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Three Dimensional Geometry

Multiple Choice Questions

101.

If the planes $\vec{r} . (2 \hat{i} - λ \hat{j} + 3 \hat{k}) = 0 and \vec{r} . (λ \hat{i} + 5 \hat{j} - \hat{k}) = 5$ are perpendicular to each other, then the value of $λ^{2} + λ$ is

102.

The cartesian form of the plane $\vec{r} = (s - 2 t) \hat{i} + (3 - t) \hat{j} + (2 s + t) \hat{k}$ is

2x - 5y - z - 15 = 0
2x - 5y + z - 15 = 0
2x - 5y - z + 15 = 0
2x + 5y - z + 15 = 0

103.

Let P(- 7, 1, - 5) be a point on a plane and let O be the origin. If OP is normal to the plane, then the equation of the plane is

7x - y + 5z + 75 = 0
7x + y - 5z + 73 = 0
7x + y + 5z + 73 = 0
7x - y - 5z + 75 = 0

104.

The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x² + y² + z² + 4x - 2y - 6z = 155 is

26
$11 \frac{4}{13}$
13
39

105.

The point in the xy-plane which is equidistant from the point (2, 0, 3), (0, 3, 2) and (0, 0, 1) is

(1, 2, 3)
(- 3, 2, 0)
(3, - 2, 0)
(3, 2, 0)

106.
The angle between the line $\frac{3 x - 1}{3} = \frac{y + 3}{- 1} = \frac{5 - 2 z}{4}$ and the plane 3x - 3y - 6z = 10 is equal to

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

$\frac{π}{2}$

Given lines and planes are

$\frac{3 x - 1}{3} = \frac{y + 3}{- 1} = \frac{5 - 2 z}{4} or \frac{x - \frac{1}{3}}{1} = \frac{y + 3}{- 1} = \frac{(z - \frac{5}{2})}{- 2} and 3 x - 3 y - 6 z = 0 \Rightarrow x - y - 2 z = 0 Here, a_{1} = 1, b_{1} = - 1, c_{1} = - 2 and a_{2} = 1, b_{2} = - 1, c_{2} = - 2 ∴ \sin (θ) = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}} = \frac{1 \times 1 + (- 1) \times (- 1) + (- 2) \times (- 2)}{\sqrt{1 + 1 + 4} \sqrt{1 + 1 + 4}} = \frac{6}{\sqrt{6} \sqrt{6}} = 1 \Rightarrow θ = \frac{π}{2}$

107.

The angle between the straight lines $\vec{r} = (2 - 3 t) \hat{i} + (1 + 2 t) \hat{j} + (2 + 6 t) \hat{k}$ and $\vec{r} = (1 + 4 s) \hat{i} + (2 - s) \hat{j} + (8 s - 1) \hat{k}$ is

$\cos^{- 1} (\frac{\sqrt{41}}{34})$
$\cos^{- 1} (\frac{21}{34})$
$\cos^{- 1} (\frac{43}{63})$
$\cos^{- 1} (\frac{34}{63})$

108.

If Q is the image of the point P(2, 3, 4) under the reflection in the plane x - 2y + 5z = 6, then the equation of the line PQ is

$\frac{x - 2}{- 1} = \frac{y - 3}{2} = \frac{z - 4}{5}$
$\frac{x - 2}{1} = \frac{y - 3}{- 2} = \frac{z - 4}{5}$
$\frac{x - 2}{- 1} = \frac{y - 3}{- 2} = \frac{z - 4}{5}$
$\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{5}$

109.

The distance of the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane x - y + z = 5 from the point (- 1, - 5, - 10)is