Rd Sharma XII Vol 2 2020 for Class 12 Science Maths Chapter 8 - Direction Cosines And Direction Ratios

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Rd Sharma XII Vol 2 2020 Solutions for Class 12 Science Maths Chapter 8 Direction Cosines And Direction Ratios are provided here with simple step-by-step explanations. These solutions for Direction Cosines And Direction Ratios are extremely popular among Class 12 Science students for Maths Direction Cosines And Direction Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma XII Vol 2 2020 Book of Class 12 Science Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma XII Vol 2 2020 Solutions. All Rd Sharma XII Vol 2 2020 Solutions for class Class 12 Science Maths are prepared by experts and are 100% accurate.

Page No 26.23:

Question 1:

If a line makes angles of 90°, 60° and 30° with the positive direction of x, y, and z-axis respectively, find its direction cosines.

Answer:

Let the direction cosines of the line be l, m, n.

Now,

$l = \cos 90^{0} = 0 m = \cos 60^{0} = \frac{1}{2} n = \cos 30^{0} = \frac{\sqrt{3}}{2} Therefore, the direction cosines of the line are 0, \frac{1}{2}, \frac{\sqrt{3}}{2} .$

Page No 26.23:

Question 2:

If a line has direction ratios 2, −1, −2, determine its direction cosines.

Answer:

$Let the direction cosines of the line be l, m, n . Now, l = \frac{2}{\sqrt{2^{2} + {(- 1)}^{2} + {(- 2)}^{2}}} = \frac{2}{3} m = \frac{- 1}{\sqrt{2^{2} + {(- 1)}^{2} + {(- 2)}^{2}}} = \frac{- 1}{3} n = \frac{- 2}{\sqrt{2^{2} + {(- 1)}^{2} + {(- 2)}^{2}}} = \frac{- 2}{3} Therefore, the direction cosines of the line are \frac{2}{3}, \frac{- 1}{3}, \frac{- 2}{3} .$

Page No 26.23:

Question 3:

Find the direction cosines of the line passing through two points (−2, 4, −5) and (1, 2, 3).

Answer:

$The direction cosines of the line passing through two points P (x_{1}, y_{1}, z_{1}) and Q (x_{2}, y_{2}, z_{2}) are \frac{x_{2} - x_{1}}{PQ}, \frac{y_{2} - y_{1}}{PQ}, \frac{z_{2} - z_{1}}{PQ} . Here, PQ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} + {(z_{2} - z_{1})}^{2}} P = (- 2, 4, - 5) Q = (1, 2, 3) ∴ PQ = \sqrt{{[1 - (- 2)]}^{2} + {(2 - 4)}^{2} + {[3 - (- 5)]}^{2}} = \sqrt{77} Thus, the direction cosines of the line joining two points are \frac{1 - (- 2)}{\sqrt{77}}, \frac{2 - 4}{\sqrt{77}}, \frac{3 - (- 5)}{\sqrt{77}}, i . e . \frac{3}{\sqrt{77}}, \frac{- 2}{\sqrt{77}}, \frac{8}{\sqrt{77}} .$

Page No 26.23:

Question 4:

Using direction ratios show that the points A (2, 3, −4), B (1, −2, 3) and C (3, 8, −11) are collinear.

Answer:

$The given points are A (2, 3, - 4), B (1, - 2, 3) and C (3, 8, - 11) . We know that the direction ratios of the line joining the points, (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} . The direction ratios of the line joining A and B are 1 - 2, - 2 - 3, 3 + 4, i . e . - 1, - 5, 7 . The direction ratios of the line joining B and C are 3 - 1, 8 + 2, - 11 - 3, i . e . 2, 10, - 14 . It is clear that the direction ratios of BC are - 2 times that of AB, i . e . they are proportional . Therefore, AB is parallel to BC . Also, point B is common in both AB and BC . Therefore, points A, B and C are collinear .$

Page No 26.23:

Question 5:

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), (−1, 1, 2) and (−5, −5, −2).

Answer:

$The vertices of ∆ ABC are A (3, 5, - 4), B (- 1, 1, 2) and C (- 5, - 5, - 2) .$

$The direction ratios of AB are (- 1 - 3), (1 - 5), [2 - (- 4)], i . e . - 4, - 4, 6 . Therefore, the direction cosines of AB are \frac{- 4}{\sqrt{{(- 4)}^{2} + {(- 4)}^{2} + {(6)}^{2}}}, \frac{- 4}{\sqrt{{(- 4)}^{2} + {(- 4)}^{2} + {(6)}^{2}}}, \frac{6}{\sqrt{{(- 4)}^{2} + {(- 4)}^{2} + {(6)}^{2}}} = \frac{- 4}{2 \sqrt{17}}, \frac{- 4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}} = \frac{2}{\sqrt{17}}, \frac{2}{\sqrt{17}}, \frac{- 3}{\sqrt{17}} The direction ratios of BC are [- 5 - (- 1)], (- 5 - 1), (- 2 - 2), i . e . - 4, - 6, - 4 . Therefore, the direction cosines of BC are \frac{- 4}{\sqrt{{(- 4)}^{2} + {(- 6)}^{2} + {(- 4)}^{2}}}, \frac{- 6}{\sqrt{{(- 4)}^{2} + {(- 6)}^{2} + {(- 4)}^{2}}}, \frac{- 4}{\sqrt{{(- 4)}^{2} + {(- 6)}^{2} + {(- 4)}^{2}}} = \frac{- 4}{2 \sqrt{17}}, \frac{- 6}{2 \sqrt{17}}, \frac{- 4}{2 \sqrt{17}} = \frac{2}{\sqrt{17}}, \frac{3}{\sqrt{7}}, \frac{2}{\sqrt{17}} The direction ratios of CA are [3 - (- 5)], [5 - (- 5)], [- 4 - (- 2)], i . e . 8, 10, - 2 . Therefore, the direction cosines of CA are \frac{8}{\sqrt{{(8)}^{2} + {(10)}^{2} + {(- 2)}^{2}}}, \frac{10}{\sqrt{{(8)}^{2} + {(10)}^{2} + {(- 2)}^{2}}}, \frac{- 2}{\sqrt{{(8)}^{2} + {(10)}^{2} + {(- 2)}^{2}}} = \frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{- 2}{2 \sqrt{42}} = \frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{- 1}{\sqrt{42}}$

Page No 26.23:

Question 6:

Find the angle between the vectors with direction ratios proportional to 1, −2, 1 and 4, 3, 2.

Answer:

$Let \vec{a} be a vector with direction ratios 1, - 2, 1 . \Rightarrow \vec{a} = \hat{i} - 2 \hat{j} + \hat{k} . Let \vec{b} be a vector with direction ratios 4, 3, 2 . \Rightarrow \vec{b} = 4 \hat{i} + 3 \hat{j} + 2 \hat{k} . Let θ be the angle between the given vectors . Now, c os θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{(\hat{i} - 2 \hat{j} + \hat{k}) . (4 \hat{i} + 3 \hat{j} + 2 \hat{k})}{|\hat{i} - 2 \hat{j} + \hat{k}| |4 \hat{i} + 3 \hat{j} + 2 \hat{k}|} = \frac{4 - 6 + 2}{\sqrt{1 + 4 + 1} \sqrt{16 + 9 + 4}} = \frac{0}{\sqrt{6} \sqrt{29}} = 0 ∴ θ = \frac{π}{2} Thus, the angle between the given vectors measures \frac{π}{2} .$

Page No 26.23:

Question 7:

Find the angle between the vectors whose direction cosines are proportional to 2, 3, −6 and 3, −4, 5.

