Answer
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Hint: Here, we will use a property of direction cosines of a line to find the value of the given expression. Since, it is given that the line makes angles $\alpha ,\beta ,\gamma $ with the axes of coordinates, so $\cos \alpha ,\cos \beta ,\cos \gamma $ are the direction cosines of the line.
Complete step-by-step solution -
In analytical geometry, the direction cosines ( or directional 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. Direction cosines are on analogous extension of the usual notion of slope to higher dimensions.
Generally, we represent the direction cosines of a line as $l,m\,and\,n$.
If a, band c are three numbers proportional to the direction cosines $l,m\,and\,n$ of a straight line, then a, b and c are called its direction ratios. They are also called direction components.
We define a, b and c as:
$a=\dfrac{l}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}\,,\,b=\dfrac{m}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}\,and\,c=\dfrac{n}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$
A property of the direction cosines of a line is that:
${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1................\left( 1 \right)$
Since, we know that:
$\begin{align}
& \cos 2x=2{{\cos }^{2}}x-1 \\
& \Rightarrow {{\cos }^{2}}x=\dfrac{1+\cos 2x}{2} \\
\end{align}$
Applying this trigonometric formula in equation (1), we get:
$\begin{align}
& \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+\dfrac{1+\cos 2\gamma }{2}=1 \\
& \Rightarrow \dfrac{1+\cos 2\alpha +1+\cos 2\beta +1+\cos 2\gamma }{2}=1 \\
& \Rightarrow 3+\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2 \\
& \Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =2-3 \\
& \Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =-1 \\
\end{align}$
So, the value of the given expression is = -1.
Hence, option (b) is the correct answer.
Note: Students should remember the property of the direction that is applied in this question. It is not necessary to use the same trigonometric identity that we have used here. Remember that the sum of squares of direction cosines of any given vector is equal to one
Complete step-by-step solution -
In analytical geometry, the direction cosines ( or directional 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. Direction cosines are on analogous extension of the usual notion of slope to higher dimensions.
Generally, we represent the direction cosines of a line as $l,m\,and\,n$.
If a, band c are three numbers proportional to the direction cosines $l,m\,and\,n$ of a straight line, then a, b and c are called its direction ratios. They are also called direction components.
We define a, b and c as:
$a=\dfrac{l}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}\,,\,b=\dfrac{m}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}\,and\,c=\dfrac{n}{\sqrt{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}}$
A property of the direction cosines of a line is that:
${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1................\left( 1 \right)$
Since, we know that:
$\begin{align}
& \cos 2x=2{{\cos }^{2}}x-1 \\
& \Rightarrow {{\cos }^{2}}x=\dfrac{1+\cos 2x}{2} \\
\end{align}$
Applying this trigonometric formula in equation (1), we get:
$\begin{align}
& \dfrac{1+\cos 2\alpha }{2}+\dfrac{1+\cos 2\beta }{2}+\dfrac{1+\cos 2\gamma }{2}=1 \\
& \Rightarrow \dfrac{1+\cos 2\alpha +1+\cos 2\beta +1+\cos 2\gamma }{2}=1 \\
& \Rightarrow 3+\cos 2\alpha +\cos 2\beta +\cos 2\gamma =2 \\
& \Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =2-3 \\
& \Rightarrow \cos 2\alpha +\cos 2\beta +\cos 2\gamma =-1 \\
\end{align}$
So, the value of the given expression is = -1.
Hence, option (b) is the correct answer.
Note: Students should remember the property of the direction that is applied in this question. It is not necessary to use the same trigonometric identity that we have used here. Remember that the sum of squares of direction cosines of any given vector is equal to one
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