KCET · Maths · Three Dimensional Geometry
If \( \cos \alpha, \cos \beta, \cos \gamma \) are the direction cosines of a vector \( \vec{a} \), then \( \cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma \) is
equal to
- A \( 12 \)
- B \( 13 \)
- C \( -1 \)
- D \( 00 \)
Answer & Solution
Correct Answer
(C) \( -1 \)
Step-by-step Solution
Detailed explanation
We know that, \(\cos 2 \alpha=2 \cos ^{2} \alpha-1\)
So, \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\)
\(=2\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)-3\)
Since, \(\cos \alpha, \cos \beta, \cos \gamma\) are the direction cosines of a vector \(\vec{a}\)
Then,
\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
Therefore,
\(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=2(1)-3=-1\)
So, \(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma\)
\(=2\left(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma\right)-3\)
Since, \(\cos \alpha, \cos \beta, \cos \gamma\) are the direction cosines of a vector \(\vec{a}\)
Then,
\(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\)
Therefore,
\(\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma=2(1)-3=-1\)
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