Answer:

$Let \vec{a} be a vector with direction ratios 2, 3, - 6 . \Rightarrow \vec{a} = 2 \hat{i} + 3 \hat{j} - 6 \hat{k} . Let \vec{b} be a vector with direction ratios 3, - 4, 5 . \Rightarrow \vec{b} = 3 \hat{i} - 4 \hat{j} + 5 \hat{k} Let θ be the angle between the given vectors . Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{(2 \hat{i} + 3 \hat{j} - 6 \hat{k}) . (3 \hat{i} - 4 \hat{j} + 5 \hat{k})}{|2 \hat{i} + 3 \hat{j} - 6 \hat{k}| |3 \hat{i} - 4 \hat{j} + 5 \hat{k}|} = \frac{6 - 12 - 30}{\sqrt{4 + 9 + 36} \sqrt{9 + 16 + 25}} = \frac{- 36}{\sqrt{49} \sqrt{50}} = \frac{- 36}{35 \sqrt{2}} Rationalising the result, we get \cos θ = - \frac{18 \sqrt{2}}{35} ∴ θ = \cos^{- 1} (- \frac{18 \sqrt{2}}{35}) Thus, the angle between the given vectors measures \cos^{- 1} (- \frac{18 \sqrt{2}}{35}) .$

Page No 26.23:

Question 8:

Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.

Answer:

$Let \vec{a} be a vector parallel to the vector with direction ratios 2, 3, 6 . \Rightarrow \vec{a} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k} . Let \vec{b} be a vector parallel to the vector with direction ratios 1, 2, 2 . \Rightarrow \vec{b} = \hat{i} + 2 \hat{j} + 2 \hat{k} Let θ be the angle between the the given vectors . Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{(2 \hat{i} + 3 \hat{j} + 6 \hat{k}) . (\hat{i} + 2 \hat{j} + 2 \hat{k})}{|2 \hat{i} + 3 \hat{j} + 6 \hat{k}| |\hat{i} + 2 \hat{j} + 2 \hat{k}|} = \frac{2 + 6 + 12}{\sqrt{4 + 9 + 36} \sqrt{1 + 4 + 4}} = \frac{20}{21} \Rightarrow θ = \cos^{- 1} (\frac{20}{21})$

Page No 26.23:

Question 9:

Show that the points (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.

Answer:

$Suppose the points are A (2, 3, 4), B (- 1 . - 2, 1) and C (5, 8, 7) . We know that the direction ratios of the line joining the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} . The direction ratios of A B are (- 1 - 2), (- 2 - 3), (1 - 4), i . e . - 3, - 5, - 3 . The direction ratios of B C are (5 - (- 1)), (8 - (- 2)), (7 - 1), i . e . 6, 10, 6 . It can be seen th at the direction ratios of B C are - 2 times that of A B, i . e . they are proportional . Therefore, A B is parallel to B C . Since point B is common in both A B and B C, points A, B, and C are collinear .$

Page No 26.23:

Question 10:

Show that the line through points (4, 7, 8) and (2, 3, 4) is parallel to the line through the points (−1, −2, 1) and (1, 2, 5).

Answer:

$We know that the direction ratios of the line passing through the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} . Let the first two points be A (4, 7, 8) and B (2, 3, 4) . Thus, the dir ection ratios of AB are (2 - 4), (3 - 7), (4 - 8), i . e . - 2, - 4, - 4 . Similarly, let the other two points be C (- 1, - 2, 1) and D (1, 2, 5) . Thus, the direction ratios of CD are [1 - (- 1)], [2 - (- 2)], (5 - 1), i . e . 2, 4, 4 . It can be seen that the direction ratios of CD are - 1 times that of AB, i . e . they are proportional . Therefore, AB and CD are parallel lines .$

Page No 26.23:

Question 11:

Show that the line through the points (1, −1, 2) and (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Answer:

$We know that two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 . The direction ratios of the line passing through the points (1, - 1, 2) and (3, 4, - 2) are (3 - 1), [4 - (- 1)], (- 2 - 2), i . e . 2, 5, - 4 . \Rightarrow a_{1} = 2, b_{1} = 5, c_{1} = - 4 Similarly, the direction ratios of the line passing through the points (0, 3, 2) and (3, 5, 6) are (3 - 0), (5 - 3), (6 - 2), i . e . 3, 2, 4 . \Rightarrow a_{2} = 3, b_{2} = 2, c_{2} = 4 ∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 \times 3 + 5 \times 2 + (- 4) \times 4 = 6 + 10 - 16 = 0$

Thus, the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Page No 26.23:

Question 12:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1) and (4, 3, −1).

Answer:

$We know that two lines with direction ratios a_{1}, b_{1}, c_{1} and a_{2}, b_{2}, c_{2} are perpendicular if a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 . The direction ratios of the line join ing the origin (0, 0, 0) to the point (2, 1, 1) are (2 - 0), (1 - 0), (1 - 0) or 2, 1, 1 . \Rightarrow a_{1} = 2, b_{1} = 1, c_{1} = 1 Simila rly, the direction ratios of the line joining the points (3, 5, - 1) and (4, 3, - 1) are (4 - 3), (3 - 5), [- 1 - (- 1)] or 1, - 2, 0 . \Rightarrow a_{2} = 1, b_{2} = - 2, c_{2} = 0 ∴ a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 2 - 2 + 0 = 0$

Therefore, the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1) and (4, 3, -1).

Page No 26.23:

Question 13:

Find the angle between the lines whose direction ratios are proportional to a, b, c and b − c, c − a, a − b.

Answer:

$Let θ be the angle between the given lines . We have a_{1} = a, b_{1} = b, c_{1} = c a_{2} = b - c, b_{2} = c - a, c_{2} = a - b Now, \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} = \frac{a (b - c) + b (c - a) + c (a - b)}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{{(b - c)}^{2} + (c - a) (a - b)}} = \frac{a b - a c + b c - a b + a c - b c}{\sqrt{a^{2} + b^{2} + c^{2}} \sqrt{{(b - c)}^{2} + (c - a) (a - b)}} = 0 \Rightarrow θ = \frac{π}{2} Thus, the angle between the given lines measures 90 ° .$

Page No 26.23:

Question 14:

If the coordinates of the points A, B, C, D are (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2), then find the angle between AB and CD.

Answer:

$The given points are A (1, 2, 3), B (4, 5, 7), C (- 4, 3, - 6) and D (2, 9, 2) . We know that the direction ratios of the line joining the points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) are x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1} . The direction ratios of A B are (4 - 1), (5 - 2), (7 - 3), i . e . 3, 3, 4 . The direction ratios of C D are [2 - (- 4)], (9 - 3), [2 - (- 6)], i . e . 6, 6, 8 . Let θ be the angle between A B and C D . We have a_{1} = 3, b_{1} = 3, c_{1} = 4 a_{2} = 6, b_{2} = 6, c_{2} = 8 ∴ \cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} = \frac{18 + 18 + 32}{\sqrt{9 + 9 + 16} \sqrt{36 + 36 + 64}} = \frac{68}{68} = 1 \Rightarrow θ = 0 ° Thus, the angle between A B and C D measures 0 ° .$

Page No 26.23:

Question 15:

Find the direction cosines of the lines, connected by the relations: l + m +n = 0 and 2lm + 2ln − mn = 0.

