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 \( \cos^{-1}\left( \frac{6\sqrt{2}}{13} \right) \).

Answer 2

Given two planes:
\[ \vec{r} \cdot (12\hat{i} + 4\hat{j} - 3\hat{k}) = 5 \] and \[ \vec{r} \cdot (5\hat{i} + 3\hat{j} + 4\hat{k}) = 7 \] Find the angle between the planes.

Step 1: The angle \( \theta \) between two planes is the angle between their normal vectors:
\[ \cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| \times |\vec{n}_2|} \] where
\[ \vec{n}_1 = (12, 4, -3), \quad \vec{n}_2 = (5, 3, 4) \]

Step 2: Calculate dot product \( \vec{n}_1 \cdot \vec{n}_2 \):
\[ 12 \times 5 + 4 \times 3 + (-3) \times 4 = 60 + 12 - 12 = 60 \]

Step 3: Calculate magnitudes:
\[ |\vec{n}_1| = \sqrt{12^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13 \] \[ |\vec{n}_2| = \sqrt{5^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5 \sqrt{2} \]

Step 4: Calculate cosine of angle:
\[ \cos \theta = \frac{60}{13 \times 5 \sqrt{2}} = \frac{60}{65 \sqrt{2}} = \frac{12}{13 \sqrt{2}} = \frac{6 \sqrt{2}}{13} \] (after rationalizing numerator and denominator)

Therefore,
\[ \boxed{\cos^{-1} \left( \frac{6 \sqrt{2}}{13} \right)} \]