Question 7:

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

Answer:

$We know that the distance of the point (x_{1}, y_{1}, z_{1}) from the plane a x + b y + c z + d = 0 is given by \frac{|a x_{1} + b y_{1} + c z_{1} + d|}{\sqrt{a^{2} + b^{2} + c^{2}}} Equation of t h e xy - plane is z = 0, which means 0 x + 0 y + z = 0 So, the required distance = \frac{|0 (2) + 0 (3) + (5)|}{\sqrt{0^{2} + 0^{2} + 1^{2}}} = \frac{5}{1} = 5 units$

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Question 8:

Find the distance of the point (3, 3, 3) from the plane $\vec{r} \cdot (5 \hat{i} + 2 \hat{j} - 7 k) + 9 = 0$

Answer:

Since for equation of plane ax + by + cz = d, d represents distance from origin.

$Hence, for \frac{2}{7} x + \frac{3}{7} y - \frac{6}{7} z = 1$

Hence, the correct answer is option A.

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Question 1:

If the plane $x + 2 y - 2 z = d, d > 0$ is at a 5 unit distance from the point $(1, - 2, 1)$ ,then d = ____________.

Answer:

For given plane, x + 2y – 2z = d
Distance of this plane from (1, –2, 1) is 5 is given
$i . e . 5 = \frac{|(1) + 2 (- 2) - 2 (1) - d|}{\sqrt{1^{2} + 2^{2} + 2^{2}}} i . e . 5 = \frac{|1 - 4 - 2 - d|}{\sqrt{9}} i . e . 15 = |- 5 - d| i . e . 15 = 5 + d i . e . d = 10$

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Question 2:

The planes $3 x - 6 y - 2 z = 7 and 2 x + y - λ z = 5$ are perpendicular. Then the value of $λ$ is __________.

Answer:

Given, planes, $3 x - 6 y - 2 z = 7 and 2 x + y - λ z = 5$ are perpendicular

Since two planes a_{1} x + b_{1} y + c_{1} z + d_{1} = 0 and a_{2} x + b_{2} y + c_{2} z + d_{2} = 0 are perpendicular If a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0 i . e . 3 (2) + (- 6) (1) + (- 2) (- λ) = 0 i . e . 6 - 6 + 2 λ = 0 i . e . 2 λ = 0 i . e . λ = 0

Since distance between (1, 1, 1) and (0, 0, 0) $= \sqrt{{(1 - 0)}^{2} + {(1 - 0)}^{2} + {(1 - 0)}^{2}} = \sqrt{3} . . . . (1)$

Distance of (1, 1, 1) from (x + y + z + k = 0)

d = \frac{|1 + 1 + 1 + k|}{\sqrt{1 + 1 + 1}} = \frac{|3 + k|}{\sqrt{3}} . . . . (2) Since (1) = \frac{1}{2} (2) is given \sqrt{3} = \frac{1}{2} \frac{|3 + k|}{\sqrt{3}} i . e . 2 \times 3 = |3 + k| i . e . \pm 6 = 3 + k i . e . k = 3 or k = - 9

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Question 7:

If the plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$ meets the coordinates axes at A, B and C, then the coordinates of the centroid of ∆ABC are ______________.

Answer:

Given plane, $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 3$

Meets x-axis when y = 0 = z , $\frac{x}{a} = 3$

i.e. x = 3a

Plane meets y-axis when x = 0 = z,

i . e . \frac{y}{b} = 3 i . e . y = 3 b

and this plane meets z-axis when x = 0 = y

$i . e . \frac{z}{c} = 3 i . e . z = 3 c$

∴ A is given by (3a, 0, 0)

B is given by (0, 3b, 0)

C is given by (0, 0, 3c)

$∴ Centroid is (\frac{3 a + 0 + 0}{3}, \frac{0 + 3 b + 0}{3}, \frac{0 + 0 + 3 c}{3})$

i.e. Centroid of ΔABC = (a,b,c)

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Question 8:

If O is the origin, then the equation of the plane passing through P(a, b, c) and perpendicular to OP is _____________.