Answer:

$Given : l + m + n = 0 . . . (1) 2 l m + 2 \ln - n m = 0 . . . (2) From (1), we get l = - m - n Substituting l = - m - n in (2), we get 2 (- m - n) m + 2 (- m - n) n - m n = 0 \Rightarrow - 2 m^{2} - 2 m n - 2 m n - 2 n^{2} - m n = 0 \Rightarrow 2 m^{2} + 2 n^{2} + 5 m n = 0 \Rightarrow (m + 2 n) (2 m + n) = 0 \Rightarrow m = - 2 n, - \frac{n}{2} If m = - 2 n, then from (1), we get l = n . If m = - \frac{n}{2}, then from (1), we get l = - \frac{n}{2} . Thus, the direction ratios of the two lines are proportional to n, - 2 n, n and - \frac{n}{2}, - \frac{n}{2}, n, i . e . 1, - 2, 1 and - \frac{1}{2}, - \frac{1}{2}, 1 . Hence, their direction cosines are \pm \frac{1}{\sqrt{6}}, \pm \frac{- 2}{\sqrt{6}}, \pm \frac{1}{\sqrt{6}} \pm \frac{- 1}{\sqrt{6}}, \pm \frac{- 1}{\sqrt{6}}, \pm \frac{2}{\sqrt{6}}$

Page No 26.23:

Question 16:

Find the angle between the lines whose direction cosines are given by the equations
(i) l + m + n = 0 and l² + m² − n² = 0
(ii) 2l − m + 2n = 0 and mn + nl + lm = 0
(iii) l + 2m + 3n = 0 and 3lm − 4ln + mn = 0
(iv) 2l + 2m − n = 0, mn + ln + lm = 0

Answer:

$(i) Given : l + m + n = 0 . . . (1) l^{2} + m^{2} - n^{2} = 0 . . . (2) From (1), we get m = - l - n Substituting m = - l - n in (2), we get l^{2} + {(- l - n)}^{2} - n^{2} \Rightarrow l^{2} + l^{2} + n^{2} + 2 \ln - n^{2} = 0 \Rightarrow 2 l^{2} + 2 \ln = 0 \Rightarrow 2 l (l + n) = 0 \Rightarrow l = 0, l = - n If l = 0, then by substituting l = 0 in (1), we get m = - n . If l = - n, then by substituting l = - n in (1), we get m = 0 . Thus, the direction ratios of the two lines are proportional to 0, - n, n and - n, 0, n or 0, - 1, 1 and - 1, 0, 1 . Vectors parallel to these lines are \vec{a} = 0 \hat{i} - \hat{j} + \hat{k} \vec{b} = - \hat{i} + 0 \hat{j} + \hat{k} If θ is the angle between the lines, then θ is also the angle between \vec{a} and \vec{b} . Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1}{\sqrt{0 + 1 + 1} \sqrt{1 + 0 + 1}} = \frac{1}{2} \Rightarrow θ = \frac{π}{3}$

   $(ii) Given : 2 l - m + 2 n = 0 . . . (1) m n + n l + l m = 0 . . . (2) From (1), we get m = 2 l + 2 n Substituting m = 2 l + 2 n in (2), we get (2 l + 2 n) n + n l + l (2 l + 2 n) = 0 \Rightarrow 2 \ln + 2 n^{2} + n l + 2 l^{2} + 2 \ln = 0 \Rightarrow 2 l^{2} + 5 l n + 2 n^{2} = 0 \Rightarrow (l + 2 n) (2 l + n) = 0 \Rightarrow l = - 2 n, - \frac{n}{2} If l = - 2 n, then by substituting l = - 2 n in (1), we get m = - 2 n . If l = - \frac{n}{2}, then by substituting l = - \frac{n}{2} in (1), we get m = n . Thus, the direction ratios of the two lines are proportional to - 2 n, - 2 n, n and - \frac{n}{2}, n, n or - 2, - 2, 1 and - \frac{1}{2}, 1, 1 . Vectors parallel these lines are \vec{a} = - 2 \hat{i} - \hat{2 j} + \hat{k} \vec{b} = - \frac{1}{2} \hat{i} + \hat{j} + \hat{k} If θ is the angle between the lines, then θ is also the angle between \vec{a} and \vec{b .} Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{1 - 2 + 1}{\sqrt{4 + 4 + 1} \sqrt{\frac{1}{4} + 1 + 1}} = 0 \Rightarrow θ = \frac{π}{2}$

$(iii) Given : l + 2 m + 3 n = 0 . . . (1) 3 l m - 4 \ln + m n = 0 . . . (2) From (1), we get l = - 2 m - 3 n Substituting l = - 2 m - 3 n in (2), we get 3 (- 2 m - 3 n) m - 4 (- 2 m - 3 n) n + m n = 0 \Rightarrow - 6 m^{2} - 9 m n + 8 m n + 12 n^{2} + m n = 0 \Rightarrow 12 n^{2} - 6 m^{2} = 0 \Rightarrow m^{2} = 2 n^{2} \Rightarrow m = \sqrt{2} n, - \sqrt{2} n If m = \sqrt{2} n, then by substituting m = \sqrt{2} n in (1), we get l = n (- 2 \sqrt{2} - 3) . If m = - \sqrt{2} n, then by substituting m = - \sqrt{2} n in (1), we get l = n (2 \sqrt{2} - 3) . Thus, the direction ratios of the two lines are proportional to n (- 2 \sqrt{2} - 3), \sqrt{2} n, n and n (2 \sqrt{2} - 3), - \sqrt{2} n, n or (- 2 \sqrt{2} - 3), \sqrt{2}, 1 and (- 2 \sqrt{2} - 3), - \sqrt{2}, 1 . Vectors parallel to these lines are \vec{a} = (- 2 \sqrt{2} - 3) \hat{i} + \sqrt{2} \hat{j} + \hat{k} \vec{b} = (2 \sqrt{2} - 3) \hat{i} - \sqrt{2} \hat{j} + \hat{k} If θ is the angle between the lines, then θ is also the angle between \vec{a} and \vec{b} . Now, \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{[(- 2 \sqrt{2} - 3) \hat{i} + \sqrt{2} \hat{j} + \hat{k}] . [(2 \sqrt{2} - 3) \hat{i} - \sqrt{2} \hat{j} + \hat{k}]}{\sqrt{8 + 9 + 12 \sqrt{2} + 2 + 1} \sqrt{8 + 9 - 12 \sqrt{2} + 2 + 1}} = \frac{- (8 - 9) - 2 + 1}{\sqrt{20 + 12 \sqrt{2}} \sqrt{20 - 12 \sqrt{2}}} = \frac{0}{\sqrt{20 + 12 \sqrt{2}} \sqrt{20 - 12 \sqrt{2}}} = 0 \Rightarrow θ = \frac{π}{2}$

(iv) The given relations are

2l + 2m − n = 0               .....(1)

mn + ln + lm = 0               .....(2)

From (1), we have

n = 2l + 2m

Putting this value of n in (2), we get

$m (2 l + 2 m) + l (2 l + 2 m) + l m = 0 \Rightarrow 2 l m + 2 m^{2} + 2 l^{2} + 2 l m + l m = 0 \Rightarrow 2 m^{2} + 5 l m + 2 l^{2} = 0 \Rightarrow (2 m + l) (m + 2 l) = 0 \Rightarrow 2 m + l = 0 or m + 2 l = 0 \Rightarrow l = - 2 m or l = - \frac{m}{2}$

When $l = - 2 m$ , we have

$n = 2 \times (- 2 m) + 2 m = - 4 m + 2 m = - 2 m$

When $l = - \frac{m}{2}$ , we have

$n = 2 \times (- \frac{m}{2}) + 2 m = - m + 2 m = m$

Thus, the direction ratios of two lines are proportional to

$- 2 m, m, - 2 m$ and $- \frac{m}{2}, m, m$

Or $- 2, 1, - 2$ and $- 1, 2, 2$

So, vectors parallel to these lines are $\vec{a} = - 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = - \hat{i} + 2 \hat{j} + 2 \hat{k}$ .

Let $θ$ be the angle between these lines, then $θ$ is also the angle between $\vec{a}$ and $\vec{b}$ .

$∴ \cos θ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{(- 2 \hat{i} + \hat{j} - 2 \hat{k}) . (- \hat{i} + 2 \hat{j} + 2 \hat{k})}{\sqrt{4 + 1 + 4} \sqrt{1 + 4 + 4}} = \frac{- 2 \times (- 1) + 1 \times 2 + (- 2) \times 2}{3 \times 3} = \frac{2 + 2 - 4}{9} = 0 \Rightarrow θ = \frac{π}{2}$

Thus, the angle between the two lines whose direction cosines are given by the given relations is $\frac{π}{2}$ .

