Question

Find the distance of the point $P(-2,3,-4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $4x+12y-3z+1=0$

• A. $\frac{17}{2}$
• B. $\frac{9}{2}$
• C. $4$
• D. $9$

Answer 1

The correct option is A $\frac{17}{2}$

Let the equation of the line passing through the point $(-2,3,-4)$ be

$\frac{x+2}{a}=\frac{y-3}{b}=\frac{z+4}{c}=k$(say)

$\Rightarrow x=ak-2,y=bk+3,z=ck-4$

Thus coordinate of any point on $L$ are $(ak-2,bk+3,ck-4)$

Equation of the given line

$L_1:\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}=\lambda$(let)

$\Rightarrow \frac{(ak-2)+2}{3}=\frac{2bk+6+3}{4}=\frac{3ck-12+4}{5}=\lambda$

$\Rightarrow \frac{ak}{3}=\frac{2bk+9}{4}=\frac{3ck-8}{5}=\lambda$ ...(2)

$\Rightarrow a=\frac{3\lambda }{k},b=\frac{4\lambda -9}{2k},c=\frac{5\lambda +8}{3k}$

Since, the line $L$ and plane $P:4x+12y-3z+1=0$ are parallel

So,

$4a+12b-3c+1=0\Rightarrow \frac{12\lambda }{k}+\frac{12(4\lambda -9)}{2k}-\frac{3(5\lambda +8)}{3k}=0$

$\Rightarrow \frac{31\lambda -62}{k}=0\Rightarrow \lambda =2\Rightarrow a=\frac{6}{k},b=-\frac{1}{2k},c=\frac{6}{k}$

Thus, equation of $L$ are $\frac{x+2}{6}=\frac{y-3}{-\frac{1}{2}}=\frac{z+4}{6}$

And

any point on $L$ has coordinate $(6k-2,-\frac{1}{2}k+3,6k-4)$

Now

$\lambda =\frac{9k}{3}=\frac{6}{k}\times \frac{k}{3}=2$

Thus

$L_1:\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}=2$

Thus

$L$ and $L_1$ intersect at $(4,\frac{5}{2},2)$

Thus,

Required distance $=\sqrt{{(4+2)}^2+{(\frac{5}{2}-3)}^2+{(2+4)}^2}=\frac{17}{2}$

Hence, option A.