Find the coordinates of the foot of the perpendicular drawn from the point
Answer
Given: Equation of line is
To find: coordinates of foot of the perpendicular from
Formula Used:
1. Equation of a line is
Cartesian form:
where
2. Distance between two points
\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}{\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}
Since this is perpendicular to the line,
3(3λ+5)+2(2λ+5)–2(–2λ+4)=0⇒9λ+15+4λ+10+4λ–8=0⇒17λ=–17⇒λ=–1\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}