A line which makes an angle of 60^o with the | vedclass

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Question

(A) Let the direction angles of the line be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given that $\beta = 60^o$ and $\gamma = 60^o$

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) Let the direction angles of the line be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given that $\beta = 60^o$ and $\gamma = 60^o$.The sum of the squares of the direction cosines is given by $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting the given values:$\cos^2 \alpha + \cos^2 60^o + \cos^2 60^o = 1$$\cos^2 \alpha + (\frac{1}{2})^2 + (\frac{1}{2})^2 = 1$$\cos^2 \alpha + \frac{1}{4} + \frac{1}{4} = 1$$\cos^2 \alpha + \frac{1}{2} = 1$$\cos^2 \alpha = 1 - \frac{1}{2} = \frac{1}{2}$$\cos \alpha = \frac{1}{\sqrt{2}}$Therefore,$\alpha = 45^o$.

$A$ line which makes an angle of $60^o$ with the $y$-axis and $z$-axis,then the angle which it makes with the $x$-axis is ......... $^o$.

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