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A vector $v$ is equally inclined to the $x$ -axis, $y$ -axis and $z$ -axis respectively, its direction cosines are
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Verified Answer
The correct answer is:
$ < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>$ or $ < -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>$
Let the vector $\mathbf{v}$ make an angle $\alpha$ with each of the three axes, then direction cosine of $\mathbf{v}$ are
$$
\begin{array}{l}
\quad < \cos \alpha, \cos \alpha, \cos \alpha> \\
\text { Also, } \quad \cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1 \\
\Rightarrow \quad \cos ^{2} \alpha=1 / 3 \\
\Rightarrow \quad \cos \alpha=\pm \frac{1}{\sqrt{3}}
\end{array}
$$
Hence, direction cosine of $\mathbf{v}$ are
$$ < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>
$$
Or
$$ < -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>
$$
$$
\begin{array}{l}
\quad < \cos \alpha, \cos \alpha, \cos \alpha> \\
\text { Also, } \quad \cos ^{2} \alpha+\cos ^{2} \alpha+\cos ^{2} \alpha=1 \\
\Rightarrow \quad \cos ^{2} \alpha=1 / 3 \\
\Rightarrow \quad \cos \alpha=\pm \frac{1}{\sqrt{3}}
\end{array}
$$
Hence, direction cosine of $\mathbf{v}$ are
$$ < \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}>
$$
Or
$$ < -\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}>
$$
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