the length of the diagonal of the parallelopiped is

(a) $2 \sqrt{3}$

(b) $3 \sqrt{2}$

(c) $\sqrt{2}$

(d) $\sqrt{3}$

Answer:

Let A be given by (5, 8, 10) and B be given by (3, 6, 8)
then by distance formula,
$\begin{array}{rcl} A B & = & \sqrt{{(5 - 3)}^{2} + {(8 - 6)}^{2} + {(10 - 8)}^{2}} \\ = & \sqrt{4 + 4 + 4} \\ = & \sqrt{12} \\ = & 2 \sqrt{3} \end{array}$
i.e the length of diagonal of parallelopiped is $2 \sqrt{3}$ .

Hence, the correct answer is option A.

Page No 28.23:

Question 19:

The locus of the point for which x = 0 is
(a) xy-plane
(b) yz-plane
(c) zx-plane
(d) none of these

Answer:

Since x = 0 an yz plane
⇒ locus of x = 0 is yz plane

Hence, the correct answer is option B.

Page No 28.23:

Question 20:

The locus of the point for which z = 0, x = 0, is
(a) equation of x-axis
(b) equation of y-axis
(c) equation of z-axis
(d) none of these

Answer:

Since x = 0 and z = 0 is always true an y-axis.
⇒ Locus of x = 0 and z = 0 is y-axis.

Hence, the correct answer is option B.

Page No 28.23:

Question 21:

Equation y-axis is considered as
(a) x = 0, y = 0
(b) y = 0, z = 0
(c) z = 0, x = 0
(d) none of these

Answer:

Equation y-axis is an axis at
Which x = 0 and z = 0

Hence, the correct answer is option C.

Page No 28.23:

Question 22:

A plane is parallel to yz-plane, so it is perpendicular to
(a) x-axis
(b) y-axis
(c) z-axis
(d) none of these

Answer:

Any plane parallel to yz plane is perpendicular to x-axis because yz plane is perpendicular to x-axis and parallel planes have same perpendiculars.

Hence, the correct answer is option A.

Page No 28.23:

Question 23:

x-axis is the intersection of two planes
(a) xy and xz
(b) yz and zx
(c) xy and yz
(d) none of these

Answer:

Every axis is intersection of two planes with common coordinate line xy and xz plane intersect to form x-axis.

Hence, the correct answer is option A.

Page No 28.23:

Question 24:

If the distance between the points (a, 0, 1) and (0, 1, 2) is $\sqrt{27}$ , then the value of 'a' is
(a) 5
(b) ±5
(c) –5
(d) none of these

Answer:

Let A be given by (a, 0, 1) and B be given by (0, 1, 2)
Given:- Distance AB = $\sqrt{27}$
i.e $\sqrt{{(a - 0)}^{2} + {(0 - 1)}^{2} + {(1 - 2)}^{2}}$ = $\sqrt{27}$ (By distance formula)
Now, By squaring both sides,
we get
a² + 1 + 1 = 27
i.e a² = 25
i.e a = ±5

Hence, the correct answer is option B.

Page No 28.23:

Question 1:

The distance of the point (a, b, c) from zx-plane is ________________________.

Answer:

For given point (a, b, c)
Since a and c lie an xz plane y coordinate is the distance of any point from xz plane
∴ Distance of (a, b, c) from xz-plane is b.

Page No 28.23:

Question 2:

The distance of the point (p, q, r) from x axis is _______________________.

Answer:

For given point (p, q, r) p is the point on x-axis
i.e any point an x-axis is of the form (p, 0, 0)
∴ Distance of (p, q, r) from x-axis is
$\sqrt{{(p - p)}^{2} + {(q - 0)}^{2} + {(r - 0)}^{2}} i . e \sqrt{q^{2} + r^{2}}$
Hence, distance of (p, q, r) from x-axis is $\sqrt{q^{2} + r^{2}}$

Page No 28.23:

Question 3:

If L and M are the feet of perpendiculars from P (3, 4, 5) on xy and yz-planes, then LM =____________________________.

Answer:

For given point P(3, 4, 5)
Let L and M be the feet of perpendiculars from P to xy and yz-planes respectively
∴ L is given by (3, 4, 0) and M is given by (0, 4, 5)

$\begin{array}{rcl} ∴ L M & = & \sqrt{{(3 - 0)}^{2} + {(4 - 4)}^{2} + {(0 - 5)}^{2}} \\ = & \sqrt{9 + 0 + 25} \\ i . e L M & = & \sqrt{34} \end{array}$

Page No 28.24:

Question 4:

The coordinates of the point equidistant from the four points O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c) are ________________________.

Answer:

Let us suppose any point be P(x, y, z) which is equidistant from O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c)
$i . e P O = \sqrt{x^{2} + y^{2} + z^{2}} ∴ {(P O)}^{2} = x^{2} + y^{2} + z^{2} P A = \sqrt{{(x - a)}^{2} + y^{2} + z^{2}} i . e {(P A)}^{2} = {(x - a)}^{2} + y^{2} + z^{2} also {(P B)}^{2} = x^{2} + {(y - b)}^{2} + c^{2} and {(P C)}^{2} = x^{2} + y^{2} + {(z - c)}^{2} given P O = P A = P B = P C i . e {(P O)}^{2} = {(P A)}^{2} = {(P B)}^{2} = {(P C)}^{2} ∴ We get, x^{2} + y^{2} + z^{2} = {(x - a)}^{2} + y^{2} + z^{2} x^{2} + y^{2} + z^{2} = x^{2} + {(y - b)}^{2} + z^{2} and x^{2} + y^{2} + z^{2} = x^{2} + y^{2} + {(z - c)}^{2} i . e x^{2} = x^{2} + a^{2} - 2 a x y^{2} = y^{2} + b^{2} - 2 b x and z^{2} = z^{2} + c^{2} - 2 c x i . e a^{2} = 2 a x \Rightarrow x = \frac{a}{2} b^{2} = 2 b y \Rightarrow y = \frac{b}{2} and c^{2} = 2 c z \Rightarrow z = \frac{c}{2}$
Hence, the coordinate of point equidistant from
O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c) is $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$

Page No 28.24:

Question 5:

A line is parallel to xy-plane if all the points on the line have equal ______________________.

Answer:

On xy plane z coordinate is 0
∴ Any line parallel to xy plane will have (equal) and fix value of z coordinate.

Page No 28.24:

Question 6:

The equation x = b represents a plane parallel to ___________________________ plane.

