3D Geometry
Angle Between Line and Plane
Grade 12
Question:
<p><strong>Ex. 61 (C):</strong> The angle between the line \(x = y = z\) and the plane \(4x - 3y + 5z = 2\) 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 \(\theta\) between a line and a plane satisfies \(\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}\) where \(\vec{d}\) is the direction vector and \(\vec{n}\) is the normal.
Solution: Direction ratios of line: \(\langle 1, 1, 1 \rangle\) Normal to plane: \(\langle 4, -3, 5 \rangle\) \(\sin\theta = \frac{|1(4) + 1(-3) + 1(5)|}{\sqrt{1^2+1^2+1^2}\sqrt{4^2+(-3)^2+5^2}}\) \(\sin\theta = \frac{|4-3+5|}{\sqrt{3}\sqrt{50}} = \frac{6}{\sqrt{150}} = \frac{6}{5\sqrt{6}} = \frac{6}{5\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{6\sqrt{6}}{30}\) Simplifying: \(\sin\theta = \frac{6}{5\sqrt{6}} = \frac{6}{\sqrt{150}}\) ∴ Answer is (p) \(\sin^{-1}\frac{6}{25}\)
Correct Answer: A