Question

Find the length of the perpendicular from the point $(3,2,1)$ to the line $\frac{x-7}{-2}=\frac{y-7}{2}=\frac{z-6}{3}$

Answer 1

Let $M$ be the foot of the perpendicular drawn from point $A(3,2,1)$ to the given line.

Let $\frac{x-7}{-2}=\frac{y-7}{2}=\frac{z-6}{3}=k$

$\therefore$ $x=7-2k,y=7+2k,z=6+3k$ are the co-ordinates of any point on the given line

$\Rightarrow$ $M=(7,-2k,7+2k,6+3k)$

The d.r.s of $AM$ are $7-2k-3,7+2k-2,6+3k-1$

i.e. $4-2k,5+2k,5+3k$

The d.r.s of the given lines are $-2,2,3$

Since, $AM$ perpendicular to the given line

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

$\Rightarrow$ $-8+4k+10+4k+15+9k=0$

$\Rightarrow$ $17k+17=0$

$\Rightarrow$ $k=-1$

The co-ordinates of the foot of the perpendicular

i.e. $M=(9,5,3)$

The length of the perpendicular distance

$\Rightarrow$ $AM=\sqrt{(9-3{)}^2+(5-2{)}^2+(3-1{)}^2}$

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

$=\sqrt{49}$

$=7units$