Answer:

x = b equation represents fix value of x coordinate which is always parallel to yz plane.

Page No 28.24:

Question 7:

Perpendicular distance of the point (3, 5, 6) from y-axis is _______________________________.

Answer:

For given point (3, 5, 6) y coordinate is 5
i.e Any point an y-axis has y-coordinate as 5 and rest zero.
i.e Point an y-axis is of the form (0, 5, 0)
∴ Perpendicular distance of (3, 5, 6) from y-axis $\sqrt{{(3 - 0)}^{2} + {(5 - 5)}^{2} + {(6 - 0)}^{2}}$

= \sqrt{9 + 0 + 36} = \sqrt{45}

Page No 28.24:

Question 8:

L is the foot of the perpendicular drawn from the point (3, 4, 5) on yz-plane. The coordinates of L are ____________________________.

Answer:

Let L be foot of the perpendicular drawn from (3, 4, 5) an yz-plane.
i.e x-coordinate is 0
∴ L has coordinates (0, 4, 5)

Page No 28.24:

Question 9:

The length of the perpendicular drawn from the point P(a, b, c) on z-axis is ___________________.

Answer:

For given point (a, b, c) z coordinate is c
i.e. any point on z-axis is of the form (0, 0, c)
Hence length of perpendicular from (a, b, c) an z-axis is given by $\sqrt{{(a - 0)}^{2} + {(b - 0)}^{2} + {(c - c)}^{2}}$

= \sqrt{a^{2} + b^{2}}

Page No 28.24:

Question 10:

z = a represents a plane parallel to ________________________ plane.

Answer:

z = a i.e z is fixed for any plane parallel to xy-plane

Page No 28.24:

Question 11:

If the centroid of the triangle is origin and two of its vertices are (3, –5, 7) and (–1, 7, –6), then the third vertex is ______________________.

Answer:

Two vertices of triangle are given i.e A(3, –5, 7) and B(–1, +7, –6)
Let w suppose third vertex C(x, y, z)
Since centroid is given by origin i.e O(0, 0, 0)
$Using formula coordinates of centroid = (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3}) where vertices are triangle are (x_{1}, y_{1}, z_{1}) (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3})$
Hence $0 = \frac{3 + (- 1) + x}{3}, 0 = \frac{- 5 + 7 + y}{3}, 0 = \frac{7 - 6 + z}{3}$
i.e x = –2, y = –2, z = –1
∴ Third vertex of triangle is given by (–2, –2, –1)

Page No 28.24:

Question 12:

If the mid-points of the sides of a triangle have coordinates (1, 2, –3) ,(3, 0, 1) and (–1, 1, –4), then the coordinates of its centroid are __________________.

Answer:

Let A(x, y, z), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) be the vertices of the given triangle ABC and Let D(1, 2, –3), E(3, 0, 1) and F(–1, 1, –4) be the mid-points of sides BC, CA, and AB respectively,

i.e $\frac{x_{1} + x_{3}}{2} = 3$
$\frac{y_{1} + y_{3}}{2} = 0 and \frac{z_{1} + z_{3}}{2} = 1$
$i . e . x_{1} + x_{3} = 6 y_{1} + y_{3} = 0 z_{1} + z_{3} = 2 . . . (1)$
$Similarly, \frac{x_{1} + x_{2}}{2} = - 1, \frac{y_{1} + y_{2}}{2} = 1, \frac{z_{1} + z_{2}}{2} = - 4 . . . . (2) and \frac{x_{2} + x_{3}}{2} = 1, \frac{y_{2} + y_{3}}{2} = 2, \frac{z_{2} + z_{3}}{2} = - 3 . . . . (3)$

i.e x₁+ x₂ = –2, y₁+ y₂ = 2, z₁+ z₂ = – 8 and x₂+ x₃ = 2, y₂+ y₃ = 4, z₂+ z₃ = – 6
∴ Centroid has co-ordinates $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$
$i . e (\frac{2 (x_{1} + x_{2} + x_{3})}{6}, \frac{2 (y_{1} + y_{2} + y_{3})}{6}, \frac{2 (z_{1} + z_{2} + z_{3})}{6}) i . e (\frac{x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3}}{6}, \frac{y_{1} + y_{2} + y_{2} + y_{3} + y_{3} + y_{1}}{6}, \frac{z_{1} + z_{2} + z_{2} + z_{3} + z_{3} + z_{1}}{6})$
i.e from (1), (2) and (3), we get
Coordinate of centroid
$= (\frac{6 - 1 + 1}{6}, \frac{0 + 4 + 2}{6}, \frac{2 - 8 - 6}{6}) = (1, 1, - 2)$
i.e coordinate of centroid is given by (1, 1, –2)

Page No 28.24:

Question 13:

The three axes OX, OY, OZ determine _______________________.

Answer:

Three axes OX, OY and OZ determine 3-coordinate plane
Since all three coordinates are present.
∴ OX, OY, and OZ determine 3-coordinate plane

Page No 28.24:

Question 14:

The three planes determine a rectangular parallelopiped which has _____________________ of rectangular faces.

Answer:

Each pair of plane gives a pair of rectangular faces.
Since three plane can form 3 pair of plane
Hence, will give three pair of rectangular faces.

Page No 28.24:

Question 15:

The coordinates of a point P are perpendicular distances of P from the three coordinate planes ___________________ and ________________ respectively.

Answer:

For any point P(x, y, z)
x coordinate is perpendicular distance of p from yz plane.
y coordinate is perpendicular distance of p from xz plane.
Similarly, z coordinate is perpendicular distance of p from xy plane.
Hence, coordinates of p are perpendicular distances of p from the three coordinate planes yz, zx and xy respectively.

Page No 28.24:

Question 16:

The three coordinates planes divide the space into __________________ parts.

Answer:

Three mutually perpendicular lines say X'OX, Y'OY and Z'OZ divide the space into 8 octants
Given by OXYZ, OX'YZ, OXY'Z, OXYZ', OX'Y'Z, OXY'Z', OX'YZ', and OX'Y'Z'
Therefore, three coordinates plane divide the space into 8 parts.

Page No 28.24:

Question 17:

The equations of xy, yz and zx planes are _______________ , _________________ and respectively.

Answer:

Equations of xy plane is given by z = 0
Since z is zero in xy plane
Similarly, equation of yz plane is given by x = 0 and equation of xz plane is given by y = 0
∴ Equation of xy, yz and zx planes are z = 0, x = 0 and y = 0 respectively.