Page No 26.24:

Question 1:

For every point P (x, y, z) on the xy-plane,
(a) x = 0
(b) y = 0
(c) z = 0
(d) x = y = z = 0

Answer:

Page No 26.24:

Question 2:

For every point P (x, y, z) on the x-axis (except the origin),
(a) x = 0, y = 0, z ≠ 0
(b) x = 0, z = 0, y ≠ 0
(c) y = 0, z = 0, x ≠ 0
(d) x = y = z = 0

Answer:

$(c) y = 0, z = 0, x \neq 0 Both Y and Z coordinates on each point of the x - axis are equal to zero . The X - coordinate on the origin is also equal to zero . Therefore, the Y and Z coordinates on each point of the x - axis, except the origin, are equal to zero, while the X - coordinate is non - zero .$

Page No 26.24:

Question 3:

A rectangular parallelopiped is formed by planes drawn through the points (5, 7, 9) and (2, 3, 7) parallel to the coordinate planes. The length of an edge of this rectangular parallelopiped is
(a) 2
(b) 3
(c) 4
(d) all of these

Answer:

$(d) all of these The given points (5, 7, 9) and (2, 3, 7) are two diagonally opposite vertices of the parallelopiped as all of their coordinates are different . ∴ Edges of the parallelopiped = |5 - 2|, |7 - 3|, |9 - 7| = 3, 4, 2$

Page No 26.24:

Question 4:

A parallelopiped is formed by planes drawn through the points (2, 3, 5) and (5, 9, 7), parallel to the coordinate planes. The length of a diagonal of the parallelopiped is
(a) 7
(b) $\sqrt{38}$
(c) $\sqrt{155}$
(d) none of these

Answer:

$(a) 7 The given points (2, 3, 5) and (5, 9, 7) are two diagonally opposite vertices of the parallelopiped as all of their coordinates are different . ∴ Edges of the parallelopiped = |2 - 5|, |3 - 9| and |5 - 7| = 3, 6 and 2 Now, Length of the diagonal of the parallelopiped = \sqrt{{(3)}^{2} + {(6)}^{2} + {(2)}^{2}} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7 Hence, length of the diagonal of the parallelopiped formed by the planes parallel to coordinate planes and drawn through points (2, 3, 5) and (5, 9, 7) is 7 units .$

Page No 26.25:

Question 5:

The xy-plane divides the line joining the points (−1, 3, 4) and (2, −5, 6)
(a) internally in the ratio 2 : 3
(b) externally in the ratio 2 : 3
(c) internally in the ratio 3 : 2
(d) externally in the ratio 3 : 2

Answer:

$(b) externally in the ratio 2 : 3 Let the X Y - plane divide the line segment joining points P (- 1, 3, 4) and Q (2, - 5, 6) in the ratio k : 1 . Using the section formula, the coordinates of the point of intersection are given by (\frac{k (2) - 1}{k + 1}, \frac{k (- 5) + 3}{k + 1}, \frac{k (6) + 4}{k + 1}) On the X Y - plane, the Z - coordinate of any point is zero . \Rightarrow \frac{k (6) + 4}{k + 1} = 0 \Rightarrow 6 k + 4 = 0 \Rightarrow k = - \frac{2}{3} Thus, the X Y - plane divides the line segment joining the given points in the ratio 2 : 3 externally .$

Page No 26.25:

Question 6:

If the x-coordinate of a point P on the join of Q (2, 2, 1) and R (5, 1, −2) is 4, then its z-coordinate is
(a) 2
(b) 1
(c) −1
(d) −2

Answer:

$(c) - 1 Suppose the point P divides the line joining the point Q (2, 2, 1) and R (5, 1, - 2) in the ratio k : 1 . Using the section formula, the coordinates of the point of intersection are given by (\frac{k (5) + 2}{k + 1}, \frac{k (1) + 2}{k + 1}, \frac{k (- 2) + 1}{k + 1}) It is given that the X - coordinate of P is 4 . \Rightarrow \frac{k (5) + 2}{k + 1} = 4 \Rightarrow 5 k + 2 = 4 (k + 1) \Rightarrow k = 2 Now, Z - coordinate of P = \frac{k (- 2) + 1}{k + 1} = \frac{2 (- 2) + 1}{2 + 1} [Substituting k = 2] = - 1$

Page No 26.25:

Question 7:

The distance of the point P (a, b, c) from the x-axis is
(a) $\sqrt{b^{2} + c^{2}}$

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

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

(d) none of these

Answer:

$(a) \sqrt{b^{2} + c^{2}} The projection of the point P (a, b, c) on the x - axis is (a, 0, 0) as both Y and Z coordinates on any point on the x - axis are equal to zero . ∴ Distance of P (a, b, c) from x - axis = Distance of P (a, b, c) from (a, 0, 0) = \sqrt{{(a - a)}^{2} + {(b - 0)}^{2} + {(c - 0)}^{2}} = \sqrt{b^{2} + c^{2}}$

Page No 26.25:

Question 8:

Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is
(a) 3 : 1 internally
(b) 3 : 1 externally
(c) 1 : 2 internally
(d) 2 : 1 externally

Answer:

$(b) 3 : 1 externally Suppose the X Y - plane divides the line segment joining the points P (1, 2, 3) and Q (4, 2, 1) in the ratio k : 1 . Using the section formula, the coordinates of the point of intersection are given by (\frac{k (4) + 1}{k + 1}, \frac{k (2) + 2}{k + 1}, \frac{k (1) + 3}{k + 1}) The Z - coordinate of any point on the X Y - plane is zero . \Rightarrow \frac{k (1) + 3}{k + 1} = 0 \Rightarrow k + 3 = 0 \Rightarrow k = - 3 = - \frac{3}{1} Thus, the X Y - plane divides the line segment joining the given points in the ratio 3 : 1 externally .$

Page No 26.25:

Question 9:

If P (3, 2, −4), Q (5, 4, −6) and R (9, 8, −10) are collinear, then R divides PQ in the ratio
(a) 3 : 2 internally
(b) 3 : 2 externally
(c) 2 : 1 internally
(d) 2 : 1 externally

Answer:

$(b) 3 : 2 externally Suppose the point R divides P Q in the ratio λ : 1 . Coordinates of R are (\frac{5 λ + 3}{λ + 1}, \frac{4 λ + 2}{λ + 1}, \frac{- 6 λ - 4}{λ + 1}) . But the coordinates of R are (9, 8, - 10) . ∴ \frac{5 λ + 3}{λ + 1} = 9, \frac{4 λ + 2}{λ + 1} = 8 and \frac{- 6 λ - 4}{λ + 1} = - 10 From each of these equation s, we get λ = - \frac{3}{2} ∴ R divides P Q in the ratio 3 : 2 externally .$

Page No 26.25:

Question 10:

A (3, 2, 0), B (5, 3, 2) and C (−9, 6, −3) are the vertices of a triangle ABC. If the bisector of ∠ABC meets BC at D, then coordinates of D are
(a) (19/8, 57/16, 17/16)
(b) (−19/8, 57/16, 17/16)
(c) (19/8, −57/16, 17/16)
(d) none of these

Answer:

$Since the bisector of ∠ ABC cannot meet BC, the solution of this question is not possible .$

Disclaimer:This question is wrong, so the solution has not been provide.

