Explanation:
A Given,
Line L- \(\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda\)
So, direction cosines \(=(1,2,3)\)
Coordinates of any point on the line is -
\(\mathrm{x}=\lambda, \mathrm{y}=2 \lambda+1, \mathrm{z}=3 \lambda+2 .\)
Let, \(\mathrm{Q}=(\lambda, 2 \lambda+1,3 \lambda+2)\)
And, \(\mathrm{P}=(1,6,3)\)
Direction cosine of \(\mathrm{PQ}=(\lambda-1,2 \lambda+1-6,3 \lambda+2-3)\) \(\therefore \quad(\lambda-1,2 \lambda-5,3 \lambda-1)\)
Sum of the dot product of the direction cosine must be zero,
\((\lambda-1)(1)+(2 \lambda-5)(2)+(3 \lambda-1)(3)=0\)
\(\Rightarrow \lambda-1+4 \lambda-10+9 \lambda-3=0\)
\(\Rightarrow \lambda=1\)
Direction cosine of \(\mathrm{PQ}=(\lambda-1,2 \lambda-5,3 \lambda-1)\) So, coordinates of the foot of the perpendicular
\(=(0,-3,2)\)
Equation of line \(\mathrm{PQ}-\)
\(\frac{\mathrm{x}-1}{0}=\frac{\mathrm{y}-3}{-3}=\frac{\mathrm{z}-5}{2}\)
On substituting the value of \(\lambda\) in \(\mathrm{Q}\), we get -
\(\mathrm{Q}=(1,3,5)\)