Question

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line

• A. $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$
• B. $\frac{x+3}{-3}=\frac{y-5}{-1}=\frac{z+2}{5}$
• C. $\frac{x-3}{3}=\frac{y+5}{1}=\frac{z-2}{-5}$
• D. $\frac{x-3}{-3}=\frac{y+5}{-1}=\frac{z-2}{5}$

Answer 1

The correct option is A $\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

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, the equation of the plane is

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

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

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

Substituting the above values in the equation of the plane we have

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

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

$\Rightarrow 6k+6=0$

$\Rightarrow k=-1$

Thus the coordinates of $M$ are

$x=2\times -1+1=-2+1=-1$

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

$z=-1+4=3$

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

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

$\therefore -1=\frac{1+x^{\prime }}{2},4=\frac{3+y^{\prime }}{2},3=\frac{4+z^{\prime }}{2}$

$\Rightarrow 1+x^{\prime }=-2,3+y^{\prime }=8,4+z^{\prime }=6$

$\Rightarrow x^{\prime }=-2-1=-3,y^{\prime }=8-3=5,z^{\prime }=6-4=2$

$\therefore \frac{x+3}{3}=\frac{y-3}{1}=\frac{z-2}{-5}$ is the image of the given line.