The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$, from the straight line…
The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$, from the straight line passing through the point $(2,-3,-4)$ and parallel to the vector $6 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ is units.
$\sqrt{\frac{340}{61}}$
$\frac{341}{61}$
$\frac{\sqrt {341}}{61}$
$\sqrt{\frac{341}{61}}$
Solution
Given equation of the line is
$\begin{aligned}
& \overline{\mathrm{r}}=\overline{\mathrm{a}}+\lambda \overline{\mathrm{b}} \\
& \therefore \quad \bar{r}=2 \hat{i}-3 \hat{j}-4 \hat{k}+\lambda(6 \hat{i}+3 \hat{j}-4 \hat{k}) \\
&
\end{aligned}$
To find: Its distance from point $\bar{\alpha}=\hat{i}-2 \hat{j}-6 \hat{k}$.
$\therefore \quad \text { Required distance }=\sqrt{|\bar{\alpha}-\bar{a}|^2-\left[\frac{(\bar{\alpha}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2}$
Here, $|\bar{\alpha}-\bar{a}|^2=6$ and $\left[\frac{(\bar{\alpha}-\bar{a}) \cdot \bar{b}}{|\bar{b}|}\right]^2=\frac{25}{61}$.
$\therefore \quad \text { Requiređ distance }=\sqrt{\frac{341}{61}}$