Question

The distance of the point $(1,-5,9)$ from the plane$x-y+z=5$ measured along a straight line $x=y=z$ is

• A. $3\sqrt{5}$
• B. $10\sqrt{3}$
• C. $5\sqrt{3}$
• D. $3\sqrt{10}$

Answer 1

The correct option is B $10\sqrt{3}$

$P(1,-5,9)$, Plane:$x-y+z-5=0$

Line: $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}=r$

From fig. $(x-1)=r$

$(y+5)=r$

$(z-9)=r$

$x=r+1$

$y=r-5$

$z=r+9$

$\therefore r+1-r+5+r+9-5=0$

$r=-10$

$\therefore (x,y,z)=(-10+1,-10-5,-10+9)$

$(x,y,z)=(-9,-15,-1)$

distance$\sqrt{(-9-1{)}^2+(-15+5{)}^2+(-1-9{)}^2}=10\sqrt{3}$

Ans:$B$