If alpha, beta, gamma are the direction angle | vedclass

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(A) The direction cosines of a vector satisfy the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given $\cos \alpha = \frac{14}{15}$ and

Vedclass · Accepted Answer

$\pm \frac{2}{15}$

Vedclass · Answer

(A) The direction cosines of a vector satisfy the identity $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. Given $\cos \alpha = \frac{14}{15}$ and $\cos \beta = \frac{1}{3}$. Substituting these values into the identity: $\left( \frac{14}{15} \right)^2 + \left( \frac{1}{3} \right)^2 + \cos^2 \gamma = 1$ $\frac{196}{225} + \frac{1}{9} + \cos^2 \gamma = 1$ $\frac{196}{225} + \frac{25}{225} + \cos^2 \gamma = 1$ $\frac{221}{225} + \cos^2 \gamma = 1$ $\cos^2 \gamma = 1 - \frac{221}{225} = \frac{4}{225}$ $\cos \gamma = \pm \sqrt{\frac{4}{225}} = \pm \frac{2}{15}$.

If $\alpha, \beta, \gamma$ are the direction angles of a vector and $\cos \alpha = \frac{14}{15}$,$\cos \beta = \frac{1}{3}$,then $\cos \gamma = $

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