3D Geometry 3 Question 59
59. Consider three planes $P _1: x-y+z=1$
| $P _2: x+y-z=-1$ | |
|---|---|
| and $\quad$ | $P _3: x-3 y+3 z=2$ |
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)
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Answer:
Correct Answer: 59. (d)
Solution:
- Given three planes are
$ \begin{aligned} 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 $\quad x-3 y+3 z=0-3 y+3(1+y)=3(\neq 2)$
So, Statement II is true.
Next, since we know that direction ratios of line of intersection of planes $a _1 x+b _1 y+c _1 z+d _1=0$
$ \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 above result,
Direction ratios of lines $L _1, L _2$ and $L _3$ are
$ 0,2,2 ; 0,-4,-4 ; 0,-2,-2 $
Since, all the three lines $L _1, L _2$ and $L _3$ are parallel pairwise.
Hence, Statement I is false.