Answer:

Since plane is perpendicular to OP and passes through (a, b, c)

Equation of plane is given by a(x − a) + b(y − b) + c(z − c) = 0

i.e. ax − a² + by − b² + cz − c² = 0

i.e. ax + by + cz = a² + b²+ c²

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Question 9:

A variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axis is $\frac{1}{2}$ . The plane passes through the point ____________.

Answer:

For a variable plane, given sum of reciprocal of its intercept is $\frac{1}{2}$ let a,b,c be the intercepts with co-ordinates axes according to given condition,
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{2} i . e . \frac{2}{a} + \frac{2}{b} + \frac{2}{c} = 1 i . e . 2 (\frac{1}{a}) + 2 (\frac{1}{b}) + 2 (\frac{1}{c}) = 1$
i.e plane passes through fixed point (2, 2, 2,)

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Question 10:

The value of $α$ for which the plane x + $α y$ + z = 5 cuts equal intercepts on the axes, is ____________.

Answer:

Equation of a plane which equal intercept on the axes is $\frac{x}{a} + \frac{4}{a} + \frac{z}{a} = 1$
Since x + αy + z = 5 cuts equal intercept

$i . e . \frac{x}{5} + \frac{α}{5} y + \frac{z}{5} = 1 i . e . \frac{1}{5} = \frac{α}{5} i . e . α = 1$

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Question 11:

If the line $\frac{2 x - 1}{2} = \frac{2 - y}{2} = \frac{z + 1}{k}$ is parallel to the plane $2 x - y + z = 3,$ then k = _______________.

Answer:

$Given line \frac{2 x - 1}{2} = \frac{2 - y}{2} = \frac{z + 1}{k} is parallel to plane 2 x - y + z = 3$
Line is of the pr $\frac{x - \frac{1}{2}}{1} = \frac{y - 2}{- 2} = \frac{z - (- 1)}{k}$

is parallel if $\vec{n}$ for plane is perpendicular to line for plane 2x – y + z = 3

Normal is \vec{n} 2 \hat{i} - \hat{j} + \hat{k} and \vec{b} = \hat{i} - 2 \hat{j} + k \hat{k} Since \vec{n} . \vec{b} = 0 i . e . 2 (1) - 1 (- 2) + 1 (k) = 0 i . e . 2 + 2 + k = 0 i . e . k = - 4

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Question 12:

If the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + 7 = 0, then α$ = _________________.

Answer:

$Given line \frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$
Lies in the plane x + 3y – αz + 7 = 0
$Whose normal \vec{n} = \hat{i} + 3 \hat{j} - α \hat{k} then \vec{b} = (3 \hat{i} - 5 \hat{j} + 2 \hat{k}) ⊥ \vec{n} = (\hat{i} + 3 \hat{j} - α \hat{k}) i . e . 3 (1) - 5 (3) + 2 (- α) = 0 i . e . 3 - 15 - 2 α = 0 i . e . - 12 - 2 α = 0 i . e . 2 α = - 12 i . e . α = - 6$

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Question 13:

If line $\frac{x + 1}{λ} = \frac{y - 1}{1} = \frac{z + 2}{- 4}$ is perpendicular to the plane $2 x + 2 y - 8 z + 5 = 0$ . Then the value of λ is _____________.

Answer:

$Given line, \frac{x + 1}{λ} = \frac{y - 1}{1} = \frac{z + 2}{- 4} is perpendicular to the plane 2 x + 2 y - 8 z + 5 = 0$

$Normal \vec{n} for the plane is \vec{n} = 2 \hat{i} + 2 \hat{j} - 8 \hat{k} and direction vector for line \vec{b} = λ \hat{i} + \hat{j} - 4 \hat{k}$

Since line is perpendicular to plane

i . e . \frac{2}{λ} = \frac{2}{1} = \frac{- 8}{- 4} i . e . \vec{n} and \vec{b} have proportional vectors i . e . \frac{2}{λ} = \frac{2}{1} i . e . λ = 1

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Question 14:

A plane passes through the points (2, 0, 0), (0, 3, 0) and (0, 0, 4). The equation of the plane is _______________.

