Question

The equation of the line through the point (0,1,2) and perpendicular to the line
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

• A. $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{3}$
• B. $\frac{x}{3}=\frac{y-1}{-4}=\frac{z-2}{3}$
• C. $\frac{x}{3}=\frac{y-1}{4}=\frac{z-2}{-3}$
• D. $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

Answer 1

Answer: (D)

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=r$
$\Rightarrow P(x,y,z)=(2r+1,3r-1,-2r+1)$
Since, $\overrightarrow{QP}\perp (2\hat{i}+3\hat{j}-2\hat{k})$
$\Rightarrow 4r+2+9r-6+4r+2=0$
$\Rightarrow r=\frac{2}{17}$

$\Rightarrow P\left(\frac{21}{17},\frac{-11}{17},\frac{13}{17}\right)$
$\Rightarrow \overrightarrow{PQ}=\frac{21\hat{i}-28\hat{j}-21\hat{k}}{17}$
So, $\overrightarrow{QP}:\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$