Question

The reflection of the point $A(1,0,0)$ in the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$ is

• A. $(3,-4,-2)$
• B. $(5,-8,-4)$
• C. $(1,-1,-10)$
• D. $(2,-3,8)$

Answer 1

Answer: (B)

$(5,-8,-4)$

A point which lies on the line can be given by:$(2k+1,-3k-1,8k-10)$

Now image of $(1,0,0)$ can be found for a certain value of k. Since line connecting image and point is perpendicular to the line:

$(2k+1-1)2+(-3k-1-0)-3+(8k-10-0)8=0$

$4k+9k+3+64k-80=0$

$77k-77=0$

$k=1$

Image will be: $(3,-4,-2)$

Reflection and object will be equidistant from line and vectors joining them will be equal.

$\therefore (x-3)\hat{i}+(y+4)\hat{j}+(z+2)\hat{k}=(3-1)\hat{i}+(-4-0)\hat{j}+(-z-0)\hat{k}$

$\therefore x-3=2$

$x=5$

$\therefore y+4=-4$

$y=-8$

$\therefore z+2=-2$

$z=-4$

The point will be $(5,-8,-4)$.