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

$\left( {3,\,\,1,\,\, - \frac{1}{2}} \right)$

JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

Let the acute angle bisector of the two planes

x - 2y - 2z + 1 = 0\,\,\,\mathrm{ and }\,\,2x - 3y - 6z + 1 = 0

be the plane

P

Then, which of the following points lies on

P

View answer and explanation

Correct option: B

\left( { - 2,\,\,0,\,\, - \frac{1}{2}} \right)

Option B

\left( { - 2,\,0,\, - \frac{1}{2}} \right)

is correct because it satisfies the equation of the acute angle bisector plane.

Practice All Chapter MCQs →

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

p x-q y+

z=5

p, q \in \mathbb{R}

x-c y-b z=0, c x-y+a z=0

b x+a y-z=0

a^2+b^2+c^2+2 a b c

2 x-3 y+9 z=0

P

(0,1,-1)

4 x-2 y+7 z+9=0

P

(3,2,0)

\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}

a x+

=0

\alpha

z=0

\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0

(3,2,1)

x, y

z

A, B

C

y z

A

z x

B

x y

C