# Find the distance of a point(2,4,-1) from the line $\large \frac{x-5}{1}\;=\frac{y-3}{4}\;=\frac{z-6}{-9}$

Toolbox:
• To get distance between a point and line get foot of $\perp$ of the point on the line and get the distance.
• If two lines are $\perp$ then the dot product of their d.r. =0
Step 1
Let $P(2, 4,-1)$ be the given point and the Line $L \rightarrow \large\frac{x-5}{1}=\large\frac{y-3}{4}=\large\frac{z-6}{-9}= \lambda$
$\Rightarrow\:d.r.$ of the line $L=(1,4,-9)$
Any point $Q(x,y,z)$ on the line L is given by $Q (\lambda+5, 4\lambda+3, -9\lambda+6)$
Let this point $Q$ be the foot of $\perp$ of the point P on the line $L$.
Step 2
d.r of $\overline{ PQ}$ the $\perp$
$=(\lambda+5-2,4\lambda+3-4,-9\lambda+6+1)$
$= (\lambda+3, 4\lambda-1, -9\lambda+7)$
Step 3
PQ is $\perp$ to the line L
$\therefore\: \overline {PQ} . (1, 4, -9)=0$
$(\lambda+3, 4\lambda-1, -9\lambda+7).(1,4,-9)=0$
$\Rightarrow ( \lambda+3)1+(4\lambda-1)4-9(-9\lambda+7)=0$
$\Rightarrow\:\lambda+3+16\lambda-4+81\lambda-63=0$
$\Rightarrow\:98 \lambda=64$
$\Rightarrow\:\lambda=\large\frac{32}{49}$
Step 4
Substituting $\lambda$ in Q we get the foot of $\perp Q$ as