Question

A straight line passes through (1, -2, 3) and perpendicular to the plane $2x+3y-z=7$.

What is the image of the point (1, -2, 3) in the plane ?

• A. (2, -1, 5)
• B. (-1, 2, -3)
• C. (5, 4, 1)
• D. None of the above

Answer 1

The correct option is C (5, 4, 1)

Let $P(1,-2,3)$ be a point

Let $M$ be the point on the plane

Equation of the plane is $2x+3y-z=7$

Thus the equation of the line is

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

Any point on the above line $PM$, is of the form,

$x=2k+1;y=3k-2;z=-k+3$

Substituting the above values in the equation of the plane, we get

$2(2k+1)+3(3k-2)-(-k+3)-7=0$

$\Rightarrow 4k+2+9k-6+k-3-7=0$

$\Rightarrow k=1$

Thus the coordinates of $M$ are:

$x=3$, $y=1$, $z=2$

Let $Q(x^{{}^{\prime }},y^{{}^{\prime }},z^{{}^{\prime }})$ be the image of $P$

Thus, $M$ is the midpoint of $PQ$

therefore,

$3=\frac{1+x^{{}^{\prime }}}{2};1=\frac{-2+y^{{}^{\prime }}}{2};2=\frac{3+z^{{}^{\prime }}}{2}$

$x^{{}^{\prime }}=5,y^{{}^{\prime }}=4,z^{{}^{\prime }}=1$

Thus, $Q=(5,4,1)$