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

The correct option is A

$\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$

Explanation of the correct option:

Given: $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$

Let $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}=\lambda$

Any point on the line $\left(2\lambda +1,3\lambda -1,-2\lambda +1\right)$

Direction ratio of given line $\overrightarrow{d_1}=\left(2,3,-2\right)$

Direction ratio of the line to be found $\overrightarrow{d_2}=\left(2\lambda +1-0,3\lambda -1-1,-2\lambda +1-2\right)=(2\lambda +1,3\lambda -2,-2\lambda -1)$

Since, $\overrightarrow{d_1.}\overrightarrow{d_2}=0$

$\Rightarrow$$4\lambda +2+9\lambda -6+4\lambda +2=0$

$\Rightarrow$ $17\lambda -2=0$

$\Rightarrow$ $\lambda =\frac{2}{17}$

Therefore,

$$\begin{gathered} \overrightarrow{d_2}=\left(2\left(\frac{2}{17}\right)+1,3\left(\frac{2}{17}\right)-2,-2\left(\frac{2}{17}\right)-1\right) \\ \Rightarrow \overrightarrow{d_2}=\left(\frac{21}{17},\frac{-28}{17},\frac{-21}{17}\right) \\ \Rightarrow \overrightarrow{d_2}=\left(-3,4,3\right) \end{gathered}$$

Therefore the equation of the line is $\frac{x}{-3}=\frac{y-1}{4}=\frac{z-2}{3}$.

Hence, Option $\left(\mathrm{A}\right)$ is the correct answer.