Direction ratios of the line perpendicular to the lines $\frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$ and $\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$ are :
Answer & explanation
Correct answer: option 4
lines $l_1, : \frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$
$l_2: \frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$
so vector $\vec{v}_1||l_1$
$\Rightarrow \vec{v}_1=2 \hat{i}-3 \hat{j}+\hat{k}$
vector $\vec{v}_2 || l_2$
$\Rightarrow \vec{v}_2=\hat{i}+2 \hat{j}-2 \hat{k}$
so $\vec p=\vec{v}_1×\vec{v}_2$ (perpendicular to both $l_1,l_2$)
$\vec{p}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & -3 & 1\\1 & 2 & -2\end{vmatrix}$
$\vec{p} = 4\hat{i} + 5\hat{j}+7\hat{k}$ (Upon solving)
so DR's(p) → $<4, 5, 7>$