Question

Assertion :

Consider three planes:

$P_1:x-y+z=1P_2:x+y-z=1P_3:x-3y+3z=2$

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.

At least two of the lines $L_1,L_2$ and $L_3$ are non-parallel. Reason: The three planes do not have a common point.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

$P_1:x-y+z=1$ ...$\left(1\right)$

$P_2:x+y-z=-1$ ...$\left(2\right)$

$P_3:x-3y+3z=2$ ...$\left(3\right)$

Line $L_1$ is intersection of $P_2,P_3$

$\therefore L_1$ is $\parallel$ to vector

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& -3& 3\end{array}\right|=-4\hat{j}-4\hat{k}$

$L_2$ is intersection of $P_3$ and $P_1$

$\therefore L_2$ is $\parallel$ to vector $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 1& -3& 3\end{array}\right|=-2\hat{j}-2\hat{k}$

and line $L_3$ is intersection of $P_1$ and$P_2$

$\therefore L_3$ is $\parallel$ to vector $\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ 1& 1& -1\end{array}\right|=2\hat{j}+2\hat{k}$

clearly lines $L_1,L_2$ and $L_3$ are $\parallel$ to each other

Also family of planes passing through the inter section of $P_1$ and $P_2$ is $P_1+\lambda P_2=0.$

If $P_3$ is represented by $P_1+\lambda P_2=0$ for some value of $\lambda ,$ then the three planes pass through the same point $P_1+\lambda P_2=0$

$x(1+\lambda )+y(\lambda -1)+z(1-\lambda )+\lambda -1=0$

This will be identical to $P_3$ if

$\frac{1+\lambda }{1}=\frac{\lambda -1}{-3}=\frac{1-\lambda }{3}=\frac{1-\lambda }{2}$ ...$\left(4\right)$

taking $\frac{1+\lambda }{1}=\frac{1-\lambda }{2}\Rightarrow \lambda =-\frac{1}{3}$

taking $\frac{1+\lambda }{1}=\frac{1-\lambda }{3}\Rightarrow \lambda =-\frac{1}{2}$

$\therefore$ there is no value of $\lambda$ which satisfies equation $\left(4\right)$.

The three planes do not have a common point

$\Rightarrow$ reason is true therefore, $D$ is correct