Question

The angle between the planes $\vec{r} \cdot (12\hat{i} + 4\hat{j} - 3\hat{k}) = 5$ and $\vec{r} \cdot (5\hat{i} + 3\hat{j} + 4\hat{k}) = 7$ is:

• A. $\cos^{-1}\left( \frac{12}{13} \right)$
• B. $\cos^{-1}\left( \frac{6\sqrt{2}}{13} \right)$
• C. $\cos^{-1}\left( \frac{3\sqrt{2}}{13} \right)$
• D. $\cos^{-1}\left( \frac{6}{13} \right)$

Hint:

The angle between two planes is based on the angle between their normal vectors. Use the dot product and magnitudes to compute the cosine of the angle.

Answer 1

The Correct Option is B

The angle between two planes is given by the formula: $$\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|},$$ where $\vec{n}_1$ and $\vec{n}_2$ are the normal vectors to the planes. The normal vector to the first plane is $\vec{n}_1 = 12\hat{i} + 4\hat{j} - 3\hat{k}$, and the normal vector to the second plane is $\vec{n}_2 = 5\hat{i} + 3\hat{j} + 4\hat{k}$. Now, calculate the dot product $\vec{n}_1 \cdot \vec{n}_2$ and the magnitudes of the normal vectors $|\vec{n}_1|$ and $|\vec{n}_2|$. After performing the calculations, we find that the cosine of the angle is: $$\cos \theta = \frac{6\sqrt{2}}{13}.$$ Thus, the correct answer is $\cos^{-1}\left( \frac{6\sqrt{2}}{13} \right)$.