Brenton Dixon

Answered

2022-07-20

Show that the planes $ax+az=c$ and $bx-by=d$, where $a,b,c,d\in \mathbb{R},a,b\ne 0$, always form an angle of $\pi /3={60}^{\circ }$

Answer & Explanation

Damarion Pierce

Expert

2022-07-21Added 11 answers

The main idea would be to note that the angle between planes is the same as the angle between their normal vectors, which are $\left(a,0,a\right)$ and $\left(b,-b,0\right)$
Now for any vectors $\stackrel{\to }{x},\stackrel{\to }{y}$ , the key to the angle ${\theta }_{xy}$ between them is in their dot product, which can be computed coordinate-wise, but is also given by
$\stackrel{\to }{x}\cdot \stackrel{\to }{y}=|\stackrel{\to }{x}||\stackrel{\to }{y}|\mathrm{cos}{\theta }_{xy}.$
You have $\stackrel{\to }{x}=\left(a,0,a\right),\stackrel{\to }{y}=\left(b,-b,0\right)$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?