Question

The acute angle between the planes 2x-y-3z=7and x+2y+2z=0 is

• A. \(\cos^{-1}(\frac{-\sqrt{14}}{14})\)
• B. \(\pi-\cos^{-1}(\frac{-\sqrt{14}}{7})\)
• C. \(\cos^{-1}(\frac{\sqrt{14}}{11})\)
• D. \(\pi-\cos^{-1}(\frac{-\sqrt{14}}{21})\)
• E. \(\pi-\cos^{-1}(\frac{\sqrt{14}}{7})\)

Answer 1

The Correct Option is B

Given: The equations of the planes are: 

\[ 2x - y - 3z = 7 \] \[ x + 2y + 2z = 0 \]

Step 1: Find normal vectors

Normal to the first plane: \( \vec{n_1} = (2, -1, -3) \)

Normal to the second plane: \( \vec{n_2} = (1, 2, 2) \)

Step 2: Find the cosine of the angle

The formula for the angle \( \theta \) between two planes is:

\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]

Compute the dot product:

\[ \vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(2) + (-3)(2) = 2 - 2 - 6 = -6 \]

Compute magnitudes:

\[ |\vec{n_1}| = \sqrt{2^2 + (-1)^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} \] \[ |\vec{n_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Compute \( \cos \theta \):

\[ \cos \theta = \frac{-6}{\sqrt{14} \times 3} = \frac{-6}{3\sqrt{14}} = \frac{-\sqrt{14}}{7} \]

Final Answer:

\[ \pi - \cos^{-1} \left( \frac{-\sqrt{14}}{7} \right) \]