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. $2$
• B. $1$
• C. $\frac{2}{3}$
• D. $\frac{3}{7}$

Answer 1

The correct option is B $1$

Let $A(1,-2,3),P_1:x-y+z=5,L_1:\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$
Let the point of intersection of line with plane be $B$
As distance from $A(1,-2,3)$ is measured parallel to $L_1$
So DR's of $AB=2,3,-6$
Equation of $AB$ line
$\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=k$
$\therefore$ Coordinates of B $=(2k+1,3k-2,-6k+3)$
Putting in plane $P_1$
$(2k+1)-(3k-2)+(-6k+3)=5$
$-7k+6=5$
$k=\frac{1}{7}$
Coordinates of $B(2\times \frac{1}{7}+1,3\times \frac{1}{7}-2,-6\times \frac{1}{7}+3)$
Coordinates of $B(\frac{9}{7},\frac{-11}{7},\frac{15}{7})$
Distance $AB=\sqrt{{\left(\frac{2}{7}\right)}^2+{\left(\frac{3}{7}\right)}^2+{\left(\frac{-6}{7}\right)}^2}=\sqrt{\left(\frac{49}{49}\right)}=1$unit