Question

The perpendicular distance from a point $P$ with position vector $5\overrightarrow{i}+\overrightarrow{j}+3\overrightarrow{k}$ to the line $\overrightarrow{r}=(3\overrightarrow{i}+7\overrightarrow{j}+\overrightarrow{k})+t(\overrightarrow{j}+\overrightarrow{k})$ is

• A. $3$
• B. $6$
• C. $9$
• D. $12$

Answer 1

Answer: (B)

$6$

Point Pis $(5,1,3)$ and line is given

$\overrightarrow{r}=\left(3\hat{i}+7\hat{j}+\hat{k}\right)+t\left(\hat{j}+\hat{k}\right)$

let the foot of the perpendicular from point +p to the given line be $\theta \left(x{,}^{1}y{,}^1,z^1\right)$ then direction ratios of $pq$ are

$\left(x^1-5,y^1-1,z^1-3\right)$

since $pq$is $\perp$ to the given line so

$b\times \left(x^1-5\right)+1\times \left(y^1-11+1\times \left(z^1-31\right)\right)=0$

$y^1+1+z^1-3=0$

$y^1+z^1=4=0$

artesian equation of the given line is

$\frac{x-3}{0}=\frac{y-7}{1}=\frac{z-1}{1}$

let $\frac{x-3}{0}=\frac{y-7}{1}=\frac{z-1}{1}=r$

$x=3,y=r+7,z=r+1$

so general point on this line is (3,r+7,r+1)

since $\left(x^1,y^1,z^1\right)$

lies on this line so $x^1=3y^1=r+7z^1=r+1$

putting values of $y^1\&$ in

$y^1+z^1=4$

$r+7+r+1=4$

$2r=-4$

$=r=-2$

thus $x^1=3,y^1=-2+7=5$

$z^1=-2+1=-1$

so point Q is $(3,5,-1)$

therefore perpendicular distance $\sqrt{{\left(5-3\right)}^2+{\left(1-5\right)}^2+{\left(3+1\right)}^2}$

$\sqrt{4+16+16}$

$\sqrt{36}$

$=6units$