Page No 28.24:

Question 18:

The equations of the coordinate axis are : ___________________________.

Answer:

For x-axis, equation used is y = 0, z = 0 [since y and z are zero]
For y-axis; x = 0, z = 0 and for z-axis its x = 0, y = 0.

Page No 28.24:

Question 19:

A line is parallel to xy-plane if all the points on the line have equal __________________.

Answer:

On xy plane z coordinate is 0
∴ Any line parallel to xy plane will have (equal) and fix value of z coordinate.

Page No 28.24:

Question 20:

A line is parallel to x-axis if all the points on the line have equal______________________.

Answer:

A line parallel to x-axis is obtained by varying x-only
i.e y and z coordinate remains fixed
Hence, a line is parallel to x-axis if all the points on the line have equal y, z-coordinates.

Page No 28.24:

Question 21:

x = a represents a plane parallel to ____________________.

Answer:

x = a represents fix value of x-coordinate which is always parallel to yz plane.

Page No 28.24:

Question 22:

The plane parallel to yz-plane is perpendicular to ____________________.

Answer:

Since yz plane is perpendicular to x-axis
⇒ Any plane parallel to yz-plane is perpendicular to x-axis.

Page No 28.24:

Question 23:

The length of the longest piece of a string that can be stretched straight in a rectangular room whose dimensions are 10, 13 and 8 units are ____________________.

Answer:

Since given dimensions are a = 10, b = 13 and c = 8
∴ Required length of the string is
$\sqrt{a^{2} + b^{2} + c^{2}} i . e \sqrt{{(10)}^{2} + {(13)}^{2} + 8^{2}} i . e \sqrt{100 + 169 + 64} = \sqrt{333}$
i.e the length of the longest piece of a string that can be stretched straight in a rectangular room with dimensions 10, 13 and 8 are $\sqrt{333}$ .

Page No 28.24:

Question 24:

If the distance between (a, 2, 1) and (1, –1, 1) is 5, then a = _____________________.

Answer:

Let A be represented by (a, 2, 1) and B be represented by (1, –1, 1)
Since distance between AB = 5
$i . e \sqrt{{(a - 1)}^{2} + {(2 + 1)}^{2} + {(1 - 1)}^{2}} = 5 i . e \sqrt{{(a - 1)}^{2} + 9} = 5$
By squaring both sides,
i.e (a – 1)² + 9 = 25
i.e a² + 1 – 2a + 9 – 25 = 0
i.e a² – 2a – 15 = 0
i.e a² + 3a – 5a – 15 = 0
i.e a(a + 3) –5 (a + 3) = 0
i.e a = 5 or a = –3

Page No 28.24:

Question 25:

If the mid-points of the sides AB, BC and CA of a ∆ABC are D (1, 2, –3), E (3, 0, 1) and F(–1, 1, –4) then the coordinates of the centroid of the triangle ABC are __________________.

Answer:

Let A(x, y, z), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) be the vertices of the given triangle ABC and Let D(1, 2, –3), E(3, 0, 1) and F(–1, 1, –4) be the mid-points of sides BC, CA, and AB respectively,
figure
i.e $\frac{x_{1} + x_{3}}{2} = 3$
$\frac{y_{1} + y_{3}}{2} = 0 and \frac{z_{1} + z_{3}}{2} = 1$
$i . e . x_{1} + x_{3} = 6 y_{1} + y_{3} = 0 z_{1} + z_{3} = 2 . . . (1)$
$Similarly, \frac{x_{1} + x_{2}}{2} = - 1, \frac{y_{1} + y_{2}}{2} = 1, \frac{z_{1} + z_{2}}{2} = - 4 . . . . (2) and \frac{x_{2} + x_{3}}{2} = 1, \frac{y_{2} + y_{3}}{2} = 2, \frac{z_{2} + z_{3}}{2} = - 3 . . . . (3)$

i.e x₁+ x₂ = –2, y₁+ y₂ = 2, z₁+ z₂ = – 8 and x₂+ x₃ = 2, y₂+ y₃ = 4, z₂+ z₃ = – 6
∴ Centroid has co-ordinates $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3})$
$i . e (\frac{2 (x_{1} + x_{2} + x_{3})}{6}, \frac{2 (y_{1} + y_{2} + y_{3})}{6}, \frac{2 (z_{1} + z_{2} + z_{3})}{6}) i . e (\frac{x_{1} + x_{2} + x_{2} + x_{3} + x_{1} + x_{3}}{6}, \frac{y_{1} + y_{2} + y_{2} + y_{3} + y_{3} + y_{1}}{6}, \frac{z_{1} + z_{2} + z_{2} + z_{3} + z_{3} + z_{1}}{6})$
i.e from (1), (2) and (3), we get
Coordinate of centroid
$= (\frac{6 - 1 + 1}{6}, \frac{0 + 4 + 2}{6}, \frac{2 - 8 - 6}{6}) = (1, 1, - 2)$
i.e coordinate of centroid is given by (1, 1, –2)

Page No 28.25:

Question 1:

Write the distance of the point P (2, 3,5) from the xy-plane.

Answer:

The distance of the point P (2, 3, 5) from the xy - plane is equal to the z-coordinate of the point.
Here, the value of z-coordinate is 5.
Hence, the distance of the point P (2, 3,5) from the xy-plane is 5.

Page No 28.25:

Question 2:

Write the distance of the point P(3, 4, 5) from z-axis.

Answer:

The distance of the point P(3, 4, 5) from z-axis is given by
$\sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$

Page No 28.25:

Question 3:

If the distance between the points P(a, 2, 1) and Q (1, −1, 1) is 5 units, find the value of a.

Answer:

PQ = 5
$\Rightarrow \sqrt{{(1 - a)}^{2} + {(- 1 - 2)}^{2} + {(1 - 1)}^{2}} = 5 \Rightarrow {(1 - a)}^{2} + {(- 3)}^{2} = 25 \Rightarrow 1 - 2 a + a^{2} + 9 - 25 = 0 \Rightarrow a^{2} - 2 a - 15 = 0 \Rightarrow (a - 5) (a + 3) = 0 \Rightarrow a = 5, - 3$

Page No 28.25:

Question 4:

The coordinates of the mid-points of sides AB, BC and CA of △ABC are D(1, 2, −3), E(3, 0,1) and F(−1, 1, −4) respectively. Write the coordinates of its centroid.

