Question

The distance of the plane $3x+4y+5z+19=0$ from the point $(1,-1,1)$ measured along a line parallel to the line with direction ratios $2,3,1$ is

• A. $\frac{23}{5\sqrt{2}}$
• B. $\frac{\sqrt{71}}{5\sqrt{2}}$
• C. $\sqrt{14}$
• D. $\sqrt{23}$

Answer 1

Answer: (C)

According to given information

Equation of lines is passing through $(1,-1,1)$ and
having DC's is $(2,3,1)$
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{1}=r$
Here, $(2r+1,3r-1,r+1)$ lie on plane.
$\therefore$ These points satisfy the equation of plane.
$3(2r+1)+4(3r-1)+5(r+1)+19=0$
$\Rightarrow 6r+3+12r-4+5r+5+19=0$
$\Rightarrow 23r+23=0\Rightarrow r=-1$
So, point is $(-2+1,-3-1,-1+1)$ i.e. $(-1,-4,0)$
Now, required distance $=\sqrt{(-2{)}^2+(-3{)}^2+1^2}$
$=\sqrt{4+9+1}=\sqrt{14}$