The vector equation of line passing through (2, -1, 3) and perpendicular to the lines $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z+2}{2}$ and $\frac{x+3}{-4}=\frac{y-5}{-3}=\frac{z+1}{2}$ is
(Here $λ$ is a parameter)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\vec r= (2\hat i-\hat j+3\hat k) +λ (8\hat i-14\hat j-5\hat k)$
Given lines in symmetric form:
$\frac{x-2}{3}=\frac{y-1}{1}=\frac{z+2}{2}\;\Rightarrow\;$ direction $\vec{v}_{1}=\langle 3,1,2\rangle$
$\frac{x+3}{-4}=\frac{y-5}{-3}=\frac{z+1}{2}\;\Rightarrow\;$ direction $\vec{v}_{2}=\langle -4,-3,2\rangle$
Required line ⟂ to both ⇒ direction $\vec{v}=\vec{v}_{1}\times\vec{v}_{2}$
$\vec{v}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&1&2\\-4&-3&2\end{vmatrix} =8\hat{i}-14\hat{j}-5\hat{k}$
Passing through $(2,-1,3)$:
$\vec{r}=(2\hat{i}-\hat{j}+3\hat{k})+\lambda\,(8\hat{i}-14\hat{j}-5\hat{k})$