Question

The distance from the point $(1,6,3)$ to the line $\vec{r}=\left(\hat{j}+2\hat{k}\right)+\lambda \left(\hat{i}+2\hat{j}+3\hat{k}\right)$ is

• A. $\sqrt{13}$
• B. $13$
• C. $2\sqrt{13}$
• D. None of these

Answer 1

Answer: (A)

Distance $PM$ = ?
$P(\alpha ,\beta ,\gamma )=(1,6,3)$
$M(x,y,z)=(\lambda ,2\lambda +1,3\lambda +2)$
dr’s of PM are $(\lambda -1,2\lambda -5,3\lambda -1)$

Dr’s of line are $(1,2,3)$
Now, $PM\perp AB$
$\therefore \lambda =1$
$\therefore M(1,3,5)$
$\therefore P(1,6,3),M(1,3,5)$
$\therefore PM=\sqrt{{\left(6-3\right)}^2+{\left(3-5\right)}^2}=\sqrt{13}$
Short Cut Method :
Using fact : The $\perp$ distance from $(\alpha ,\beta ,\gamma )$ on the line
$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is given by
$\sqrt{{\left(\alpha -x_1\right)}^2+{\left(\beta -y_1\right)}^2+{\left(\gamma -z_1\right)}^2-{\left\{l\left(\alpha -x_1\right)+m\left(\beta -y_1\right)+n\left(\gamma -z_1\right)\right\}}^2}$
Now, $(\alpha ,\beta ,\gamma )=(1,6,3)$,
$(x_1,y_1,z_1)=(0,1,2)$,
$(l,m,n)=\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)$
$\therefore$ Required distance
$=\sqrt{{\left(1-0\right)}^2+{\left(6-1\right)}^2+{\left(3-2\right)}^2-{\left[\frac{1}{\sqrt{14}}\left(1+2\left(5\right)+3\left(1\right)\right)\right]}^2}$
$=\sqrt{27-14}=\sqrt{13}$