Page No 26.25:

Question 11:

If O is the origin, OP = 3 with direction ratios proportional to −1, 2, −2 then the coordinates of P are
(a) (−1, 2, −2)
(b) (1, 2, 2)
(c) (−1/9, 2/9, −2/9)
(d) (3, 6, −9)

Answer:

$(a) (- 1, 2, - 2) Let the coordinates of P be (x, y, z) . Then, Direction ratios of O P = Coordinates of P - Coordinates of O - 1, 2, 2 = (x - 0), (y - 0), (z - 0) Thus, coordinates of P are (- 1, 2, - 2) .$

Page No 26.25:

Question 12:

The angle between the two diagonals of a cube is
(a) 30°

(b) 45°

(c) $\cos^{- 1} (\frac{1}{\sqrt{3}})$

(d) $\cos^{- 1} (\frac{1}{3})$

Answer:

(d) $\cos^{- 1} (\frac{1}{3})$

$Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure . Clearly, O P, A R, B S, and C Q are the diagonals of the cube . Consider the diagonals O P and A R . Direction ratios of O P and A R are proportional to a - 0, a - 0, a - 0 and 0 - a, a - 0, a - 0, i . e . a, a, a and - a, a, a, respectively . Let θ be the angle between O P and A R . Then, \cos θ = \frac{a \times - a + a \times a + a \times a}{\sqrt{a^{2} + a^{2} + a^{2}} \sqrt{{(- a)}^{2} + a^{2} + a^{2}}} \Rightarrow \cos θ = \frac{- a^{2} + a^{2} + a^{2}}{\sqrt{3 a^{2}} \sqrt{3 a^{2}}} \Rightarrow \cos θ = \frac{1}{3} \Rightarrow θ = \cos^{- 1} (\frac{1}{3}) Similarly, the angles between other pairs of the diagonals are equal to \cos^{- 1} (\frac{1}{3}) as the angle between any two diagonals of a cube is \cos^{- 1} (\frac{1}{3}) .$

Page No 26.25:

Question 13:

If a line makes angles α, β, γ, δ with four diagonals of a cube, then cos² α + cos² β + cos² γ + cos² δ is equal to

(a) $\frac{1}{3}$

(b) $\frac{2}{3}$

(c) $\frac{4}{3}$

(d) $\frac{8}{3}$

Answer:

$(c) \frac{4}{3}$

$Let a be the length of an edge of the cube and let one corner be at the origin as shown in the figure . Clearly, O P, A R, B S and C Q are the diagonals of the cube . The direction ratios of O P, A R, B S and C Q are a - 0, a - 0, a - 0, i . e . a, a, a 0 - a, a - 0, a - 0, i . e . - a, a, a a - 0, 0 - a, a - 0, i . e . a, - a, a a - 0, a - 0, 0 - a, i . e . a, a, - a Let the direction ratios of a line be proportional to l, m and n . Suppose this line makes angles α, β, γ and δ with O P, A R, B S and C Q, respectively . Now, α is the angle between O P and the line whose direction ratios are proportional to l, m and n . \cos α = \frac{a . l + a . m + a . n}{\sqrt{a^{2} + a^{2} + a^{2}} \sqrt{l^{2} + m^{2} + n^{2}}} \Rightarrow \cos α = \frac{l + m + n}{\sqrt{3} \sqrt{l^{2} + m^{2} + n^{2}}} Since β is the angle between A R and the line with direction ratios proportional to l, m and n, we get \cos β = \frac{- a . l + a . m + a . n}{\sqrt{a^{2} + a^{2} + a^{2}} \sqrt{l^{2} + m^{2} + n^{2}}} \Rightarrow \cos β = \frac{- l + m + n}{\sqrt{3} \sqrt{l^{2} + m^{2} + n^{2}}} Similarly, \cos γ = \frac{a . l - a . m + a . n}{\sqrt{a^{2} + a^{2} + a^{2}} \sqrt{l^{2} + m^{2} + n^{2}}} \Rightarrow \cos γ = \frac{l - m + n}{\sqrt{3} \sqrt{l^{2} + m^{2} + n^{2}}} \cos δ = \frac{a . l + a . m - a . n}{\sqrt{a^{2} + a^{2} + a^{2}} \sqrt{l^{2} + m^{2} + n^{2}}} \Rightarrow \cos δ = \frac{l + m - n}{\sqrt{3} \sqrt{l^{2} + m^{2} + n^{2}}} \cos^{2} α + \cos^{2} β + \cos^{2} γ + \cos^{2} δ = \frac{{(l + m + n)}^{2}}{3 (l^{2} + m^{2} + n^{2})} + \frac{{(- l + m + n)}^{2}}{3 (l^{2} + m^{2} + n^{2})} + \frac{{(I - m + n)}^{2}}{3 (l^{2} + m^{2} + n^{2})} + \frac{{(l + m - n)}^{2}}{\sqrt{3} \sqrt{l^{2} + m^{2} + n^{2}}} = \frac{1}{3 (l^{2} + m^{2} + n^{2})} \{{(l + m + n)}^{2} + {(- l + m + n)}^{2} + {(I - m + n)}^{2} + {(l + m - n)}^{2}\} = \frac{1}{3 (l^{2} + m^{2} + n^{2})} 4 (l^{2} + m^{2} + n^{2}) = \frac{4}{3}$

Page No 26.25:

Question 14:

The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are
(a) (2, 0, 0)
(b) (0, 5, 0)
(c) (0, 0, 7)
(d) (0, 5, 7)

Answer:

For given point (2, 5, 7)
Co-ordinates of foot of perpendicular an x-axis means y = 0, z = 0
i.e co-ordinates are given by (2, 0, 0)

Hence, the correct answer is option A.

Page No 26.25:

Question 15:

P is a point on the line segment joining the points (3, 2, –1) and (6, 2, –2). If x-coordinate of P is 5, then its y-coordinates is
(a) 2
(b) 1
(c) –1
(d) –2

Answer:

Line joining (3, 2, –1) and (6, 2, –2) has same y-coordinate
∴ Any point on the line joining (3, 2, –1) and (6, 2, –2) will have same y-coordinate i.e 2

Hence, the correct answer is option A.

Page No 26.25:

Question 16:

The distance of the point $(α, β, γ)$ from y-axis is
$(a) β$
$(b) |β|$
$(c) |β| + |γ|$
$(d) \sqrt{α^{2} + γ^{2}}$

Answer:

For any point (α, β, γ)
Let (0, β, 0) be a point on y-axis
∴ distance between (α, β, γ) and (0, β, 0) is $\sqrt{{(α - 0)}^{2} + {(β - β)}^{2} + {(γ - 0)}^{2}}$

i.e $\sqrt{α^{2} + 0^{2} + γ^{2}}$
i.e $\sqrt{α^{2} + γ^{2}}$

Hence, the correct answer is option D.

Page No 26.25:

Question 17:

The direction cosines of a line are k, k, k, then
(a) k > 0
(b) 0 < k < 1
(c) k = 1
(d) $k = \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}$

Answer:

Given direction cosines of a line are k, k and k.
Say l = k, m = k, n = k
Since direction cosines are such that
l² + m² + n² = 1
∴ k² + k² + k² = 1
i.e 3k² = 1
i.e k^₂ = $\frac{1}{3}$
$i . e k = \pm \frac{1}{\sqrt{3}}$

Hence, the correct answer is option D.

Page No 26.26:

Question 1:

The distance of the point (a, b, c) from y-axis is ____________.

Answer:

For given point (a, b, c)
Point perpendicular on y-axis is (0, b, 0)
∴ Distance of (a, b, c) from y axis is given by
$\sqrt{{(a - 0)}^{2} + {(b - b)}^{2} + {(c - 0)}^{2}} = \sqrt{a^{2} + 0^{2} + c^{2}} = \sqrt{a^{2} + c^{2}}$
i.e distance of (a, b, c) from y-axis is $\sqrt{a^{2} + c^{2}}$

Page No 26.26:

Question 2:

The distance of the point (a, b, c) from z-axis is ____________.

Answer:

For given point (a, b, c)
Point on z-axis will be (0, 0, c)
∴ Distance between (a, b, c) and z-axis, is given by
$\sqrt{{(a - 0)}^{2} + {(b - 0)}^{2} + {(c - c)}^{2}} = \sqrt{a^{2} + b^{2}}$
Hence, distance of (a, b, c) from z-axis is $\sqrt{a^{2} + b^{2}}$ .

Page No 26.26:

Question 3:

If a line makes angles $\frac{π}{2}, \frac{3 π}{4} and \frac{π}{4}$ with x, y, z axes respectively, then its direction cosines are _____________.

