Question

Let the lines $\frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda }$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P?

• A. $(0,-2,-2)$
• B. $(-5,0,-1)$
• C. $(3,-1,0)$
• D. $(0,4,5)$

Answer 1

Answer: (D)

Given, $L_1:\frac{x-1}{\lambda }=\frac{y-2}{1}=\frac{z-3}{2}$
and $L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda }$ are coplanar
$\Rightarrow \left|\begin{array}{ccc}27& 20& 31\\ \lambda & 1& 2\\ -2& 3& \lambda \end{array}\right|=0$
$\Rightarrow \lambda =3$
Now, normal of plane $P$, which contains $L_1$ and $L_2$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& 2\\ -2& 3& 3\end{array}\right|$
$=-3\hat{i}-13\hat{j}+11\hat{k}$
$\Rightarrow$Equation of required plane $P$
$3x+13y-11z+4=0$
$(0,4,5)$ does not lie on plane $P$.