The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is:

(A) $2\sqrt 6$
(B) $\sqrt {26}$
(C) $2\sqrt 5 $
(D) $4\sqrt 2 $

Solution

We have, $3x-z+4=0$ or $z=3x+4$
& $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$

Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$

Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$

Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$

Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0.

Answer: (A)

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P?

(A) $(0,2,-4)$
(B) $(4,0,-2)$
(C) $(-2,0,-\frac {1}{2})$
(D) $(3,1,-\frac {1}{2})$

Solution

Bisectors are given by,

$\frac{{x – 2y – 2z + 1}}{3} =  \pm \frac{{2x – 3y – 6z + 1}}{7}$

$ \Rightarrow 7x – 14y – 14z + 7 =  \pm (6x – 9y – 18z + 3)$

Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$

Let $\theta $ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$

$\cos \theta  = \frac{{|1 + 10 – 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$

$\theta > 45^\circ $

So, $x-5y+4z+4=0$ is the obtuse angle bisector.

$\therefore $ Acute angle bisector $P \equiv 13x – 23y – 32z + 10 = 0$

The point $(-2,0,-\frac {1}{2})$ satisfies P.

Answer: (C)