Question

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

• A. $\frac{x-1}{2}=\frac{y}{-1}=\frac{z-2}{7}$
• B. $\frac{x-1}{1}=\frac{y}{2}=\frac{z-2}{7}$
• C. $\frac{x-1}{1}=\frac{y}{-2}=\frac{z-2}{7}$
• D. $\frac{x-1}{2}=\frac{y}{1}=\frac{z-2}{-7}$

Answer 1

The correct option is C $\frac{x-1}{1}=\frac{y}{-2}=\frac{z-2}{7}$

Let $P\equiv (1,0,2)$
Given, $L:\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}=r\left(\text{say}\right)$
$\therefore$ Any point on the line $L$ is
$Q\equiv (3r-1,2-2r,-r-1)$

Direction ratios of $PQ$ are $(3r-2,2-2r,-r-3)$
Direction ratios of line $L$ are $(3,-2,-1)$

Since, the line $PQ$ is perpendicular to the given line $L.$
$\therefore 3(3r-2)-2(2-2r)-1(-r-3)=0$
$\Rightarrow 14r=7\Rightarrow r=\frac{1}{2}$

Direction ratio of $PQ$ are $-\frac{1}{2},1,-\frac{7}{2}$ or $1,-2,7$

Hence, equation of line $PQ$ is
$\frac{x-1}{1}=\frac{y}{-2}=\frac{z-2}{7}$