Answer:

Suppose equation of plane is a(x − x₁) + b(y − y₁) + c(z − z₁) = 0
Where plane passes through (x₁, y₁, z₁)
Given plane passes through (2, 0, 0), (0, 3, 0) and (0, 0, 4)
i.e. a(x − 2) + b(y − 0) + c(z − 0) = 0 ...(1)

Since equation (1) passes through (0, 3, 0) and (0, 0, 4)
We get, a(0 − 2) + b(3 − 0) + c(0 − 0) = 0
i.e. −2a + 3b = 0 and a(0 − 2) + b(0 − 0) + c(4 − 0) = 0
i.e. −2a + 4c = 0
i.e. a = 2c and 3b = 2a = 4c
∴ equation of plane is 2c(x − 2) + $\frac{4}{3}$ c (y − 0) + c(z − 0) = 0
2(x − 2) + $\frac{4}{3}$ y + z = 0
i.e. 2x + $\frac{4}{3}$ y + z − 4 × 3 = 0
i.e. 6x + 4y + 3z − 12 = 0 is the equation of the plane

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Question 15:

The cartesian equation of the plane $\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 2$ is _____________.

Answer:

Answer:

$Let α, β and γ be the angles made by \vec{n} with x, y and z -axes, respectively. It is given that α = β = γ \Rightarrow \cos α = \cos β = \cos γ \Rightarrow l = m = n, where l, m, n are direction cosines of \vec{n} . But l^{2} + m^{2} + n^{2} = 1 \Rightarrow l^{2} + l^{2} + l^{2} = 1 \Rightarrow 3 l^{2} = 1 \Rightarrow l^{2} = \frac{1}{3} \Rightarrow l = \frac{1}{\sqrt{3}} So, l = m = n = \frac{1}{\sqrt{3}} It is given that the length of the perpendicular of the plane from the origin, p = 5 \sqrt{3} The normal form of the plane is l x + m y + n z = p \Rightarrow \frac{1}{\sqrt{3}} x + \frac{1}{\sqrt{3}} y + \frac{1}{\sqrt{3}} z = 5 \sqrt{3} \Rightarrow x + y + z = 5 \sqrt{3} (\sqrt{3}) \Rightarrow x + y + z = 15$

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Question 24:

Find the acute angle between the planes $\vec{r} \cdot (\hat{i} - 2 \hat{j} - 2 \hat{k}) = 1 and \vec{r} \cdot (3 \hat{i} - 6 \hat{j} + 2 \hat{k})$ = 0.

Answer:

For given plane
$\vec{r} . (\hat{i} - 2 \hat{j} - 2 \hat{k}) = 1 and \vec{r} . (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = 0 Using formula for plane \vec{r} . {\vec{n}}_{1} = d_{1} and \vec{r} . {\vec{n}}_{2} = d_{2} \cos θ = \frac{{\vec{n}}_{1} . {\vec{n}}_{2}}{|{\vec{n}}_{1}| |{\vec{n}}_{2}|} where θ is angle between two given planes i . e . \cos θ = \frac{(\hat{i} - 2 \hat{j} - 2 \hat{k}) . (3 \hat{i} - 6 \hat{j} + 2 \hat{k})}{|\hat{i} - 2 \hat{j} - 2 \hat{k}| |3 \hat{i} - 6 \hat{j} + 2 \hat{k}|} i . e . \cos θ = \frac{3 + 12 - 4}{\sqrt{1 + 4 + 4} \sqrt{9 + 36 + 4}} i . e . \cos θ = \frac{11}{\sqrt{9} \sqrt{49}} = \frac{11}{3 \times 7} i . e . \cos θ = \frac{11}{21} i . e . θ = \cos^{- 1} (\frac{11}{21})$

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