MHT CET · Maths · Three Dimensional Geometry
A line makes angles \(\propto, \beta, \gamma\) with the co-ordinate axes and \(\propto+\beta=90^{\circ}\), then \(\gamma=\)
- A \(60^{\circ}\)
- B \(90^{\circ}\)
- C \(45^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(B) \(90^{\circ}\)
Step-by-step Solution
Detailed explanation
We know that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
It is given that \(\alpha+\beta=90^{\circ} \Rightarrow \alpha=90^{\circ}-\beta \Rightarrow \cos \alpha=\cos \left(90^{\circ}-\beta\right)\)
\(\therefore \cos \alpha=\sin \beta \Rightarrow \cos ^{2} \alpha=\sin ^{2} \beta=1-\cos ^{2} \beta \Rightarrow\) \(\cos ^{2} \alpha+\cos ^{2} \beta=1\)
Thus \(1+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=0 \Rightarrow \gamma=90^{\circ}\)
It is given that \(\alpha+\beta=90^{\circ} \Rightarrow \alpha=90^{\circ}-\beta \Rightarrow \cos \alpha=\cos \left(90^{\circ}-\beta\right)\)
\(\therefore \cos \alpha=\sin \beta \Rightarrow \cos ^{2} \alpha=\sin ^{2} \beta=1-\cos ^{2} \beta \Rightarrow\) \(\cos ^{2} \alpha+\cos ^{2} \beta=1\)
Thus \(1+\cos ^{2} \gamma=1 \Rightarrow \cos ^{2} \gamma=0 \Rightarrow \gamma=90^{\circ}\)
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