Question

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

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

Answer 1

The correct option is A $\sqrt{13}$

Using fact the $\perp$ distance from $(\alpha ,\beta ,\gamma )$ in the line

$\frac{x-x_1}{1}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is given by

$\sqrt{(\alpha -x_1{)}^2+(\beta -y_1{)}^2+(\gamma -z_1{)}^2}$

$\sqrt{-\left[l\right(\alpha -x_1)+m(\beta -y_1)+n(\gamma -z_1){]}^2}$

Now $(\alpha ,\beta ,\gamma )=(1,6,3),(x_1,y_1.z_1)=0,1,2,l,m,n=\frac{1}{\sqrt{14}},\frac{1}{\sqrt{14}},\frac{1}{\sqrt{14}}$

$\therefore$ Required distance

$=\sqrt{(1-0{)}^2+(6-1{)}^2+(3-2{)}^2-{\left[\frac{1}{\sqrt{14}}\right(1+2\left(5\right)+3\left(1\right)]}^2}=\sqrt{27-14}=\sqrt{13}$