Question

The distance of the point $P(3,8,2)$ from the line $\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}$ measured parallel to the plane $3x+2y-2z+17=0$ is

• A. $2$
• B. $3$
• C. $5$
• D. $7$

Answer 1

The correct option is D $7$

The equation of plane is $3x+2y-2z+17=0$

$\Rightarrow$ $3x+2y-2z+c=0$ ---- ( 1 )

Substituting Point $P(3,8,2)$ in above equation, we get,

$\Rightarrow$ $3\left(3\right)+2\left(8\right)-2\left(2\right)+c=0$

$\Rightarrow$ $9+16-4+c=0$

$\Rightarrow$ $c=-21$

Substituting value of $c$ in equation ( 1 ) we get,

$\Rightarrow$ $3x+2y-2z-21=0$ ----- ( 2 )

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

So we get,

$\Rightarrow$ $x=2\lambda +1$

$\Rightarrow$ $y=4\lambda +3$

$\Rightarrow$ $z=3\lambda +2$

Substituting above value in equation ( 2 ) we get,

$\Rightarrow$ $3(2\lambda +1)+2(4\lambda +3)-2(3\lambda +2)-21=0$

$\Rightarrow$ $6\lambda +3+8\lambda +6-6\lambda -4-21=0$

$\Rightarrow$ $8\lambda =16$

$\therefore$ $\lambda =2$

Now,

$x=2\lambda +1=2\left(2\right)+1=5$

$y=4\lambda +3=4\left(2\right)+3=11$

$z=3\lambda +2=3\left(2\right)+2=8$

Now, we have to find the distance between points $(3,8,2)$ and $(5,11,8)=\sqrt{(5-3{)}^2+(11-8{)}^2+(8-2{)}^2}$

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

$=\sqrt{49}$

$=7$