Question

The perpendicular distance of the point $(2,4,-1)$ from line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$ is

• A. $3$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (A)

$3$

Drs of $PQ=(k-7,4k-7,-9k+7)$

Drs of line $=(1,4,-9)$

$PQ\perp r$ line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$

$\therefore (k-7)1+4(4k-7)-9(-9k+7)=0$

$\Rightarrow k-7+16k-28+81k-63=0$

$\Rightarrow 98k-98=0$

$\Rightarrow k=1$

$\therefore$ root of tr point $Q(-4,1,-3)$

$PQ=\sqrt{(2+4{)}^2+(4-1{)}^2+(-1+3{)}^2}$

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

$=\sqrt{49}$

$\therefore PR=7$