Consider the planes 3x –6y + 2z + 5 = 0 and 4x –12y + 3z = 3. The plane 67x –162y + 47z + 44 = 0 bisects the angle between the given planes which

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

is obtuse

is acute

right angle

contains origin

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

3x –6y +2z +5 = 0 ...(i); –4x +12y–3z+3=0 ...(ii)
Bisectors are
$\frac{3 x - 6 y + 2 z + 5}{\sqrt{9 + 36 + 4}} = \pm \frac{- 4 x + 12 y - 3 z + 3}{\sqrt{16 + 144 + 9}}$

The plane which bisects the angle between the planes
that contains the origin
$\begin{aligned} 13 (3 x - 6 y + 2 z + 5) = 7 (- 4 x + 12 y - 3 z + 3) \\ \Rightarrow 3 \times (- 4) + (- 6) (12) + 2 \times (- 3) < 0 \dots \dots (iii) \end{aligned}$
Further, $3 \times (- 4) + (- 6) (12) + 2 \times (- 3) < 0$
Hence, the origin lies in the acute angle.