Question

The distance of the point (1,-2,3) from the plane x-y+z=5, measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

• A. $\frac{1}{7}$
• B. 1
• C. 7
• D. None of these

Answer 1

The correct option is B 1

Point is $(1,-2,3)$

Equation of plane = $x-y+z=5$

Equation of parallel line is $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$

$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda$

$x=2\lambda +1,y=3\lambda -2,z=-6\lambda +3$

Since,

$x-y+z=5$

$2\lambda +1-3\lambda +2-6\lambda +3=5$

$-7\lambda +6=5$

$\lambda =\frac{1}{7}$

Thus,

$x=\frac{9}{7},y=\frac{-11}{7},z=\frac{15}{7}$

Therefore,

Distance = $\sqrt{(\frac{2}{7}{)}^2+(\frac{3}{7}{)}^2+(\frac{-6}{7}{)}^2}$ = $\sqrt{\frac{49}{49}}$ = $1$ unit