Question

Let the lines\( \begin{array}{l} \frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}\ \text{and}\ \frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}\end{array}\)be coplanar and P be the plane containing these two lines. Then which of the following points does NOT lie on P?

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

Answer 1

The Correct Option is D

Given,\(\begin{array}{l} L_1:\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2},\end{array}\)through a point\(\begin{array}{l} \overrightarrow{a}_1\equiv \left(1, 2, 3\right)\end{array}\)parallel to \(\begin{array}{l} \overrightarrow{b}_1\equiv \left(\lambda, 1, 2\right)\end{array}\)

and \(\begin{array}{l} L_2:\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda} \end{array}\)through a point\(\begin{array}{l} \overrightarrow{a}_2=\left(-26,-18,-28\right)\end{array}\)parallel to\(\begin{array}{l} \overrightarrow{b}_2=\left(-2,3,1\right) \end{array}\)

If lines are coplanar then, \(\begin{array}{l} \left(\overrightarrow{a}_2-\overrightarrow{a}_1\right)\cdot\overrightarrow{b}_1\times\overrightarrow{b}_2=0\end{array}\)

\(\begin{array}{l} \Rightarrow\ \begin{vmatrix}27 & 20 & 31 \\\lambda & 1 & 2 \\-2 & 3 & \lambda \\\end{vmatrix}=0\Rightarrow \lambda =3\end{array}\)

Vector normal to the required plane \(\begin{array}{l} \overrightarrow{n}=\overrightarrow{b}_1\times \overrightarrow{b}_2\end{array}\)

\(\begin{array}{l} \Rightarrow\ \overrightarrow{n}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & 1 & 2 \\-2 & 3 & 3 \\\end{vmatrix}=-3\hat{i}-13\hat{j}+11\hat{k}\end{array}\)

Equation of plane\(\begin{array}{l} \equiv\left(\left(x-1\right),\left(y-2\right),\left(z-3\right)\right)\cdot\left(-3,-13,11\right)=0\end{array}\)

\(\begin{array}{l} \Rightarrow 3x+13y-11z+4=0 \end{array}\)

From given option (0, 4, 5) does not lie on the plane.