Answer:

Since direction cosines are given by
l = cosα, m = cosβ and n = cosγ
where α is angle with x-axis, β is angle with y-axis and γ is angle with z-axis
Given, a line makes angles $\frac{π}{2}, \frac{3 π}{4} and \frac{π}{4}$ with x, y and z axes respectively.
$\begin{array}{rcl} ∴ l & = & \cos \frac{π}{2} = 0 \\ m & = & \cos \frac{3 π}{4} = - \frac{1}{\sqrt{2}} [∵ \cos \frac{3 π}{4} = \cos (π - \frac{π}{4})] \\ n & = & \cos \frac{π}{4} = \frac{1}{\sqrt{2}} \end{array}$

Page No 26.26:

Question 4:

If a line makes angles α, β, γ with positive directions of the coordinate axes, then the value of cos 2α + cos 2β + cos 2γ is __________.

Answer:

Suppose a line makes angles α, β and γ with positive directions of the coordinate axes, then l = cosα, m = cosβ and n = cosγ
Here cos2α + cos2β + cos2γ
= 2cos²α – 1 + 2cos²β – 1 + 2cos²γ – 1
= 2(cos²α + cos²β + cos²γ) – 3
Since cos²α + cos²β + cos²γ
= l² + m² + n²
= 1
∴ cos2α + cos2β + cos2γ
= 2(cos²α + cos²β + cos²γ) – 3
= 2(1) – 3
i.e cos2α + cos2β + cos2γ = –1

Page No 26.26:

Question 5:

If a line angles α, β, γ with positive directions of the coordinate axes, then the value of sin²α + sin²β + sin² γ is __________.

Answer:

Given, a line makes angles α, β and  γ with positive directions of x-axis
Then l = cosα, m = cosβ and n = cosγ
Here  sin²α + sin²β + sin²γ
= 1 – cos²α + 1 – cos²β + 1 – cos²γ
     = 3 – (cos²α + cos²β + cos²γ)
    = 3 – (l² + m² + n²)
    = 3 – 1
i.e sin²α + sin²β + sin²γ = 2

Page No 26.26:

Question 6:

If a line makes an angle with $\frac{π}{4}$ each of y and z-axis, then the angle which it makes with x-axis, is _____________.

Answer:

Since a line makes an angle $\frac{π}{4}$ with y and z-axis
Then l = cos $α$ , m = cos $\frac{π}{4}$ , n = cos $\frac{π}{4}$
Since l² + m² + n² = 1
$i . e \cos^{2} α + {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1 i . e \cos^{2} α + \frac{1}{2} + \frac{1}{2} = 1 i . e \cos^{2} α = 0 i . e \cos α = 0 i . e α = \frac{π}{2}$
∴ The angle it makes with x-axis is $\frac{π}{2}$ .

Page No 26.26:

Question 7:

The direction cosines of the vector $2 \hat{i} + 2 \hat{j} - \hat{k}$ are ______________.

Answer:

Direction cosines of the vector $2 \hat{i} + 2 \hat{j} - \hat{k}$ is given by $\frac{2}{\sqrt{2^{2} + 2^{2} + 1^{2}}}, \frac{2}{\sqrt{2^{2} + 2^{2} + 1^{2}}}, \frac{- 1}{\sqrt{2^{2} + 2^{2} + 1}}$
i.e $\frac{2}{3}, \frac{2}{3}, \frac{- 1}{3}$ are the direction cosines.

Page No 26.26:

Question 8:

A unit vector making angle $\frac{π}{4}$ with x-axis, $\frac{π}{3}$ with y-axis and an acute angle with z-axis is ______________.

Answer:

Given, a unit vector makes angle $\frac{π}{4}$ with x-axis, $\frac{π}{3}$ with y-axis
i.e l = cos $\frac{π}{4}$ , m = cos $\frac{π}{3}$ , n = cos $γ$ (say)
Since l² + m² + n² = 1

$i . e {(\cos \frac{π}{4})}^{2} + {(\cos \frac{π}{3})}^{2} + \cos^{2} γ = 1 i . e {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{2})}^{2} + \cos^{2} γ = 1$
$\begin{array}{rcl} i . e \cos^{2} γ & = & 1 - \frac{1}{2} - \frac{1}{4} \\ = & \frac{4 - 2 - 1}{4} = \frac{1}{4} \\ i . e \cos γ & = & \pm \frac{1}{2} \end{array}$
Since γ is given to be acute
$∴ \cos γ = \frac{1}{2} ∴ vector is \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \frac{1}{2} \hat{k}$

Page No 26.26:

Question 9:

If the projections of a line segment on the coordinates axes are 3, 4 and 5 then its length is equal to ____________.

Answer:

Since coordinates of a line segment from coordinate axes are 3, 4 and 5 then length is given by
$\sqrt{{(3 - 0)}^{2} + {(4 - 0)}^{2} + {(5 - 0)}^{2}}$
$= \sqrt{9 + 16 + 25} = \sqrt{25 + 25} = \sqrt{50} = \sqrt{5 \times 2 \times 5}$
$i . e length is 5 \sqrt{2}$

Page No 26.26:

Question 10:

A vector of magnitude 21 having direction ratios proportional to 2, –3, 6 is ____________.

Answer:

Let us suppose vector is given by $\vec{r} = a \hat{i} + b \hat{j} + c \hat{k}$
$Since \frac{a}{2} = \frac{b}{- 3} = \frac{c}{6} = h (say) Then \vec{r} = 2 h \hat{i} - 3 h \hat{j} + 6 h \hat{k} = 2 h \hat{i} - 3 h \hat{j} + 6 h \hat{k} i . e \vec{r} = h (2 \hat{i} - 3 \hat{j} + 6 \hat{k}) and |\vec{r}| = 21 ∴ 4 h^{2} + 9 h^{2} + 36 h^{2} = {(21)}^{2} i . e 49 h^{2} = 21 \times 21 i . e h^{2} = 3 \times 3 i . e h = 3 ∴ vector \vec{r} = 6 \hat{i} - 9 \hat{j} + 18 \hat{k}$

Page No 26.26:

Question 11:

The direction cosines of the line joining points (4, 3, –5) and (–2, 1, –8) are ______________.

Answer:

For points given by (4, 3, –5) and (–2, 1, –8) line joining these points is $6 \hat{j} + 2 \hat{j} + 3 \hat{k}$

$∴ Magnitude is give by \sqrt{36 + 4 + 9} = \sqrt{49} = 7$
Hence direction cosines are given by $\frac{6}{7}, \frac{2}{7}, \frac{3}{7}$

Page No 26.26:

Question 12:

If $\frac{1}{c}, \frac{1}{c}, \frac{1}{c}$ are direction cosines of a line, then the values of c are _______________.

Answer:

Let us suppose $l = \frac{1}{c}, m = \frac{1}{c} and n = \frac{1}{c}$ be the given direction cosines of a line
$i . e l^{2} + m^{2} + n^{2} = 1 i . e {(\frac{1}{c})}^{2} + {(\frac{1}{c})}^{2} + {(\frac{1}{c})}^{2} = 1 i . e \frac{1}{c^{2}} + \frac{1}{c^{2}} + \frac{1}{c^{2}} = 1 i . e 3 \times \frac{1}{c^{2}} = 1 i . e c^{2} = 3 i . e c = \pm \sqrt{3}$

Page No 26.26:

Question 13:

If O is the origin and OP = 6 with direction ratios proportional to –1, 2, –2 then the coordinates of P are _____________.

Answer:

Let us suppose co-ordinate of P are (a, b, c)
Now, since direction ratios are proportional to –1, 2, –2.
$∴ \frac{a}{- 1} = \frac{b}{2} = \frac{c}{- 2} = λ$
i.e a = –λ, b = 2λ, c = –2λ
Since OP = 6

$∴ \sqrt{{(- λ - 0)}^{2} + {(2 λ - 0)}^{2} + {(- 2 λ - 0)}^{2}} = 6$

i.e λ² + 4λ² + 4λ² = 36
i.e 9λ² = 36
i.e λ = ±2 ∴ Co-ordinates of P are ( –2, 4, –4)

Page No 26.26:

Question 14:

The angle between the vectors with direction ratios proportional to 1, 1, 2 and $\sqrt{3} - 1, - \sqrt{3} - 1, 4$ is ________________.

