Question

The image of the point $(1,3,4)$ in the plane $2x-y+z+3=0$ is

• A. $-1,4,3$
• B. $-3,5,2$
• C. $1,-4,-3$
• D. $3,-5,-2$

Answer 1

The correct option is B $-3,5,2$

Let $P(1,3,4)$ be a point

Let M be the point on the plane

Equation of the plane is $2x-y+z+3=0$

Thus equation of the plane is;

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=k$

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

$x=2k+1;y=-k+3;z=k+4$

Substituting the above value in the equation

of the place we have

$2(2k+1)-(-k+3)+(k+4)+3=0$

$\Rightarrow 4k+2+k-3+k+4+3=0$

$\Rightarrow 6k+6=0$

$k=-1$

Thus coordinates on M are

$x=2(-1)+1=-1$

$y=-(-1)+3=4$

$z=(-1)+4=3$

Let $Q(x_1,y_1,z_1)$ be the image of P

Thus M is the midpoint of PQ

Therefore,

$-1=\frac{1+x_1}{2};4=\frac{3+y_1}{2};3=\frac{4+z_1}{2}$

$\Rightarrow x_1=-3,y_1=5,z_1=2$