A smooth plane is described by the equation $x_1+x_2+x_3=3$. A light ray hits the plane along the direction $(−1,−1,0)$. Find the normalized direction of the reflection. Enter the third component of the direction.
My solution (normalized): (0.66, 0.66, 0.35)
I'm pretty sure I have to normalize this. Can somebody please confirm my answer for me? I am unsure if I am correct.
Your answer is wrong. The reflected ray will be in the direction given by $$ \vec{r} = \vec{d} - 2 \, \text{proj}_{\vec{n}}\vec{d} $$ where $\vec{d}$ is the direction of the incident ray, $\vec{n}$ is a normal to the plane and $\text{proj}_{\vec{n}}\vec{d}$ is the orthogonal projection of $\vec{d}$ on $\vec{n}$. $$ \vec{r} = (-1,-1,0) - 2 \frac{(-1,-1,0)\cdot (1,1,1)}{(1,1,1)\cdot (1,1,1)}(1,1,1) =\left(\frac{1}{3},\frac{1}{3},\frac{4}{3}\right) $$ so $\vec{r}$ normalized is $$ \left(\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}},\frac{4}{3\sqrt{2}}\right) $$