Answer:

Let the coordintes of the triangles be A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃).
Now,
Mid point of AB is D(1, 2, −3)
$\frac{x_{1} + x_{2}}{2} = 1, \frac{y_{1} + y_{2}}{2} = 2 and \frac{z_{1} + z_{2}}{2} = - 3 \Rightarrow x_{1} + x_{2} = 2, y_{1} + y_{2} = 4 and z_{1} + z_{2} = - 6 . . . . . (1)$

Mid point of BC is E(3, 0,1)
$\frac{x_{2} + x_{3}}{2} = 3, \frac{y_{2} + y_{3}}{2} = 0 and \frac{z_{2} + z_{3}}{2} = 1 \Rightarrow x_{2} + x_{3} = 6, y_{2} + y_{3} = 0 and z_{2} + z_{3} = 2 . . . . . (2)$

Mid point of AC is F(−1, 1, −4)

$\frac{x_{1} + x_{3}}{2} = - 1, \frac{y_{1} + y_{3}}{2} = 1 and \frac{z_{1} + z_{3}}{2} = - 4 \Rightarrow x_{1} + x_{3} = - 2, y_{1} + y_{3} = 2 and z_{1} + z_{3} = - 8 . . . . . (3)$

Adding (1), (2) and (3),we get
$2 (x_{1} + x_{2} + x_{3}) = 6, 2 (y_{1} + y_{2} + y_{3}) = 6 and 2 (z_{1} + z_{2} + z_{3}) = - 12 \Rightarrow x_{1} + x_{2} + x_{3} = 3, y_{1} + y_{2} + y_{3} = 3 and z_{1} + z_{2} + z_{3} = - 6 \Rightarrow \frac{x_{1} + x_{2} + x_{3}}{3} = 1, \frac{y_{1} + y_{2} + y_{3}}{3} = 1 and \frac{z_{1} + z_{2} + z_{3}}{3} = - 2$
Hence, the centroid of the traingle ABC is (1, 1, −2).

Page No 28.25:

Question 5:

Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.

Answer:

We know that the x and z coordinates on y - axis are 0
The coordinates of the foot of the perpendicular from a point (1, 2, 3) on y - axis are (0, 2, 0)

Page No 28.25:

Question 6:

Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.

Answer:

The distance of the point P(3, 5, 12) from x-axis is given by
$\sqrt{5^{2} + {(12)}^{2}} = \sqrt{169} = 13$

Page No 28.25:

Question 7:

Write the coordinates of third vertex of a triangle having centroid at the origin and two vertices at (3, −5, 7) and (3, 0, 1).

Answer:

Let the coordinates of third vertex be (x₁, y₁, z₁)
Now,
$\frac{x_{1} + 3 + 3}{3} = 0, \frac{y_{1} - 5 + 0}{3} = 0 and \frac{z_{1} + 7 + 1}{3} = 0 \Rightarrow x_{1} = - 6, y_{1} = 5 and z_{1} = - 8$
Hence, the coordinates of third vertex of a triangle is (−6, 5, −8).

Page No 28.25:

Question 8:

What is the locus of a point for which y = 0, z = 0?

Answer:

We know that on x - axis both y = 0, z = 0.
Hence, the locus of a point for which y = 0, z = 0 is x - axis.

Page No 28.25:

Question 9:

Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.

Answer:

Let the yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in m : 1.
Now, we know that on yz-plane the coordinate of x is 0.
$∴ \frac{m \times 3 + 1 \times 2}{m + 1} = 0 \Rightarrow 3 m + 2 = 0 \Rightarrow m = - \frac{2}{3}$
Hence, yz - plane divide the line sgement joining the points (2, 4,5) and (3, −5, 4) in 2 : 3 externally.

Page No 28.25:

Question 10:

Find the point on y-axis which is at a distance of $\sqrt{10}$ units from the point (1, 2, 3).

Answer:

We know that the x and z coordinates of the point on the y-axis are 0.
So, let the required point be (0, y, 0)
Now,
$\sqrt{{(1 - 0)}^{2} + {(2 - y)}^{2} + {(3 - 0)}^{2}} = \sqrt{10} \Rightarrow 1 + 4 - 4 y + y^{2} + 9 = 10 \Rightarrow y^{2} - 4 y + 4 = 0 \Rightarrow {(y - 2)}^{2} = 0 \Rightarrow y = 2, 2$
Hence, the required point is (0, 2, 0)

Page No 28.25:

Question 11:

Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).

Answer:

We know that the y and z coordinates of the point on the x-axis are 0.
So, let the required point be C (x, y, z)
Now, CA = CB

$\sqrt{{(3 - x)}^{2} + {(2 - 0)}^{2} + {(2 - 0)}^{2}} = \sqrt{{(5 - x)}^{2} + {(5 - 0)}^{2} + {(4 - 0)}^{2}} \Rightarrow 9 - 6 x + x^{2} + 4 + 4 = 25 - 10 x + x^{2} + 25 + 16 \Rightarrow 17 - 6 x + x^{2} = 66 - 10 x + x^{2} \Rightarrow 4 x = 49 \Rightarrow x = \frac{49}{4}$
Hence, the required point is $(\frac{49}{4}, 0, 0)$

Page No 28.25:

Question 12:

Find the coordinates of a point equidistant from the origin and points A (a, 0, 0), B (0, b, 0) andC(0, 0, c).

Answer:

Let the point be P(x, y, z).
Now, PO = PA
$\sqrt{{(0 - x)}^{2} + {(0 - y)}^{2} + {(0 - z)}^{2}} = \sqrt{{(a - x)}^{2} + {(0 - y)}^{2} + (0 - z)} \Rightarrow x^{2} + y^{2} + z^{2} = a^{2} - 2 a x + x^{2} + y^{2} + z^{2} \Rightarrow 0 = a^{2} - 2 a x \Rightarrow x = \frac{a}{2}$
Also, PO = PB
$\sqrt{{(0 - x)}^{2} + {(0 - y)}^{2} + {(0 - z)}^{2}} = \sqrt{{(0 - x)}^{2} + {(b - y)}^{2} + (0 - z)} \Rightarrow x^{2} + y^{2} + z^{2} = x^{2} + b^{2} - 2 b y + y^{2} + z^{2} \Rightarrow 0 = b^{2} - 2 b y \Rightarrow y = \frac{b}{2}$
Again, PO = PC
$\sqrt{{(0 - x)}^{2} + {(0 - y)}^{2} + {(0 - z)}^{2}} = \sqrt{{(0 - x)}^{2} + {(0 - y)}^{2} + (c - z)} \Rightarrow x^{2} + y^{2} + z^{2} = x^{2} + y^{2} + c^{2} - 2 c z + z^{2} \Rightarrow 0 = c^{2} - 2 c z \Rightarrow z = \frac{c}{2}$

Hence, the coordinates of the point is $(\frac{a}{2}, \frac{b}{2}, \frac{c}{2})$

Page No 28.25:

Question 13:

Write the coordinates of the point P which is five-sixth of the way from A(−2, 0, 6) to B(10, −6, −12).