Answer:

Let $\vec{a}$ be a vector parallel to the vector having direction ratios 1, 1 and 2 and $\vec{b}$ be the vector parallel to the vector having direction ratios $+ \sqrt{3} - 1, - \sqrt{3} - 1, 4$

$∴ \vec{a} = \hat{i} + \hat{j} + 2 \hat{k} \vec{b} = (\sqrt{3} - 1) \hat{i} + (- \sqrt{3} - 1) \hat{j} + 4 \hat{k} then cosθ = \frac{\vec{a} . \vec{b}}{|\vec{a}| |\vec{b}|} = (\hat{i} + \hat{j} + 2 \hat{k}) . ((\sqrt{3} - 1) \hat{i} - (\sqrt{3} + 1) \hat{j} + 4 \hat{k}) where θ is angle between \vec{a} and \vec{b} i . e cosθ = \frac{(\sqrt{3} - 1) . 1 - (\sqrt{3} + 1) . 1 + 8}{\sqrt{1^{2} + 1^{2} + 2^{2}} \sqrt{{(\sqrt{3} - 1)}^{2} + {(\sqrt{3} + 1)}^{2} + 16}} = \frac{6}{\sqrt{6} \times \sqrt{3 + 1 - 2 \sqrt{3} + 3 + 1 + 2 \sqrt{3} + 16}} = \frac{6}{\sqrt{6} \times \sqrt{8}} i . e cosθ = \frac{\sqrt{6}}{2 \sqrt{2}} = \frac{\sqrt{3}}{2} i . e θ = \frac{π}{3}$

Page No 26.26:

Question 15:

If $\frac{1}{2}, \frac{1}{3}, n$ are the direction cosines of a line, then the values of n are ________________.

Answer:

Let us suppose, $l = \frac{1}{2}, m = \frac{1}{3}$ and n = n are the direction cosines of a line then l² + m² + n²= 1
$i . e {(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1 i . e \frac{1}{4} + \frac{1}{9} + n^{2} = 1 i . e n^{2} = 1 - \frac{1}{4} - \frac{1}{9} i . e n^{2} = \frac{36 - 9 - 4}{36} i . e n^{2} = \frac{23}{36} i . e n = \pm \frac{\sqrt{23}}{6} i . e n = \pm \frac{\sqrt{23}}{6}$

Page No 26.26:

Question 16:

If a line a makes angles $α, β, γ$ with x, y and z axes respectively such that $α + β = \frac{π}{2}$ , then $γ$ = _____________ .

Answer:

Given; a line makes angles α, β and γ with x, y and z axes respectively.
So, Let us suppose cosα, cosβ and cosγ represent direction cosines

$i . e \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 . . . (1) Since α + β = \frac{π}{2} i . e β = \frac{π}{2} - α i . e \cos β = \cos (\frac{π}{2} - α) i . e \cos β = \sin α i . e \cos^{2} β = \sin^{2} α ∴ equation (1) reduces to, \cos^{2} α + \sin^{2} α + \cos^{2} γ = 1 i . e 1 + \cos^{2} γ = 1 i . e \cos^{2} γ = 0 i . e γ = \frac{π}{2}$

Page No 26.26:

Question 17:

The total number of straight lines equally inclined with the coordinate axis is ____________.

Answer:

for any equally inclined line, α = β = γ (i.e direction cosines are same)
i.e cosα = cosβ = cosγ
i.e cos²α + cos²β + cos²γ = 1
i.e 3 cos²α = 1
$i . e \cos^{2} α = \frac{1}{3} i . e \cos α = \pm \frac{1}{\sqrt{3}} \cos β = \pm \frac{1}{\sqrt{3}} and \cos γ = \pm \frac{1}{3}$
∴ Total 8 straight line segments are possible which are equally inclined to coordinate axes.

Page No 26.26:

Question 18:

zx-plane divides the line segment joining (2, 3, 1) and (6, 7, 1) in the ratio ______________.

Answer:

Given points are (2, 3, 1) and (6, 7, 1)

Since zx-plane divides the line segment joining (2, 3, 1) and (6, 7, 1) in the ratio say K:1

Sine y co-ordinate is 0 for zx-plane

i . e \frac{3 p + 7}{p + 1} = 0 i . e p = - \frac{7}{3}

i.e ratio is 3:7 externally.

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Question 1:

Define direction cosines of a directed line.

Answer:

$The direction cosines of a directed line segment are the cosines of the direction angles of the line segment . Let two points A (x_{1}, y_{1}, z_{1}) and B (x_{2}, y_{2}, z_{2}) define the directed line segment AB . The direction cosines of AB are given by \cos α = \frac{x_{2} - x_{1}}{d} \cos β = \frac{y_{2} - y_{1}}{d} c o s γ = \frac{z_{2} - z_{1}}{d} Here, d is the distance between A and B .$

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Question 2:

What are the direction cosines of X-axis?

Answer:

$The x - axis makes angles 0 °, 90 ° and 90 ° with x, y and z axes, respectively . Therefore, the direction cosines of x - axis are \cos 0 °, \cos 90 °, \cos 90 °, i . e . 1, 0, 0 .$

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Question 3:

What are the direction cosines of Y-axis?

Answer:

$The y - axis makes angles 90 °, 0 ° and 90 ° with x, y and z axes, respectively . Therefore, the direction cosines of y - axis are \cos 90 °, \cos 0 °, \cos 90 °, i . e . 0, 1, 0 .$

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Question 4:

What are the direction cosines of Z-axis?

Answer:

$The z - axis makes angles 90 °, 90 ° and 0 ° with x, y and z axes, respectively . Therefore, the direction cosines of z - axis are \cos 90 °, \cos 90 °, \cos 0 °, i . e . 0, 0, 1 .$

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Question 5:

Write the distances of the point (7, −2, 3) from XY, YZ and XZ-planes.

Answer:

$The distance of a general point P (x, y, z) from X Y - plane is z . Thus, distance of (7, - 2, 3) from X Y - plane is 3 . Similarly, the distance of P (x, y, z) from Y Z - plane is x . Thus, distance of (7, - 2, 3) from Y Z - plane is 7 . The distance of P (x, y, z) from X Z - plane is y . Thus, distance of (7, - 2, 3) from X Z - plane is 2 .$

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Question 6:

Write the distance of the point (3, −5, 12) from X-axis?

Answer:

$The distance of a general point (x, y, z) from x - axis is \sqrt{y^{2} + z^{2}} . ∴ Distance of the point (3, - 5, 12) from x - axis = \sqrt{{(- 5)}^{2} + 12^{2}} = \sqrt{169} = 13 units$

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Question 7:

Write the ratio in which YZ-plane divides the segment joining P (−2, 5, 9) and Q (3, −2, 4).

Answer:

$Let the Y Z - plane divide the line segment joining points P (- 2, 5, 9) and Q (3, - 2, 4) in the ratio k : 1 . Using the section formula, the coordinates of the point of intersection are given by (\frac{k (3) - 2}{k + 1}, \frac{k (- 2) + 5}{k + 1}, \frac{k (4) + 9}{k + 1}) On the Y Z - plane, the X - coordinate of any point is zero . ∴ \frac{k (3) - 2}{k + 1} = 0 \Rightarrow 3 k - 2 = 0 \Rightarrow k = \frac{2}{3} Thus, the Y Z - plane divides the line segment formed by joining the given points in the ratio 2 : 3 internally .$

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Question 8:

A line makes an angle of 60° with each of X-axis and Y-axis. Find the acute angle made by the line with Z-axis.

