P is any point on the plane l x + m y + n z = p . A point Q taken on the O P (where O is the origin) such that O P · O Q = p 2 . Show that the locus of Q is p l x + m y + n z = x 2 + y 2 + z 2

Let Q α , β , γ . . Equation of line O P is x - 0 α = y - 0 β = z - 0 γ = λ Therefore, any point on line O P is α λ , β λ , γ λ Let co-ordinates of P ≡ α λ , β λ , γ λ P lies on plane l x + m y + n z = p , therefore λ l α + m β + n γ = p . . . . 1 Also, O P · O Q = p 2 ⇒ λ 2 α 2 + β 2 + γ 2 α 2 + β 2 + γ 2 = p 2 ⇒ λ α 2 + β 2 + γ 2 = p 2 . . . 2 By 1 & 2 , p l α + m β + n γ = α 2 + β 2 + γ 2

Find the foot and hence the length of the perpendicular from the point ( 5 , 7 , 3 ) to the line x - 15 3 = y - 29 8 = 5 - z 5 . Also find the equation of the plane in which the perpendicular and the given straight line lie.

Given: x - 15 3 = y - 29 8 = z - 5 - 5 = λ Any point on the given line is Q 3 λ + 15 , 8 λ + 29 , - 5 λ + 5 Direction ratio of P Q are proportional to 3 λ + 10 , 8 λ + 22 , - 5 λ + 2 Now, P Q → · 3 i ^ + 8 j ^ - 5 k ^ = 0 ⇒ 3 ( 3 λ + 10 ) + 8 ( 8 λ + 22 ) - 5 ( 2 - 5 λ ) = 0 ⇒ ( 9 + 64 + 25 ) λ + ( 30 + 176 - 10 ) = 0 ⇒ λ = - 2 ∴ Q ≡ 9 , 13 , 15 So, P Q = 16 + 36 + 144 = 196 ⇒ P Q = 14 units And equation of required plane is x - 15 y - 29 z - 5 3 8 - 5 4 6 12 = 0 ⇒ 9 x - 4 y - z = 14

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Beta Question Bank for Engineering: Mathematics>Chapter 22 - Three Dimensional Geometry>EXERCISE-3>Q 28

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Given three point on $x - y$ plane as $O (0, 0), A (1, 0) & B (- 1, 0)$ . Point $P$ is moving on the plane satisfying the condition $(\vec{P A} \cdot \vec{P B}) + 3 (\vec{O A} \cdot \vec{O B}) = 0$ . If the maximum & minimum values of $|\vec{P A}| |\vec{P B}|$ are $M & m$ respectively then the value of $M^{2} + m^{2}$ is

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Important Questions on Three Dimensional Geometry

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Beta Question Bank for Engineering: Mathematics>Chapter 22 - Three Dimensional Geometry>EXERCISE-4>Q 1

Find the angle between the two straight lines whose direction cosines

l, m, n

are given by

2 l + 2 m - n = 0

and

m n + n l + l m = 0

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Beta Question Bank for Engineering: Mathematics>Chapter 22 - Three Dimensional Geometry>EXERCISE-4>Q 2

A variable plane is at a constant distance

p

form the origin and meet the coordinate axes in points

A, B

and

C

respectively. through these points, planes are drawn parallel to the coordinates planes. Find the locus of their point of intersection

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Beta Question Bank for Engineering: Mathematics>Chapter 22 - Three Dimensional Geometry>EXERCISE-4>Q 3

P

is any point on the plane

l x + m y + n z = p

.

A

point

Q

taken on the

O P

(where

O

is the origin) such that

O P \cdot O Q = p^{2}

. Show that the locus of

Q

is

p (l x + m y + n z) = x^{2} + y^{2} + z^{2}

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Beta Question Bank for Engineering: Mathematics>Chapter 22 - Three Dimensional Geometry>EXERCISE-4>Q 4

The plane $I x + m y = 0$ is rotated about its line of intersection with the plane $z = 0$ through angle measure $α$ . Prove that the equation of the plane in new position is $l x + m y \pm z \sqrt{l^{2} + m^{2}} \tan α = 0$ .