Question

The distance of the point $(3,8,2)$ from the line $\frac{1}{2}(x-1)=\frac{1}{4}(y-3)=\frac{1}{3}(z-2)$ measured parallel to the plane $3x+2y-2z+15=0$ is

• A. $7$
• B. $9$
• C. $\sqrt{7}$
• D. $49$

Answer 1

The correct option is A $7$

Let any point on the line

$\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$

is $A(2k+1,4k+3,3k+2)$

$\therefore \overrightarrow{PA}=(2k-2)\hat{r}+(4k-5)\hat{j}+3k\hat{k}$

$\overrightarrow{PA}\perp \hat{n}$

$\therefore (2k-2).3+(4k-5).2+(-2)\left(3k\right)=0$

$\Rightarrow 6k-6+8k-10-6k=0$

$\Rightarrow 8k=16$

$\therefore k=2$

$\therefore A(5,11,8)$

$\therefore Pa=\sqrt{(5-3{)}^2+(11-8{)}^2+(8-2{)}^2}$

$=\sqrt{4+9+36}$

$=\sqrt{49}$

$\therefore PA=7$