Originally Posted by

**shawsend** I corrected my work from an earlier attempt. Still a little unsure and thinking about a 500 million space walk but I digress. Here goes:

This is what I have so far:

I can calculate the vector angle between the incident light ray and the normal of the surface (the blue and red lines) at the impact point of $\displaystyle (4,2,3)$:

$\displaystyle \cos(t)=\frac{<-1,3,0>.<8,4,-6>}{|-1,3,0| |8,4,6|}=\sqrt{2/145}$

Now, let the reflection ray be at the point $\displaystyle (10,y,z)$. Then it will satisfy the following two equations:

$\displaystyle -18(10-4)-6(y-2)-28(z-3)=0$

$\displaystyle \frac{<10,y,z>.<8,4,6>}{|10,y,z| |8,4,-6|}=\sqrt{2/145}$

and the (numerical) solution is $\displaystyle (10,-11.9,2.12)$

Do you know what the correct answer is?