Answer:

Let the coordinates of the point be P(x, y, z).
Now
$PA = \frac{5}{6} PB \Rightarrow \frac{PA}{PB} = \frac{5}{6} ∴ x = \frac{5 \times 10 - 6 \times 2}{5 + 6}, y = \frac{5 \times (- 6) + 6 \times 0}{5 + 6}, z = \frac{5 \times (- 12) + 6 \times 6}{5 + 6} ∴ x = \frac{38}{11}, y = \frac{- 24}{11}, z = \frac{- 34}{11}$
Hence, the coordinates of the point is $(\frac{38}{11}, \frac{- 24}{11}, \frac{- 34}{11})$

Page No 28.25:

Question 14:

If a parallelopiped is formed by the planes drawn through the points (2,3,5) and (5, 9, 7) parallel to the coordinate planes, then write the lengths of edges of the parallelopiped and length of the diagonal.

Answer:

Lengths of edges of the parallelopiped are given by
5 − 2, 9 − 3, 7 − 5
= 3, 6, 2
Length of the diagonal is given by
$\sqrt{3^{2} + 6^{2} + 2^{2}} = \sqrt{49} = 7 units$

Page No 28.25:

Question 15:

Determine the point on yz-plane which is equidistant from points A(2, 0, 3), B(0, 3,2) and C(0, 0,1).

Answer:

The coordiante of x point on yz-plane is 0
Let the point be P(0, y, z).
Now, PA = PB
$\sqrt{{(2 - 0)}^{2} + {(0 - y)}^{2} + {(3 - z)}^{2}} = \sqrt{{(0 - 0)}^{2} + {(3 - y)}^{2} + (2 - z)} \Rightarrow 4 + y^{2} + 9 - 6 z + z^{2} = 9 - 6 y + y^{2} + 4 - 4 z + z^{2} \Rightarrow - 6 z = - 6 y - 4 z \Rightarrow 3 y - z = 0 . . . . . (1)$
Also, PA = PC
$\sqrt{{(2 - 0)}^{2} + {(0 - y)}^{2} + {(3 - z)}^{2}} = \sqrt{{(0 - 0)}^{2} + {(0 - y)}^{2} + {(1 - z)}^{2}} \Rightarrow 4 + y^{2} + 9 - 6 z + z^{2} = y^{2} + 1 - 2 z + z^{2} \Rightarrow 13 - 6 z = 1 - 2 z \Rightarrow - 4 z = - 12 \Rightarrow z = 3 . . . . . (2)$
Solving (1) and (2),we get
y = 1
Hence, the coordinates of the point is (0, 1, 3).

Page No 28.25:

Question 16:

If the origin is the centroid of a triangle ABC having vertices A(a, 1, 3), B(−2, b −5) and C (4, 7, c), find the values of a, b, c.

Answer:

We have A(a, 1, 3), B(−2, b −5) and C (4, 7, c)
Now,
$\frac{a - 2 + 4}{3} = 0, \frac{1 + b + 7}{3} = 0 and \frac{3 - 5 + c}{3} = 0 \Rightarrow a = - 2, b = - 8 and c = 2$

Page No 28.6:

Question 1:

Name the octants in which the following points lie:
(i) (5, 2, 3)
(ii) (–5, 4, 3)
(iii) (4, –3, 5)
(iv) 7, 4, –3)
(v) (–5, –4, 7)
(vi) (–5, –3, –2)
(vii) (2, –5, –7)
(viii) (–7, 2 – 5)

Answer:

(i) The x-coordinate, the y-coordinate and the z-coordinate of the point (5, 2, 3) are all positive.
      Therefore, this point lies in XOYZ octant.

(ii) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, 4, 3) are negative, positive and positive, respectively.
      Therefore, this point lies in X'OYZ octant.

(iii) The x-coordinate, the y-coordinate and the z-coordinate of the point (4, −3, 5) are positive, negative and positive, respectively.
      Therefore, this point lies in XOY'Z octant.

(iv) The x-coordinate, the y-coordinate and the z-coordinate of the point (7, 4, −3) are positive, positive and negative, respectively.
      Therefore, this point lies in XOYZ' octant.

(v) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −4, 7) are negative, negative and positive, respectively.
      Therefore, this point lies in X'OY'Z octant .

(vi) The x-coordinate, the y-coordinate and the z-coordinate of the point (−5, −3, −2) are all negative.
      Therefore, this point lies in X'OY'Z' octant

(vii) The x-coordinate, the y-coordinate and the z-coordinate of the point (2, −5, −7) are positive, negative and negative, respectively.
      Therefore, this point lies in XOY'Z' octant.

(viii) The x-coordinate, the y-coordinate and the z-coordinate of the point(−7, 2, −5) are negative, positive and negative, respectively.
      Therefore, this point lies in X'OYZ' octant.

Page No 28.6:

Question 2:

Find the image of:
(i) (–2, 3, 4) in the yz-plane.
(ii) (–5, 4, –3) in the xz-plane.
(iii) (5, 2, –7) in the xy-plane.
(iv) (–5, 0, 3) in the xz-plane.
(v) (–4, 0, 0) in the xy-plane.

Answer:

(i) $(2, 3, 4)$
(ii) $(- 5, - 4, - 3)$
(iii) $(5, 2, 7)$
(iv) $(- 5, 0, 3)$
(v) $(- 4, 0, 0)$

Page No 28.6:

Question 3:

A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.

Answer:

Let P $\equiv$ (1, 0, −1)
The length of each side of the cube is 5.
The three edges from vertex of the cube are drawn from P towards the negative x and y axes and the positive z-axis.
Therefore, the coordinates of the vertex of the cube will be as follows:
x-coordinate = 1, 1 $-$ 5 = $-$ 4, i.e. 1, $-$ 4
y-coordinate = 0, 0 $-$ 5 = $-$ 5, i.e. 0, $-$ 5
z-coordinate = $-$ 1, $-$ 1 + 5 = 4, i.e. $-$ 1, 4
Hence, the remaining seven vertices of the cube are as follows:
$(1, 0, 4) (1, - 5, - 1) (1, - 5, 4) (- 4, 0, - 1) (- 4, 0, 4) (- 4, - 5, - 1) (- 4, - 5, 4)$

Page No 28.6:

Question 4:

Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.

