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(A) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.We know that the sum of the squares of the d

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$2$

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(A) The direction cosines of a line are given by $l = \cos \alpha$,$m = \cos \beta$,and $n = \cos \gamma$.We know that the sum of the squares of the direction cosines is $l^2 + m^2 + n^2 = 1$.Therefore,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Using the identity $\sin^2 	heta = 1 - \cos^2 	heta$,we can write:$(1 - \sin^2 \alpha) + (1 - \sin^2 \beta) + (1 - \sin^2 \gamma) = 1$.$3 - (\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma) = 1$.$\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - 1 = 2$.

If $\alpha, \beta, \gamma$ are the angles made by a line with the positive directions of the $x, y, z$ axes respectively,then $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = \dots$

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