Question

Find the equation of the plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+1}{2}$ and parallel to the line $\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+4}{5}$.

Answer 1

line 1:$\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+1}{2};\overrightarrow{b_1}=3\hat{i}+4\hat{j}+2\hat{k}$

line 2:$\frac{x-2}{2}=\frac{y-1}{-3}=\frac{z+4}{5};\overrightarrow{b_2}=2\hat{i}+5\hat{k}-3\hat{j}$

Plane is parallel to line 2 & contains line $\perp$.

$\therefore$ its normal is $\perp$ to both lines .

$\overrightarrow{n}$ be normal to plane.

$\therefore \overrightarrow{n}=\overrightarrow{b_1}\times \overrightarrow{b_2}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{\hat{k}}\\ 3& 4& 2\\ 2& -3& 5\end{array}\right|=26\hat{i}-11\hat{j}-17\hat{k}$

$(1,-6,-1)$ is a point on line 1 & will also be on the plane.

$\therefore$ Equation of plane having directions of normals $(26,-11,-17)$ & passing through $(1,-6,-1)$ will be,

$26(x-1)-11(y+6)-17(z+1)=026x-11y-17z-109=0$