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Find the angle of intersection in radians of the plane $4x+4y−1z=0$ with the plane $−4x−2y+3z=0$.

Attempt:

Write $\overrightarrow{n_1} = (4,4,-1)$ and $\overrightarrow{n_2} = (-4, -2, 3)$ and then $\displaystyle \theta = \cos^{-1}\Big(\frac{\overrightarrow{n_1}\cdot\overrightarrow{n_2}}{|\overrightarrow{n_1}|\overrightarrow{n_2}|}\Big)=\cos^{-1}\big(\frac{-27}{|\sqrt{33}|\sqrt{29}|}\big)$.

However, this was the incorrect answer.

rae306
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Jalen Knox
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1 Answers1

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Perhaps they wanted the acute angle instead of the obtuse angle. Make the dot product positive first: $$ \cos^{-1}\left(\frac{|-27|}{\sqrt{33}\sqrt{29}}\right) \approx 0.5099 \text{ rad} $$

Adriano
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  • It seems as if they did want the acute angle, however it was not specified. Thank you, and how does adding absolute value to the numerator make something an acute angle? – Jalen Knox Sep 05 '14 at 17:39
  • The range of $y = \cos^{-1}(x)$ is $[0,\pi]$. The acute portion of this range is attained when $x \in (0, 1)$. – Adriano Sep 05 '14 at 17:43