Answer
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Hint: Use the formula $\cos 2x=2{{\cos }^{2}}x-1$ to simplify $\cos 2\alpha +\cos 2\beta +\cos 2\gamma $. You will get $\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)-3$. Next, use the formula for direction cosines that ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$ to get required answer.
Complete step-by-step answer:
In this question, we are given that a line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes.
Using this information, we need to find the value of $\cos 2\alpha +\cos 2\beta +\cos 2\gamma $.
We know that cosine of twice of an angle x is equal to the difference of twice of cosine of angle x and 1.
i.e. $\cos 2x=2{{\cos }^{2}}x-1$
Using the above formula, we can rewrite the given expression as the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2{{\cos }^{2}}\alpha -1+2{{\cos }^{2}}\beta -1+2{{\cos }^{2}}\gamma -1$
We will now rearrange these terms. After rearranging the terms in the above expression, we will get the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)-3$
Now, we know that if a line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes then its direction cosines are $\cos \alpha ,\cos \beta ,\cos \gamma $ respectively.
We also know the property of direction cosines that the sum of the squares of all the direction cosines is equal to 1.
i.e. ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$
Substituting this in the above expression, we will get the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)-3$
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2-3=-1$
So, $\cos 2\alpha +\cos 2\beta +\cos 2\gamma =-1$
Hence, option (d) is the correct answer.
Note: In this question, it is very important to know about direction cosines. In analytic geometry, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes. Equivalently, they are the contributions of each component of the basis to a unit vector in that direction.
Complete step-by-step answer:
In this question, we are given that a line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes.
Using this information, we need to find the value of $\cos 2\alpha +\cos 2\beta +\cos 2\gamma $.
We know that cosine of twice of an angle x is equal to the difference of twice of cosine of angle x and 1.
i.e. $\cos 2x=2{{\cos }^{2}}x-1$
Using the above formula, we can rewrite the given expression as the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2{{\cos }^{2}}\alpha -1+2{{\cos }^{2}}\beta -1+2{{\cos }^{2}}\gamma -1$
We will now rearrange these terms. After rearranging the terms in the above expression, we will get the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)-3$
Now, we know that if a line makes angles $\alpha ,\beta ,\gamma $ with the coordinate axes then its direction cosines are $\cos \alpha ,\cos \beta ,\cos \gamma $ respectively.
We also know the property of direction cosines that the sum of the squares of all the direction cosines is equal to 1.
i.e. ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$
Substituting this in the above expression, we will get the following:
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)-3$
$\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2-3=-1$
So, $\cos 2\alpha +\cos 2\beta +\cos 2\gamma =-1$
Hence, option (d) is the correct answer.
Note: In this question, it is very important to know about direction cosines. In analytic geometry, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes. Equivalently, they are the contributions of each component of the basis to a unit vector in that direction.
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