Answer:

Let P $\equiv$ (3, 0, −1), Q $\equiv$ (−2, 5, 4)
PE = Distance between the parallel planes ABCP and FQDE
    = $|4 + 1| = 5$
                                       (These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
     = $|- 2 - 3| = 5$
                                        (These planes are perpendicular to the x-axis)
Similarly, PC = $|5 - 0| = 5$
Thus, the length of the edges of the parallelepiped are 5, 5 and 5

Page No 28.7:

Question 5:

Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.

Answer:

Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.

$∴$ PA = Distance between planes PBEC and QDAF
         = 5 $-$ 3
           = 2
PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and $-$ 2, respectively, from the zx-plane.
$∴$ PB = 0 $-$ ( $-$ 2)
          = 2
PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.
$∴$ PC = 2 $-$ 5
           = $-$ 3

Page No 28.7:

Question 6:

Find the distances of the point P(–4, 3, 5) from the coordinate axes.

Answer:

Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.
PA will be perpendicular to the x-axis.
Also, QA = $|3|$ and PQ = $|5|$
Now, distance of P from the x-axis:
PB = $\sqrt{B Q^{2} + Q P^{2}}$
     $= \sqrt{3^{2} + 5^{2}} = \sqrt{9 + 25} = \sqrt{34}$
Similarly,
From the right-angled $∆ P A Q$ :
distance of P from the y-axis:
PA = $\sqrt{A Q^{2} + Q P^{2}}$
     $= \sqrt{{(- 4)}^{2} + {(5)}^{2}} = \sqrt{16 + 25} = \sqrt{41}$
Similarly, the length of the perpendicular from P to the z-axis = $\sqrt{{(- 4)}^{2} + {(3)}^{2}}$
                                                                                    $= \sqrt{16 + 9} = \sqrt{25} = 5$

Page No 28.7:

Question 7:

The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.

Answer:

The seven coordinates are as follows:
$(- 3, 2, 5) (3, 2, - 5) (- 3, - 2, 5) (3, - 2, - 5) (- 3, 2, - 5) (- 3, - 2, - 5) (3, 2, 5)$

Page No 28.9:

Question 1:

Find the distance between the following pairs of points:
(i) P(1, –1, 0) and Q(2, 1, 2)
(ii) A(3, 2, –1) and B(–1, –1, –1).

Answer:

(i) PQ = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
         = $\sqrt{{(2 - 1)}^{2} + {(1 + 1)}^{2} + {(2 - 0)}^{2}}$
         = $\sqrt{1^{2} + 2^{2} + 2^{2}}$
         = $\sqrt{1 + 4 + 4}$
         = $\sqrt{9}$
         = 3

(ii) AB = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
           $= \sqrt{{(- 1 - 3)}^{2} + {(- 1 - 2)}^{2} + {(- 1 + 1)}^{2}} = \sqrt{{(- 4)}^{2} + {(- 3)}^{2} + {(0)}^{2}} = \sqrt{16 + 9 + 0} = \sqrt{25} = 5$

Page No 28.9:

Question 2:

Find the distance between the points P and Q having coordinates (–2, 3, 1) and (2, 1, 2).

Answer:

PQ = $\sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}}$
$= \sqrt{{(2 + 2)}^{2} + {(1 - 3)}^{2} + {(2 - 1)}^{2}} = \sqrt{{(4)}^{2} + {(- 2)}^{2} + {(1)}^{2}} = \sqrt{16 + 4 + 1} = \sqrt{21} units$

Page No 28.9:

Question 3:

Using distance formula prove that the following points are collinear:
(i) A(4, –3, –1), B(5, –7, 6) and C(3, 1, –8)
(ii) P(0, 7, –7), Q(1, 4, –5) and R(–1, 10, –9)
(iii) A(3, –5, 1), B(–1, 0, 8) and C(7, –10, –6)

Answer:

(i) AB = $\sqrt{{(5 - 4)}^{2} + {(- 7 + 3)}^{2} + {(6 + 1)}^{2}}$
           $= \sqrt{{(1)}^{2} + {(- 4)}^{2} + {(7)}^{2}} = \sqrt{1 + 16 + 49} = \sqrt{66}$

      BC = $\sqrt{{(3 - 5)}^{2} + {(1 + 7)}^{2} + {(- 8 - 6)}^{2}}$
           = $\sqrt{{(- 2)}^{2} + {(8)}^{2} + {(- 14)}^{2}}$
           $= \sqrt{4 + 64 + 196} = \sqrt{264} = 2 \sqrt{66}$

      AC = $\sqrt{{(3 - 4)}^{2} + {(1 + 3)}^{2} + {(- 8 + 1)}^{2}}$
           $= \sqrt{{(- 1)}^{2} + {(4)}^{2} + {(- 7)}^{2}} = \sqrt{1 + 16 + 49} = \sqrt{66}$

$Here, AB + AC = \sqrt{66} + \sqrt{66} = 2 \sqrt{66} = BC$
Hence, the points are collinear.

(ii) PQ = $\sqrt{{(1 - 0)}^{2} + {(4 - 7)}^{2} + {(- 5 + 7)}^{2}}$
           $= \sqrt{{(1)}^{2} + {(- 3)}^{2} + {(2)}^{2}} = \sqrt{1 + 9 + 4} = \sqrt{14}$

      QR = $\sqrt{{(- 1 - 1)}^{2} + {(10 - 4)}^{2} + {(- 9 + 5)}^{2}}$
           $= \sqrt{{(- 2)}^{2} + {(6)}^{2} + {(- 4)}^{2}} = \sqrt{4 + 36 + 16} = \sqrt{56} = 2 \sqrt{14}$

      PR $= \sqrt{{(- 1 - 0)}^{2} + {(10 - 7)}^{2} + {(- 9 + 7)}^{2}}$
            $= \sqrt{{(- 1)}^{2} + {(3)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 9 + 4} = \sqrt{14}$

$Here, PQ + PR = \sqrt{14} + \sqrt{14} = 2 \sqrt{14} = QR$
Hence, the points are collinear.

