Question

Let the foot of the perpendicular from the point $(1,2,4)$ on the line $\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x+4y+12z+23=0$

• A. 5
• B. $\frac{50}{13}$
• C. $4$
• D. $\frac{63}{13}$

Answer 1

Answer: (A)

$\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}=\lambda$
$(x,y,z)=(4\lambda -2,2\lambda +1,3\lambda -1)$
$\overrightarrow{AP}=(4\lambda -3)\hat{i}+(2\lambda -1)\hat{j}+(3\lambda -5)\hat{k}$
$\vec{b}=4\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{AP}\cdot \vec{b}=0$
$4(4\lambda -3)+2(2\lambda -1)+3(3\lambda -5)=0$
$29\lambda =12+2+15=29$
$\lambda =1$
$P=(2,3,2)$
$3x+4y+12z+23=0$
$d=\left|\frac{6+12+24+23}{\sqrt{9+16+144}}\right|$
$d=\left|\frac{65}{13}\right|=5$