Find the coordinates of the foot of the perpendicular drawn from the point to the line . Also, find the length of the perpendicular from the given point to the line.
Answer
Given: Equation of line is .
To find: coordinates of foot of the perpendicular from to the line. And find the length of the perpendicular.
Formula Used:
1. Equation of a line is
Cartesian form:
where is a point on the line and is the direction ratios of the line.
2. Distance between two points and is

$\sqrt{{\left({\mathrm{x}}_1–{\mathrm{x}}_2\right)}^2+{\left({\mathrm{y}}_1–{\mathrm{y}}_2\right)}^2+{\left({\mathrm{z}}_1–{\mathrm{z}}_2\right)}^2}$
\sqrt{\left(\mathrm{x}_{1}-\mathrm{x}_{2}\right)^{2}+\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)^{2}+\left(\mathrm{z}_{1}-\mathrm{z}_{2}\right)^{2}}

Explanation:
Let

$\frac{x–6}{\mathrm{\backslash not}3}=\frac{y–7}{3}=\frac{z–7}{–3}=\lambda$
\frac{x-6}{\not 3}=\frac{y-7}{3}=\frac{z-7}{-3}=\lambda

So the foot of the perpendicular is
Direction ratio of the line is
Direction ratio of the perpendicular is

$\begin{array}{l}\Rightarrow (3\lambda +6–1):(2\lambda +7–2):(–2\lambda +7–3)\\ \Rightarrow (3\lambda +5):(2\lambda +5):(–2\lambda +4)\end{array}$
\begin{array}{l}
\Rightarrow(3 \lambda+6-1):(2 \lambda+7-2):(-2 \lambda+7-3) \\
\Rightarrow(3 \lambda+5):(2 \lambda+5):(-2 \lambda+4)
\end{array}

Since this is perpendicular to the line,

$\begin{array}{l}3(3\lambda +5)+2(2\lambda +5)–2(–2\lambda +4)=0\\ \Rightarrow 9\lambda +15+4\lambda +10+4\lambda –8=0\\ \Rightarrow 17\lambda =–17\\ \Rightarrow \lambda =–1\end{array}$
\begin{array}{l}
3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4)=0 \\
\Rightarrow 9 \lambda+15+4 \lambda+10+4 \lambda-8=0 \\
\Rightarrow 17 \lambda=-17 \\
\Rightarrow \lambda=-1
\end{array}

So the foot of the perpendicular is

$\begin{array}{l}\text{ Distance }=\sqrt{(3–1{)}^2+(5–2{)}^2+(9–3{)}^2}\\ =\sqrt{4+9+36}\\ =7\text{ units }\end{array}$
\begin{array}{l}
\text { Distance }=\sqrt{(3-1)^{2}+(5-2)^{2}+(9-3)^{2}} \\
=\sqrt{4+9+36} \\
=7 \text { units }
\end{array}