# Three Dimensional Geometry

Three Dimensional Geometry

## The equation of planes bisecting the angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is / are

Q: The equation of planes bisecting the angle between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is / are

(A) 5x – y – 4z – 45 = 0

(B) 5x – y – 4z – 3 = 0

(C) 23x – 13y + 32z + 45 = 0

(D) 23x – 13y + 32z + 5 = 0

Sol. Equation of planes bisecting the angle between given planes are

$\large \frac{2x-y+ 2z+3}{\sqrt{4+1+4}} = \pm \frac{3x-2y+6z+8}{\sqrt{9+4+36}}$

⇒ 5x – y – 4z – 3 = 0 and 23x – 13y + 32z + 45 = 0 are required planes.

Hence (B) and (C) are the correct answers.

## The equations of the lines of shortest distance between the lines …

Q: The equations of the lines of shortest distance between the lines $\large \frac{x}{2} = \frac{y}{-3} = \frac{z}{1}$ and $\large \frac{x-2}{3} = \frac{y-1}{-5} = \frac{z+2}{2}$ are

(A) 3(x – 21) = 3y + 92 = 3z – 32

(B) $\large \frac{x-62/3}{1/3} = \frac{y+31}{1/3} = \frac{z-31}{1/3}$

(C) $\large \frac{x-21}{1/3} = \frac{y + 92/3}{1/3} = \frac{z-32/3}{1/3}$

(D) $\large \frac{x-2}{1/3} = \frac{y+3}{1/3} = \frac{z-1}{1/3}$

Sol. Let P(2r1 , – 3r1, r1) and Q(3r2 + 2, – 5r2 + 1, 2r2 – 2) be the points on the given lines so that PQ is the line of shortest distance

d.r.s of PQ 2r1 – 3r2 – 2 , – 3r1 + 5r2 – 1, r1 – 2r2 + 2

Since it is perpendicular to given lines

2(2r1 – 3r2 – 2) – 3(3r1 + 5r2 – 1) + (r1 – 2r2 + 2) = 0

and (2r1 – 3r2 – 2) – 5(– 3r1 + 5r2 – 1) + 2(r1 – 2r2 + 2) = 0

r1 = 31/3, r2 = 19/3

P is (62/3 , -31 , 31/3) and Q is (21 , -92/3 , 32/3)

d.r.s of PQ is $(\frac{1}{3} , \frac{1}{3} , \frac{1}{3})$

Hence (A), (B) and (C) are the correct answers.

## The lines $\frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

Q: The lines  $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0}$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3}$

(A) are non coplanar

(B) are coplanar

(C) intersecting at (4, 0, – 1)

(D) intersecting at (1, 1, – 1)

Sol: Let $\large \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} = \lambda_1$ and $\large \frac{x-4}{2} = \frac{y+0}{0} = \frac{z+1}{3} = \lambda_2$

Then 1 + 3λ1 = 4 + 2λ2 …(1)

1 – λ1 = 0 …(2)

– 1 = 3λ2 – 1 …(3)

λ1 = 1 , λ2 = 0 satisfies (1)

lines are intersecting hence coplanar and point of intersection (4, 0, – 1).

Hence (B) and (C) are the correct answers.

## The coordinate of the points on the line $\frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance 3√2 from the point (1, 2, 3)

Q: The coordinate of the points on the line $\large \frac{x+2}{3} = \frac{y+1}{2} = \frac{z-3}{2}$ which are at a distance 3√2 from the point (1, 2, 3)

(A) (– 2, – 1, 3)

(B) (2, 2, 4)

(C) $( \frac{56}{17} , \frac{43}{17} , \frac{111}{17} )$

(D) $( \frac{47}{11} , \frac{42}{11} , \frac{56}{11} )$

Sol. Any point on the line (3λ – 2, 2λ – 1, 2λ + 3), then

(3λ – 3)2 + (2λ – 3)2 + (2λ)2 = 18

λ = 0, 13/17

points are (– 2, – 1, 3) and $( \frac{56}{17} , \frac{43}{17} , \frac{111}{17} )$

Hence (A) and (C) are the correct answers.

## The equation of the straight line through the origin parallel to the line (b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

Q: The equation of the straight line through the origin parallel to the line
(b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

(A) $\large \frac{x}{b^2-c^2} = \frac{y}{c^2-a^2} = \frac{z}{a^2-b^2}$

(B) $\large \frac{x}{a} = \frac{y}{b} = \frac{z}{c}$

(C) $\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab}$

(D) none of these

Sol. Equation of straight line through origin is

$\large \frac{x-0}{l} = \frac{y-0}{m} = \frac{z-0}{n}$

where l((b+c) + m(c+a) + n(a + b) = 0

and l(b – c) + m(c -a) + n(a – b) = 0

On solving,

$\large \frac{l}{2(a^2-bc)} = \frac{m}{2(b^2-ca)} = \frac{n}{2(c^2-ab)}$

Equation of the straight line is

$\large \frac{x}{a^2-bc} = \frac{y}{b^2-ca} = \frac{z}{c^2-ab}$

Hence (C) is the correct answer.