Question

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:

• A. \(\displaystyle \cos^2\theta=\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big)\)
• B. \(\displaystyle \cos^2\theta=\sec^2\beta\big(\cos^2\alpha+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma\big)\)
• C. \(\displaystyle \sin^2\theta=\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big)\)
• D. \(\displaystyle \sin^2\theta=\sec^2\beta\big(\cos^2\alpha+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma\big)\)

Hint:

Angle between a vector and a plane is best handled using the scalar triple product. Always square the expression to simplify radicals.

Answer 1

The Correct Option is C

Step 1: Angle between a vector and a plane If \(\theta\) is the angle made by vector \(\vec a\) with the plane containing \(\vec b\) and \(\vec c\), then \[ \sin\theta=\frac{|\vec a\cdot(\vec b\times\vec c)|}{|\vec a|\,|\vec b\times\vec c|}. \] Step 2: Express magnitudes using angles \[ |\vec b\times\vec c|=|\vec b|\,|\vec c|\sin\beta. \] Also, \[ \vec a\cdot(\vec b\times\vec c) =|\vec a|\,|\vec b|\,|\vec c| \sqrt{\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma}. \] Step 3: Compute \(\sin^2\theta\) \[ \sin^2\theta =\frac{\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma}{\sin^2\beta} \] \[ =\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big). \] \[ \boxed{\sin^2\theta=\cosec^2\beta\big(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\big)} \]