The length of perpendicular drawn from the point $2 \hat{i}-\hat{j}+5 \hat{k}$ to the line…
The length of perpendicular drawn from the point $2 \hat{i}-\hat{j}+5 \hat{k}$ to the line $\overline{\mathrm{r}}=(11 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-8 \hat{\mathrm{k}})+\lambda(10 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-11 \hat{\mathrm{k}})$ is
$\sqrt{14}$ units
14 units
237 units
$\sqrt{237}$ units
Solution
Let $\mathrm{P}=(2,-1,5)$ and co-ordinates of any point on the given line be
$\begin{aligned}
& \mathrm{Q} \equiv(10 \lambda+11,-4 \lambda-2,-11 \lambda-13) \\
& \text { d.r. of } \mathrm{PQ} \text { are }(10 \lambda+9,-4 \lambda-1,-11 \lambda-13) \\
& \text { d.r. of given line are }(10,-4,-11) \\
& \therefore(10 \lambda+9)(10)+(-4 \lambda-1)(-4)+(-11 \lambda-13)(-11)=0 \\
& \therefore 100 \lambda+90+16 \lambda+4+121 \lambda+143=0 \Rightarrow \lambda=-1 \\
& \therefore \mathrm{Q} \equiv(1,2,3) \text { and } \mathrm{d}(\mathrm{PQ})=\sqrt{1^2+3^2+2^2}=\sqrt{14} \text { units }
\end{aligned}$