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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
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equal to
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Verified Answer
The correct answer is:
\( -1 \)
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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