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BITSAT 2024Paper 2
Exam
BITSAT
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2024 · Paper 2 · Mathematics · Three Dimensional Geometry

Let the acute angle bisector of the two planes x-2 y-2 z+1=0 and 2...

Let the acute angle bisector of the two planes x−2y−2z+1=0x-2 y-2 z+1=0 and 2x−3y−6z+1=02 x-3 y-6 z+1=0 be the plane PP. Then which of the following points lies on P?

Options

  1. A(3,1,−12)\left(3,1,-\frac{1}{2}\right)
  2. B(−2,0,−12)\left(-2,0,-\frac{1}{2}\right)
  3. C(0,2,−4)(0,2,-4)
  4. D(4,0,−2)(4,0,-2)

Correct answer

B. (−2,0,−12)\left(-2,0,-\frac{1}{2}\right)

Step-by-step solution

Given that P1:x−2y−2z+1=0P2:2x−3y−6z+1=0P_1: x-2 y-2 z+1=0P_2: 2 x-3 y-6 z+1=0 Equation of plane bisectors ∣x−2y−2z+11+4+4∣=∣2x−3y−6z+122+32+62∣x−2y−2z+13=±2x−3y−6z+17\begin{aligned} & \left|\frac{x-2 y-2 z+1}{\sqrt{1+4+4}}\right|=\left|\frac{2 x-3 y-6 z+1}{\sqrt{2^2+3^2+6^2}}\right| \\ & \frac{x-2 y-2 z+1}{3}= \pm \frac{2 x-3 y-6 z+1}{7} \end{aligned} Since a1a2+b1b2+c1c2=20>0∴\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{ b}_2+\mathrm{c}_1 \mathrm{c}_2=20>0\therefore Negative sign will be taken for acute bisector. ⇒7x−14y−14z+7=−[6x−9y−18z+3]⇒13x−23y−32z+10=0(−2,0,−12) satisfy it \begin{aligned} & \Rightarrow 7 x-14 y-14 z+7=-[6 x-9 y-18 z+3] \\ & \Rightarrow 13 x-23 y-32 z+10=0 \\ & \left(-2,0,-\frac{1}{2}\right) \text{ satisfy it } \end{aligned}
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