3D Geometry Ques 68
68. Consider three planes $P_{1}: x-y+z=1$……(i)
| | $P_{2}: x+y-z=-1$ |…..(ii) |—|—| | and $\quad$ | $P_{3}: x-3 y+3 z=2$ |……(iii)
Let $L_{1}, L_{2}, L_{3}$ be the lines of intersection of the planes $P_{2}$ and $P_{3}, P_{3}$ and $P_{1}, P_{1}$ and $P_{2}$, respectively.
Statement I Atleast two of the lines $L_{1}, L_{2}$ and $L_{3}$ are non-parallel.
Statement II The three planes do not have a common point.
(2008, 3M)
Show Answer
Answer:
Correct Answer: 68.(d)
Solution:
Formula:
- Given three planes are parallel
$ \begin{alignedat} P_{1}: x-y+z & =1 \quad …..(1) \\ P_{2}: x+y-z & =-1 \quad …..(2) \\ \text { and } \quad P_{3}: x-3 y+3 z & =2 \quad …..(3) \end{aligned} $
On solving Eqs. (i) and (ii), we get
$ x=0, z=1+y $
which does not satisfy Eq. (iii).
As $\quad x-3 y+3 z=0-3 y+3(1+y)=3(\neq 0)$
So, Statement II is true.
Next, since we know that direction ratios of line of intersection of planes $a_{1} x+b_{1} y+c_{1} z+d_{1}=0$ and $a_{2} x+b_{2} y+c_{2} z+d_{2}=0$
$ \text { and } \quad a_{2} x+b_{2} y+c_{2} z+d_{2}=0 \text { is } $
$b_{1} c_{2}-b_{2} c_{1}, c_{1} a_{2}-a_{1} c_{2}, a_{1} b_{2}-a_{2} b_{1}$
Using the above result,
Direction ratios of lines $L_{1}, L_{2}$ and $L_{3}$ are given by
$ 0,2,2 ; 0,-4,-4 ; 0,-2,-2 $
Since all the three lines $L_{1}, L_{2}$ and $L_{3}$ are pairwise parallel.
Hence, Statement I is false.
BETA