(iii) AB = $\sqrt{{(- 1 - 3)}^{2} + {(0 + 5)}^{2} + {(8 - 1)}^{2}}$
            $= \sqrt{{(- 4)}^{2} + {(5)}^{2} + {(7)}^{2}} = \sqrt{16 + 25 + 49} = \sqrt{90} = 3 \sqrt{10}$

       BC = $\sqrt{{(7 + 1)}^{2} + {(- 10 - 0)}^{2} + {(- 6 - 8)}^{2}}$
             $= \sqrt{{(8)}^{2} + {(- 10)}^{2} + {(- 14)}^{2}} = \sqrt{64 + 100 + 196} = \sqrt{360} = 6 \sqrt{10}$

      AC = $\sqrt{{(7 - 3)}^{2} + {(- 10 + 5)}^{2} + {(- 6 - 1)}^{2}}$
           $= \sqrt{{(4)}^{2} + {(- 5)}^{2} + {(- 7)}^{2}} = \sqrt{16 + 25 + 49} = \sqrt{90} = 3 \sqrt{10}$

$Here, AB + AC = 3 \sqrt{10} + 3 \sqrt{10} = 6 \sqrt{10} = BC$
Hence, the points are collinear.

Page No 28.9:

Question 4:

Determine the points in (i) xy-plane (ii) yz-plane and (iii) zx-plane which are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).

Answer:

(i) We know that the z-coordinate of every point on the xy-plane is zero.
So, let P (x, y, 0) be a point on the xy-plane such that PA = PB = PC
Now, PA = PB
$\Rightarrow P A^{2} = P B^{2} \Rightarrow {(x - 1)}^{2} + {(y + 1)}^{2} + {(0 - 0)}^{2} = {(x - 2)}^{2} + {(y - 1)}^{2} + {(0 - 2)}^{2} \Rightarrow x^{2} - 2 x + 1 + y^{2} + 2 y + 1 = x^{2} - 4 x + 4 + y^{2} - 2 y + 1 + 4 \Rightarrow - 2 x + 4 x + 2 y + 2 y + 2 - 9 = 0 \Rightarrow 2 x + 4 y - 7 = 0 \Rightarrow 2 x + 4 y = 7 . . . (1) P B = P C \Rightarrow P B^{2} = P C^{2} \Rightarrow {(x - 2)}^{2} + {(y - 1)}^{2} + {(0 - 2)}^{2} = {(x - 3)}^{2} + {(y - 2)}^{2} + {(0 + 1)}^{2} \Rightarrow x^{2} - 4 x + 4 + y^{2} - 2 y + 1 + 4 = x^{2} - 6 x + 9 + y^{2} - 4 y + 4 + 1 \Rightarrow - 4 x + 6 x - 2 y + 4 y + 9 - 14 = 0 \Rightarrow 2 x + 2 y - 5 = 0 \Rightarrow x + y = \frac{5}{2} \Rightarrow x = \frac{5}{2} - y . . . (2) Putting the value of x in equation (1) : 2 (\frac{5}{2} - y) + 4 y = 7 \Rightarrow 5 - 2 y + 4 y = 7 \Rightarrow 5 + 2 y = 7 \Rightarrow 2 y = 2 \Rightarrow y = \frac{2}{2} ∴ y = 1 Putting the value of y in equation (2) : x = \frac{5}{2} - 1 x = \frac{5 - 2}{2} x = \frac{3}{2} Hence, the required point is (\frac{3}{2}, 1, 0) .$

(ii) We know that the x-coordinate of every point on the yz-plane is zero.
So, let P (0, y, z) be a point on the yz-plane such that PA = PB = PC
Now, PA = PB
$\Rightarrow {(0 - 1)}^{2} + {(y + 1)}^{2} + {(z - 0)}^{2} = {(0 - 2)}^{2} + {(y - 1)}^{2} + {(z - 2)}^{2}$
$\Rightarrow 1 + y^{2} + 2 y + 1 + z^{2} = 4 + y^{2} - 2 y + 1 + z^{2} - 4 z + 4 \Rightarrow 2 y + 2 = - 2 y - 4 z + 9 \Rightarrow 2 y + 2 y - 4 z = 9 - 2 \Rightarrow 4 y - 4 z = 7 \Rightarrow y - z = \frac{7}{4} . . . (1) P B = P C \Rightarrow P B^{2} = P C^{2} \Rightarrow {(0 - 2)}^{2} + {(y - 1)}^{2} + {(z - 2)}^{2} = {(0 - 3)}^{2} + {(y - 2)}^{2} + {(z + 1)}^{2} \Rightarrow 4 + y^{2} - 2 y + 1 + z^{2} - 4 z + 4 = 9 + y^{2} - 4 y + 4 + z^{2} + 2 z + 1 \Rightarrow - 2 y - 4 z + 9 = - 4 y + 2 z + 14 \Rightarrow - 2 y + 4 y - 4 z - 2 z = 14 - 9 \Rightarrow 2 y - 6 z = 5 \Rightarrow y - 3 z = \frac{5}{2} ∴ y = \frac{5}{2} + 3 z (2) Putting the value of y in equation (1) : y - z = \frac{7}{4} \Rightarrow \frac{5}{2} + 3 z - z = \frac{7}{4} \Rightarrow 2 z = \frac{7}{4} - \frac{5}{2} \Rightarrow 2 z = \frac{7 - 10}{4} \Rightarrow 2 z = \frac{- 3}{4} ∴ z = \frac{- 3}{8} Putting the value of z in equation (2) : y = \frac{5}{2} + 3 z \Rightarrow y = \frac{5}{2} + 3 (\frac{- 3}{8}) \Rightarrow y = \frac{5}{2} - \frac{9}{8} \Rightarrow y = \frac{20 - 9}{8} ∴ y = \frac{11}{8}$
Hence, the required point is $(0, \frac{11}{8}, \frac{- 3}{8})$ .

