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JEE Advanced 2010Paper 1
Exam
JEE Advanced
Year
2010
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Advanced 2010 · Paper 1 · Mathematics · Three Dimensional Geometry

If the distance between the plane A x-2 y+z=d and the plane...

If the distance between the plane Ax−2y+z=dA x-2 y+z=d and the plane containing the lines x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad and x−23=y−34=z−45\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5} is 6\sqrt{6}, then ∣d∣|d| is

Correct answer

6

Step-by-step solution

Equation of plane containing the given lines is \left|x−1y−2z−3234345\begin{array}{ccc}x-1 & y-2 & z-3 \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=0\begin{array}{rr} \Rightarrow& (x-1)(-1)-(y-2)(-2) \\ & +(z-3)(-1)=0 \\ \Rightarrow & -x+1+2 y-4-z+3=0 \\ \Rightarrow & -x+2 y-z=0 \quad \ldots(i) \end{array}Given plane is x-2 y+z=d Eqs. (i)$ and (ii) are parallel. Clearly,\quad A=1$ Now, distance between plane\begin{aligned}& =\left|\frac{d}{\sqrt{1+4+1}}\right|=\sqrt{6}\\ & \Rightarrow \quad|d|=6 \\ & \end{aligned}$$
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