The distance between a point $P$ whose position vector is $5 \hat{i}+\hat{j}+3 \hat{k}$ and the line…
The distance between a point $P$ whose position vector is $5 \hat{i}+\hat{j}+3 \hat{k}$ and the line $\mathbf{r}=(3 \hat{i}+7 \hat{j}+\hat{k})+t(\hat{j}+\hat{k})$ is
$3$
$4$
$5$
$6$
Solution
$\mathbf{P}=5 \hat{i}+\hat{j}+3 \hat{k} \Rightarrow P \equiv(5,1,3)$
$\because \quad \mathbf{r}=(3 \hat{i}+7 \hat{j}+\hat{k})+(\hat{j}+\hat{k})$
$\Rightarrow$ Line passes through the point $Q(3,7,1)$.
$\Rightarrow \mathbf{P Q}=-2 \hat{i}+6 \hat{j}-2 \hat{k}$
Direction vector of line is $(\hat{j}+\hat{k})$.
Distance $=\frac{|\mathbf{P Q} \times(\hat{j}+\hat{k})|}{|\hat{j}+\hat{k}|}=\frac{1}{\sqrt{2}}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 6 & -2 \\ 0 & 1 & 1\end{array}\right|$
$\begin{aligned} & =\left|\frac{1}{\sqrt{2}}[\hat{i}(6+2)-\hat{j}(-2)+\hat{k}(-2)]\right| \\ & =\left|\frac{1}{\sqrt{2}}(8 \hat{i}+2 \hat{j}-2 \hat{k})\right| \\ & =\sqrt{2}|4 \hat{i}+\hat{j}-\hat{k}|=\sqrt{2} \sqrt{4^2+1^2+1^2} \\ d & =\sqrt{2} \sqrt{18}=6\end{aligned}$