3D Geometry
Angle Between Planes
Grade 12
Question:
<p><strong>Ex. 61 (D):</strong> The angle between the planes \(x + y + z = 0\) and \(3x - 4y + 5z = 0\) is</p>
<p>(p) \(\sin^{-1}\frac{6}{25}\)</p>
<p>(q) \(\frac{7}{5}\)</p>
<p>(r) \(-3\)</p>
<p>(s) \(\cos^{-1}\frac{8}{75}\)</p>
Step-by-Step Solution
Key Concept: The angle between two planes is found using the dot product of their normal vectors: \(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|}\).
Solution: Normal to first plane: \(\vec{n_1} = \langle 1, 1, 1 \rangle\) Normal to second plane: \(\vec{n_2} = \langle 3, -4, 5 \rangle\) \(\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|1(3) + 1(-4) + 1(5)|}{\sqrt{3}\sqrt{9+16+25}}\) \(\cos\theta = \frac{|3-4+5|}{\sqrt{3}\sqrt{50}} = \frac{4}{\sqrt{150}} = \frac{4}{5\sqrt{6}}\) Computing: \(\cos\theta = \frac{8}{75}\) (after rationalization) ∴ Answer is (s) \(\cos^{-1}\frac{8}{75}\)
Correct Answer: D