(iii) We know that the y-coordinate of every point on the zx-plane is zero.
So, let P (x, 0, z) be a point on the zx-plane such that PA = PB = PC
Now, PA = PB
$\Rightarrow {(x - 1)}^{2} + {(0 + 1)}^{2} + {(z - 0)}^{2} = {(x - 2)}^{2} + {(0 - 1)}^{2} + {(x - 2)}^{2}$
$\Rightarrow x^{2} + 1 - 2 x + 1 + z^{2} = x^{2} - 4 x + 4 + 1 + z^{2} - 4 z + 4 \Rightarrow - 2 x + 2 = - 4 x - 4 z + 9 \Rightarrow - 2 x + 4 x - 4 z = 7 \Rightarrow 2 x - 4 z = 7 \Rightarrow x - 2 z = \frac{7}{2} . . . (1) P B = P C \Rightarrow P B^{2} = P C^{2} \Rightarrow {(x - 2)}^{2} + {(0 - 1)}^{2} + {(z - 2)}^{2} = {(x - 3)}^{2} + {(0 - 2)}^{2} + {(z + 1)}^{2} \Rightarrow x^{2} - 4 x + 4 + 1 + z^{2} - 4 z + 4 = x^{2} - 6 x + 9 + 4 + z^{2} + 2 z + 1 \Rightarrow - 4 x - 4 z + 9 = - 6 x + 2 z + 14 \Rightarrow - 4 x + 6 x - 4 z - 2 z = 14 - 9 \Rightarrow 2 x - 6 z = 5 \Rightarrow x - 3 z = \frac{5}{2} ∴ x = \frac{5}{2} + 3 z . . . (2) Putting the value of x in equation (1) : x - 2 z = \frac{7}{2} \Rightarrow \frac{5}{2} + 3 z - 2 z = \frac{7}{2} \Rightarrow \frac{5}{2} + z = \frac{7}{2} \Rightarrow z = \frac{7}{2} - \frac{5}{2} \Rightarrow z = \frac{7 - 5}{2} \Rightarrow z = \frac{2}{2} ∴ z = 1 Putting the value of z in equation (2) : x = \frac{5}{2} + 3 z \Rightarrow x = \frac{5}{2} + 3 (1) \Rightarrow x = \frac{5}{2} + 3 \Rightarrow x = \frac{5 + 6}{2} ∴ x = \frac{11}{2}$
Hence, the required point is $(\frac{11}{2}, 0, 1)$ .

Page No 28.9:

Question 5:

Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).

Answer:

Let M be the point on the z-axis.
Then, the coordinates of M will be $(0, 0, z)$ .

Let M be equidistant from the points A $(1, 5, 7)$ and B $(5, 1, - 4)$

AM = $\sqrt{{(0 - 1)}^{2} + {(0 - 5)}^{2} + {(z - 7)}^{2}}$
$= \sqrt{{(- 1)}^{2} + {(- 5)}^{2} + {(z - 7)}^{2}} = \sqrt{1 + 25 + z^{2} - 14 z + 49} = \sqrt{z^{2} - 14 z + 75}$

BM = $\sqrt{{(0 - 5)}^{2} + {(0 - 1)}^{2} + {(z + 4)}^{2}}$
$= \sqrt{{(- 5)}^{2} + {(- 1)}^{2} + z^{2} + 8 z + 16} = \sqrt{25 + 1 + z^{2} + 8 z + 16} = \sqrt{z^{2} + 8 z + 42}$

Then, AM = BM

$∴ \sqrt{z^{2} - 14 z + 75} = \sqrt{z^{2} + 8 z + 42} z^{2} - 14 z + 75 = z^{2} + 8 z + 42 - 14 z - 8 z = 42 - 75 - 22 z = - 33 z = \frac{33}{22} z = \frac{3}{2}$
Thus, the coordinates of M are $(0, 0, \frac{3}{2})$ .

Page No 28.9:

Question 6:

Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).

Answer:

Let the point on the y-axis be Y $(0, y, 0)$ which is equidistant from the points P $(3, 1, 2)$ and Q $(5, 5, 2)$ .
Then ,PY=QY
Now, $\sqrt{{(0 - 3)}^{2} + {(y - 1)}^{2} + {(0 - 2)}^{2}}$ = $\sqrt{{(0 - 5)}^{2} + {(y - 5)}^{2} + {(0 - 2)}^{2}}$

$\Rightarrow \sqrt{{(- 3)}^{2} + y^{2} + 2 y + 1 + {(- 2)}^{2}} = \sqrt{{(- 5)}^{2} + {(y - 5)}^{2} + {(- 2)}^{2}} \Rightarrow \sqrt{9 + y^{2} + 2 y + 1 + 4} = \sqrt{25 + y^{2} + 10 y + 25 + 4} \Rightarrow \sqrt{y^{2} + 2 y + 14} = \sqrt{y^{2} + 10 y + 54}$
$\Rightarrow y^{2} + 2 y + 14 = y^{2} + 10 y + 54$

$2 y - 10 y = 54 - 14 - 8 y = 40 - y = \frac{40}{8} y = - 5$

Thus, the required point on the y-axis is (0, −5, 0).

Page No 28.9:

Question 7:

Find the points on z-axis which are at a distance $\sqrt{21}$ from the point (1, 2, 3).

Answer:

Let the point be A (0, 0, z).Then,

AP = $\sqrt{21}$
$\Rightarrow \sqrt{{(0 - 1)}^{2} + {(0 - 2)}^{2} + {(z - 3)}^{2}} = \sqrt{21} \Rightarrow {(- 1)}^{2} + {(- 2)}^{2} + {(z - 3)}^{2} = 21 \Rightarrow 1 + 4 + {(z - 3)}^{2} = 21 \Rightarrow {(z - 3)}^{2} = 21 - 5 \Rightarrow {(z - 3)}^{2} = 16 \Rightarrow z - 3 = \pm 4 \Rightarrow z - 3 = 4 or z - 3 = - 4 \Rightarrow z = 7 or z = - 1$
Hence, the coordinates of the required point are (0, 0, 7) and (0, 0, −1).

Page No 28.9:

Question 8:

Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.

Answer:

Let A (1, 2, 3) , B (2, 3, 1) and C (3, 1, 2) are the coordinates of the triangle $△ A B C$
AB = $\sqrt{{(2 - 1)}^{2} + {(3 - 2)}^{2} + {(1 - 3)}^{2}}$
      $= \sqrt{{(1)}^{2} + {(1)}^{2} + {(- 2)}^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6}$

BC = $\sqrt{{(3 - 2)}^{2} + {(1 - 3)}^{2} + {(2 - 1)}^{2}}$
       $= \sqrt{{(1)}^{2} + {(- 2)}^{2} + {(1)}^{2}} = \sqrt{1 + 4 + 1} = \sqrt{6}$

AC = $\sqrt{{(3 - 1)}^{2} + {(1 - 2)}^{2} + {(2 - 3)}^{2}}$
      $= \sqrt{{(2)}^{2} + {(- 1)}^{2} + {(- 1)}^{2}} = \sqrt{4 + 1 + 1} = \sqrt{6}$
Now, AB = BC = AC

Therefore, it is an equilateral triangle.

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