Show that the two planes $x\sin\alpha+ y\cos\alpha + z = 3$ and $x\cos\alpha+y\sin\alpha- z = 5$ are not perpendicular $(\alpha \in \mathbb{R})$

Question

Show that the two planes $$\begin{align} x\sin\alpha + y\cos\alpha + z = 3 \\ x\cos\alpha + y\sin\alpha - z = 5 \end{align}$$ are not perpendicular $(\alpha \in \mathbb{R})$.

The normals for both planes are of course -

• $\mathbf{i}\sin\alpha+ \mathbf{j}\cos\alpha + \mathbf{k}$

• $\mathbf{i}\cos\alpha+ \mathbf{j}\sin\alpha - \mathbf{k}$

I know that if these normal vectors are perpendicular, then their dot product would equal $0$. So I took the dot product and ended up with -

$$\sin2\alpha - 1$$

I'm not sure how to prove that this does not equal to $0$.

Answer 1

You could state that, in general, the expression sin(2$\alpha$) - 1 doesn't evaluate to zero, except for particular values of $\alpha$ when the above expression becomes zero, i.e., when $\alpha = \frac{\pi}{4},\frac{5\pi}{4}, \frac{9\pi}{4}, \cdots$. That should suffice.