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If $\alpha, \beta, \gamma$ be the direction angles of a vector and $\cos \alpha=\frac{14}{15}, \quad \cos \beta=\frac{1}{3}$ then $\cos \gamma=$
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The correct answer is:
$\pm \frac{2}{15}$
$\begin{aligned}
\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma & =1 \\
& \Rightarrow \cos \gamma=\sqrt{1-\left(\frac{14}{15}\right)^2-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}-\left(\frac{196}{225}\right)}= \pm \frac{2}{15}
\end{aligned}$
\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma & =1 \\
& \Rightarrow \cos \gamma=\sqrt{1-\left(\frac{14}{15}\right)^2-\left(\frac{1}{3}\right)^2}=\sqrt{\frac{8}{9}-\left(\frac{196}{225}\right)}= \pm \frac{2}{15}
\end{aligned}$
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