[Math] How to find the cosine of the angle between the plane through $𝑃=(3,0,0), 𝑄=(0,7,0)$, and $𝑅=(0,0,6)$ and the $𝑦𝑧$-plane

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I'm trying to find the cosine of the angle between the plane through $𝑃=(3,0,0), 𝑄=(0,7,0)$, and $𝑅=(0,0,6)$ and the $𝑦𝑧$-plane, defined as the angle between their normal vectors.

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Here's my attempt. First I find vectors:

$\vec{PQ} = \left\langle-3,7,0\right\rangle$

$\vec{PR} = \left\langle-3,0,6\right\rangle$

Then I find the cross product of the two above vectors to get the normal vector: \begin{pmatrix}42&18&21\end{pmatrix}

Then I calculate the magnitude of it:

$3\sqrt{281}$

$cos\left(\theta \right)=3\sqrt{281}$

Idk what to do now. This isn't right. Thanks in advance.

Best Answer

In your work so far, it is not correct to set $\cos( \theta)$ equal to the magnitude of that vector you found.

You found a normal vector to that plane: $\langle 42,18,21\rangle\sim\langle 14,6,7\rangle$.

A normal vector to the other plane is $\langle1,0,0\rangle$.

So now find the cosine of the angle between these two vectors using the formula $\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\lVert\vec{u}\rVert\lVert\vec{v}\rVert}$.

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