Question

A straight line passes through the point $(2,-1,-1)$. It is parallel to the plane $4x+y+z+2=0$ and perpendicular to the line $\frac{x}{1}=\frac{y}{-2}=\frac{z-5}{1}$. Then its equation is

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

Answer 1

Answer: (C)

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

$\text{Given, a point}P(2,-1,-1)\text{on a line}$

$\text{is a parallel to}4x+y+z+2=0$

and perpendicular to the line $\frac{x}{1}=\frac{y}{-2}=\frac{2-5}{1}$

$\begin{array}{cc}\Rightarrow & \because \text{line is parallel to}4x+y+z+2=0\\ & \text{Then DRs of the line will be,}(4,1,1)\end{array}$

$4l+m+n=0$

$\text{and}\ell -2m+n=0$ [where l,m & n are DCs]

$\Rightarrow \frac{l}{3}=\frac{m}{-3}=\frac{n}{-9}$

$\Rightarrow \frac{d}{-1}=\frac{m}{+1}=\frac{n}{3}$

Therefore equation of a straight line is

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

Option C is correct.