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(C) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.We know the fundamental identity for directi

Vedclass · Accepted Answer

$\pm 2/15$

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(C) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.We know the fundamental identity for direction cosines is $l^2 + m^2 + n^2 = 1$.Given $l = \cos \alpha = 14/15$ and $m = \cos \beta = 1/3$.Substituting these values into the identity:$(14/15)^2 + (1/3)^2 + \cos^2 \gamma = 1$$196/225 + 1/9 + \cos^2 \gamma = 1$To add the fractions,find a common denominator,which is $225$:$196/225 + 25/225 + \cos^2 \gamma = 1$$221/225 + \cos^2 \gamma = 1$$\cos^2 \gamma = 1 - 221/225$$\cos^2 \gamma = (225 - 221) / 225$$\cos^2 \gamma = 4/225$Taking the square root on both sides:$\cos \gamma = \pm \sqrt{4/225} = \pm 2/15$.

If a line makes angles $\alpha, \beta, \gamma$ with the coordinate axes and $\cos \alpha = 14/15, \cos \beta = 1/3$,then $\cos \gamma = \dots$

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