Question

The length of the perpendicular distance of the point $(-1,4,0)$ from the line $\frac{x}{1}=\frac{y}{3}=\frac{z}{1}$ is equal to

• A. $\sqrt{6}$
• B. $\sqrt{5}$
• C. $2$
• D. $1$

Answer 1

Answer: (A)

Let L be the foot of the perpendicular drawn from the point
$P(-1,4,0)$ to the given line.
The coordinated of a general point on $\frac{x-0}{1}=\frac{y-0}{3}=\frac{z-0}{1}$ are given by
$\frac{x}{1}=\frac{y}{3}=\frac{z}{1}=r$
i.e., $x=r,y=3r,z=r$
Let the coordinate of L be $(r,3r,r)$ .. (i)
Direction ratios of PL are $=r+1,3r-4,r$
Direction ratios of the given line are $1,3,1$
Since, PL perpendicular to the given, line,
$\therefore$ $(r+1)+(3r-4).2+r.1=0$
$\Rightarrow$ $r+1.1+(3r-4).2+r.1=0$
$\Rightarrow$ $11r=11$
$\Rightarrow$ $r=1$
So, the coordinate of L is
$(1,3,1)$
$\therefore$ $PL=\sqrt{{(1+1)}^2+{(3-4)}^2+{(1-0)}^2}$
$=\sqrt{4+1+1}=\sqrt{6}$