Answer:

$It is given that a line makes an angle of 60 ° with both x - a x i s and y - a x i s . Suppose the line makes an angle of α with the z - axis . \Rightarrow l = \cos 60 ° = \frac{1}{2} m = \cos 60 ° = \frac{1}{2} n = \cos α We know l^{2} + m^{2} + n^{2} = 1 \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{2} + {(\cos α)}^{2} = 1 \Rightarrow \frac{1}{4} + \frac{1}{4} + \cos^{2} α = 1 \Rightarrow \cos α = \frac{1}{\sqrt{2}} \Rightarrow α = 45 ° Thus, the line makes an angle of 45 ° with the z - axis .$

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Question 9:

If a line makes angles α, β and γ with the coordinate axes, find the value of cos 2α + cos 2β + cos 2γ.

Answer:

$It is given that the the line makes angles α, β, γ with the coordinate axis . ∴ l = \cos α, m = \cos β and n = \cos γ \Rightarrow l^{2} + m^{2} + n^{2} = 1 \Rightarrow \cos^{2} α + \cos^{2} β + \cos^{2} γ = 1 . . . (1) Now, \cos 2 α + \cos 2 β + \cos 2 γ = (2 \cos^{2} α - 1) + (2 \cos^{2} β - 1) + (2 \cos^{2} γ - 1) = 2 (\cos^{2} α + \cos^{2} β + \cos^{2} γ) - 3 = 2 (1) - 3 [From (1)] = - 1$

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Question 10:

Write the ratio in which the line segment joining (a, b, c) and (−a, −c, −b) is divided by the xy-plane.

Answer:

$Suppose the line segment joining the points (a, b, c) and (- a, - c, - b) is divided by the X Y - plane at a point R in the ratio λ : 1 . Coordinates of R are (\frac{λ (- a) + 1 (a)}{λ + 1}, \frac{λ (- c) + 1 (b)}{λ + 1}, \frac{λ (- b) + 1 (c)}{λ + 1}) Since R lies on X Y - plane, Z - coordinate of R must be zero . \Rightarrow \frac{λ (- b) + 1 (c)}{λ + 1} = 0 = \frac{c}{b} Thus, the required ratio is \frac{c}{b : 1 or c : b} . Hence, the X Y - plane divides the line in the ratio c : b .$

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Question 11:

Write the inclination of a line with Z-axis, if its direction ratios are proportional to 0, 1, −1.

Answer:

$We know that if a line has direction ratio (a, b, c), then the cosine of its angle with the z - axis is given by \cos γ = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} Suppose the inclination of the line with direction ratio (0, 1, - 1) with z - axis is γ . Now, \cos λ = \frac{- 1}{\sqrt{0 + 1 + 1}} = - \frac{1}{\sqrt{2}} \Rightarrow λ = \frac{3 π}{4} Hence, the inclination of the line with z - axis is \frac{3 π}{4} .$

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Question 12:

Write the angle between the lines whose direction ratios are proportional to 1, −2, 1 and 4, 3, 2.

Answer:

$The direction ratios of the first line are 1, - 2, 1 and the direction ratios of the second line are 4, 3, 2 . Let θ be the angle between these two lines . Now, \cos θ = |\frac{1 (4) + (- 2) (3) + 1 (2)}{\sqrt{{(1)}^{2} + {(- 2)}^{2} + {(1)}^{2}} \sqrt{{(4)}^{2} + {(3)}^{2} + {(2)}^{2}}}| = |\frac{4 - 6 + 2}{\sqrt{1 + 4 + 1} \sqrt{16 + 9 + 4}}| = \frac{0}{\sqrt{6} \sqrt{29}} = 0 \Rightarrow θ = \frac{π}{2} Hence, the required angle is \frac{π}{2} .$

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Question 13:

Write the distance of the point P (x, y, z) from XOY plane.

Answer:

$The distance of the point P (x, y, z) from the X O Y plane is |z| .$

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Question 14:

Write the coordinates of the projection of point P (x, y, z) on XOZ-plane.

Answer:

The projection of the point P (x, y, z) on XOZ-plane is (x, 0, z) as Y-coordinates of any point on XOZ-plane are equal to zero.

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Question 15:

Write the coordinates of the projection of the point P (2, −3, 5) on Y-axis.

Answer:

The coordinates of the projection of the point P ( 2, -3, 5) on the y-axis are ( 0, $-$ 3, 0) as both X and Z coordinates of each point on the y-axis are equal to zero.

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Question 16:

Find the distance of the point (2, 3, 4) from the x-axis.

Answer:

$A general point (x, y, z) is at a distance of \sqrt{y^{2} + z^{2}} from the x - axis . ∴ Distace of the point (2, 3, 4) from x - axis = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5 units$

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Question 17:

If a line has direction ratios proportional to 2, −1, −2, then what are its direction consines?

Answer:

$If a line has direction ratios proportional to 2, - 1, and - 2, then its direction cosines are \frac{2}{\sqrt{{(2)}^{2} + (- 1) + {(- 2)}^{2}}}, \frac{- 1}{\sqrt{{(2)}^{2} + (- 1) + {(- 2)}^{2}}}, \frac{- 2}{\sqrt{{(2)}^{2} + (- 1) + {(- 2)}^{2}}} = \frac{2}{3}, - \frac{1}{3}, - \frac{2}{3} Thus, the direction cosines are \frac{2}{3}, - \frac{1}{3}, - \frac{2}{3} .$

Page No 26.27:

Question 18:

Write direction cosines of a line parallel to z-axis.

Answer:

$A line parallel to z - axis, makes an angle of 90 °, 90 ° and 0 ° with the x, y and z axes, respectively . Thus, the direction cosines are given by l = \cos 90 ° = 0 m = \cos 90 ° = 0 n = \cos 0 = 1 Therefore, direction cosines of a line parallel to the z - axis are 0, 0, 1 .$

Page No 26.27:

Question 19:

If a unit vector $\vec{a}$ makes an angle $\frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j}$ and an acute angle θ with $\hat{k}$ , then find the value of θ.

Answer:

$Since a unit vector makes an angle of \frac{π}{3} with \hat{i}, \frac{π}{4} with \hat{j} and an acute angle θ with \hat{k}, l = \cos \frac{π}{3} or \frac{1}{2}, m = \cos \frac{π}{4} or \frac{1}{\sqrt{2}} and n = \cos θ . We know l^{2} + m^{2} + n^{2} = 1 \Rightarrow {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + \cos^{2} θ = 1 \Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = \frac{1}{4} \Rightarrow \cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{3} Thus, the vecto r \vec{a} makes an angle of \frac{π}{3} with \hat{k} .$

Page No 26.27:

Question 20:

Answer each of the following questions in one word or one sentence or as per exact requirement of the question:

Write the distance of a point P(a, b, c) from x-axis.

Answer:

We know that a general point (x, y, z) has distance $\sqrt{y^{2} + z^{2}}$ from the x-axis.

∴ Distance of a point P(a, b, c) from x-axis = $\sqrt{b^{2} + c^{2}}$

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Question 21:

If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.

Answer:

Let the direction cosines of the line be l, m and n.
We know that l² + m² + n² = 1.
Let the line make angle θ with the positive direction of the z-axis.

$α = 90 °, β = 60 °, γ = θ So, \cos^{2} 90 ° + \cos^{2} 60 ° + \cos^{2} θ = 1 \Rightarrow 0 + {(\frac{1}{2})}^{2} + \cos^{2} θ = 1 \Rightarrow \cos^{2} θ = 1 - \frac{1}{4} \Rightarrow \cos^{2} θ = \frac{3}{4} \Rightarrow \cos θ = \pm \frac{\sqrt{3}}{2} \Rightarrow θ = 30 ° or 150 °$

cos2α+cos2β+cos2γ=1Here α=60 and β=45 and γ= θSo cos260+cos245+cos2θ=1cos2θ=1−14−12=14cosθ=±12So θ= 60 degree or 120.and here it is given that we have to find the angle made by negative z −axisSo cosθ=−12θ=120 degree

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Question 22:

If a line makes angles 90^o, 135^o, 45^o with the x, y and z axes respectively, find its direction cosines.

Answer:

ans

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