Question:medium
Let \(\alpha,\beta,\gamma\ (0<\alpha,\beta,\gamma<\tfrac{\pi}{2})\) be the angles between non–zero vectors
\(\vec a\) and \(\vec b\), \(\vec b\) and \(\vec c\), \(\vec c\) and \(\vec a\) respectively.
If \(\theta\) is the angle that the vector \(\vec a\) makes with the plane containing \(\vec b\) and \(\vec c\),
then:
Let \(\alpha,\beta,\gamma\ (0<\alpha,\beta,\gamma<\tfrac{\pi}{2})\) be the angles between non–zero vectors
\(\vec a\) and \(\vec b\), \(\vec b\) and \(\vec c\), \(\vec c\) and \(\vec a\) respectively.
If \(\theta\) is the angle that the vector \(\vec a\) makes with the plane containing \(\vec b\) and \(\vec c\),
then:
Show Hint
Angle between a vector and a plane is best handled using the scalar triple product.
Always square the expression to simplify radicals.
Updated On: Mar 5, 2026
- \(\displaystyle \cos^2\theta=\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big)\)
- \(\displaystyle \cos^2\theta=\sec^2\beta\big(\cos^2\alpha+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma\big)\)
- \(\displaystyle \sin^2\theta=\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big)\)
- \(\displaystyle \sin^2\theta=\sec^2\beta\big(\cos^2\alpha+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma\big)\)
Show Solution
The Correct Option is C
Solution and Explanation
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