Question

Show that the lines $\frac{x-4}{5}=\frac{y-3}{-2}=\frac{z-2}{-6}$ and $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z-1}{-7}$ are coplanar. Find their point of intersection and the equation of the plane in which they lie.

Answer 1

$l_1:\frac{x-4}{5}=\frac{y-3}{-2}=\frac{z-2}{-6}$

$l_2:\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z-1}{7}$

let us assume $(x,y,z)=(5r+4,-2r+3,-6r+2)$ be point of intersection of $l_1$ &$l_2$

it must satisfy $l_2$

$\frac{5r+4-3}{4}=\frac{-2r+3-2}{-3}=\frac{-6r+2-1}{7}$

$\therefore r=-1$

therefore point of intersection is $(-1,5,8)$. Since $l_1$ and $l_2$ intersects and hence they are coplanar. The normal of the plane in which $l_1$ and $l_2$ lies will be perpendiclar t the direction ratios of $l_1$ and $l_2$

$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 5& -2& -6\\ 4& -3& -7\end{array}\right|=(-4)\hat{i}+11\hat{j}+(-7)\hat{k}$

$\therefore$ equation of plane is $-4(x+1)+11(y-5)-7(z-8)=0$