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Question

(D) Let the angles made by the line with $X, Y,$ and $Z-$ axes be $\alpha, \beta,$ and $\gamma$ respectively.Given $\alpha = 30^o$ and $\beta = 45^o$.

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the angles made by the line with $X, Y,$ and $Z-$ axes be $\alpha, \beta,$ and $\gamma$ respectively.Given $\alpha = 30^o$ and $\beta = 45^o$.The direction cosines of the line are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.We know that $l^2 + m^2 + n^2 = 1$.Substituting the values: $\cos^2 30^o + \cos^2 45^o + \cos^2 \gamma = 1$.$(\frac{\sqrt{3}}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2 \gamma = 1$.$\frac{3}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.$\frac{3+2}{4} + \cos^2 \gamma = 1 \implies \frac{5}{4} + \cos^2 \gamma = 1$.$\cos^2 \gamma = 1 - \frac{5}{4} = -\frac{1}{4}$.Since $\cos^2 \gamma$ cannot be negative,such a line does not exist in real space.Therefore,the correct option is $D$.

If a line makes angles of $30^o$ and $45^o$ with $X-$ axis and $Y-$ axis,then the angle made by it with $Z-$ axis is

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