Let P , Q , R and S be the points on the plane with position vectors - 2 i ^ - j ^ , 4 i ^ , 3 i ^ + 3 j ^ and - 3 j ^ + 2 j ^ , respectively. The quadrilateral P Q R S must be a

parallelogram, which is neither a rhombus nor a rectangle.

If the distance between the plane A x - 2 y + z = d and the plane containing the lines x - 1 2 = y - 2 3 = z - 3 4 and x - 2 3 = y - 3 4 = z - 4 5 is 6 , then | d | is

We have, L 1 : x - 1 2 = y - 2 3 = z - 3 4 L 2 : x - 2 3 = y - 3 4 = z - 4 5 Equation of the plane containing these lines is a x - 2 + b y - 3 + c z - 4 = 0 … 1 where, 3 a + 4 b + 5 c = 0 … 2 2 a + 3 b + 4 c = 0 … 3 (using direction ratios of the lines) And, plane is containing L 1 , therefore a 1 - 2 + b 2 - 3 + c 3 - 4 = 0 i.e., a + b + c = 0 … 4 From equations 2 & 3 , a 1 = b - 2 = c 1 , which satisfy equation 4 Plane through lines is x - 2 y + z = 0 . Given plane is A x - 2 y + z = d Now, distance between these planes is 6 Since, direction ratios of these two planes are proportional i.e., A 1 = - 2 - 2 = 1 1 . ∴ Planes must be parallel. So, A = 1 Now, as distance between planes is 6 ∴ d - 0 1 2 + - 2 2 + 1 2 = 6 (Because distance between parallel planes a x + b y + c z + d 1 = 0 and a x + b y + c z + d 2 = 0 is given by D = d 1 - d 2 a 2 + b 2 + c 2 ) ⇒ d 6 = 6 ⇒ d = 6

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If the $π$ -plane $7 x + (α + 4) y + 4 z - r = 0$ passing through the points of intersection of the planes $2 x + 3 y - z + 1 = 0$ and $x + y - 2 z + 3 = 0$ are perpendicular to the plane $3 x - y - 2 z = 4$ and $(\frac{12}{β}, \frac{- 78}{β}, \frac{57}{β})$ is an image of point $(1, 1, 1)$ in $π$ -plane, then

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Important Questions on Three Dimensional Geometry

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 1

A line with positive direction cosines passes through the point

P (2, - 1,2)

and makes equal angles with the coordinate axes. The line meets the plane

2 x + y + z = 9

at point

Q

. The length of the line segment

P Q

equals

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 2

Let

P (3, 2, 6)

be a point in space and

Q

be a point on the line

\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (- 3 \hat{i} + \hat{j} + 5 \hat{k})

. Then, the value of

μ

for which the vector

\vec{P Q}

is parallel to the plane

x - 4 y + 3 z = 1

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 3

Let

P, Q, R

and

S

be the points on the plane with position vectors

- 2 \hat{i} - \hat{j}, 4 \hat{i}, 3 \hat{i} + 3 \hat{j}

and

- 3 \hat{j} + 2 \hat{j}

, respectively. The quadrilateral

P Q R S

must be a

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 4

Equation of the plane containing the straight line

\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2}

and

\frac{x}{4} = \frac{y}{2} = \frac{z}{3}

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 5

If the distance between the plane

A x - 2 y + z = d

and the plane containing the lines

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

and

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

\sqrt{6},

then

| d |

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 6

If the distance of the point

P (1, - 2, 1)

from the plane

x + 2 y - 2 z = α

, where

α > 0,

5

then the foot of the perpendicular from

P

to the plane is

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 8

The point

P

is the intersection of the straight line joining the points

Q (2, 3, 5)

and

R (1, - 1, 4)

with the plane

5 x - 4 y - z = 1

. If

S

is the foot of the perpendicular drawn from the point

T (2, 1, 4)

Q R

, then the length of the line segment

P S

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Alpha Question Bank for Engineering: Mathematics>Chapter 21 - Three Dimensional Geometry>Exercise-3>Q 9

The equation of a plane passing through the line of intersection of the planes

x + 2 y + 3 z = 2

and

x - y + z = 3

and at a distance

\frac{2}{\sqrt{3}}

from the point

(3, 1, - 1)

If the π-plane 7x+α+4y+4z-r=0 passing through the points of intersection of the planes 2x+3y-z+1=0 and x+y-2z+3=0 are perpendicular to the plane 3x-y-2z=4 and 12β,-78β,57β is an image of point 1,1,1 in π-plane, then

Important Questions on Three Dimensional Geometry

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