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 $(a,0,a)$ and $(b,-b,0)$

Now for any vectors $\overrightarrow{x},\overrightarrow{y}$ , 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

$\overrightarrow{x}\cdot \overrightarrow{y}=\left|\overrightarrow{x}\right|\left|\overrightarrow{y}\right|\mathrm{cos}{\theta}_{xy}.$

You have $\overrightarrow{x}=(a,0,a),\overrightarrow{y}=(b,-b,0)$

Now for any vectors $\overrightarrow{x},\overrightarrow{y}$ , 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

$\overrightarrow{x}\cdot \overrightarrow{y}=\left|\overrightarrow{x}\right|\left|\overrightarrow{y}\right|\mathrm{cos}{\theta}_{xy}.$

You have $\overrightarrow{x}=(a,0,a),\overrightarrow{y}=(b,-b,0)$

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