Question:
Let \( \alpha,\beta,\gamma\left(0<\alpha,\beta,\gamma<\frac{\pi}{2}\right) \) 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 \( \alpha,\beta,\gamma\left(0<\alpha,\beta,\gamma<\frac{\pi}{2}\right) \) 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
Take unit vectors; \( \cos\theta \) is the length of the projection of \( \vec a \) on the plane of \( \vec b,\vec c \). Write the projection as \( x\vec b+y\vec c \) and use dot products.
Updated On: Nov 12, 2026
- \( \cos^{2}\theta=\operatorname{cosec}^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \)
- \( \cos^{2}\theta=\sec^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma+2\cos\alpha\cos\beta\cos\gamma\right) \)
- \( \sin^{2}\theta=\operatorname{cosec}^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \)
- \( \sin^{2}\theta=\sec^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma+2\cos\alpha\cos\beta\cos\gamma\right) \)
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The Correct Option is A
Solution and Explanation
Take \( \vec a,\vec b,\vec c \) as unit vectors, so \( \vec a\cdot\vec b=\cos\alpha \), \( \vec b\cdot\vec c=\cos\beta \), \( \vec c\cdot\vec a=\cos\gamma \).
The projection of \( \vec a \) on the plane is \( \vec p=x\vec b+y\vec c \), and \( \cos\theta=|\vec p| \). Since \( \vec a-\vec p \) is perpendicular to the plane, \( \vec p\cdot\vec b=\cos\alpha \) and \( \vec p\cdot\vec c=\cos\gamma \):
\( x+y\cos\beta=\cos\alpha \), \( x\cos\beta+y=\cos\gamma \).
Solving: \( x=\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin^2\beta} \), \( y=\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin^2\beta} \).
\( |\vec p|^2=\vec p\cdot(x\vec b+y\vec c)=x\cos\alpha+y\cos\gamma=\frac{\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma}{\sin^2\beta} \).
Hence \( \cos^2\theta=\operatorname{cosec}^2\beta\left(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \).
Answer: (1) \( \cos^{2}\theta=\operatorname{cosec}^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \)
The projection of \( \vec a \) on the plane is \( \vec p=x\vec b+y\vec c \), and \( \cos\theta=|\vec p| \). Since \( \vec a-\vec p \) is perpendicular to the plane, \( \vec p\cdot\vec b=\cos\alpha \) and \( \vec p\cdot\vec c=\cos\gamma \):
\( x+y\cos\beta=\cos\alpha \), \( x\cos\beta+y=\cos\gamma \).
Solving: \( x=\frac{\cos\alpha-\cos\beta\cos\gamma}{\sin^2\beta} \), \( y=\frac{\cos\gamma-\cos\alpha\cos\beta}{\sin^2\beta} \).
\( |\vec p|^2=\vec p\cdot(x\vec b+y\vec c)=x\cos\alpha+y\cos\gamma=\frac{\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma}{\sin^2\beta} \).
Hence \( \cos^2\theta=\operatorname{cosec}^2\beta\left(\cos^2\alpha+\cos^2\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \).
Answer: (1) \( \cos^{2}\theta=\operatorname{cosec}^{2}\beta\left(\cos^{2}\alpha+\cos^{2}\gamma-2\cos\alpha\cos\beta\cos\gamma\right) \)
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