In three-dimensional space,a line AB makes an | vedclass

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(C) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given $\alpha = 45^\circ$ and $\beta = 1

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction angles of the line $AB$ be $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively.Given $\alpha = 45^\circ$ and $\beta = 120^\circ$.The direction cosines are $l = \cos \alpha, m = \cos \beta, n = \cos \gamma$.We know that $l^2 + m^2 + n^2 = 1$.Substituting the values: $\cos^2(45^\circ) + \cos^2(120^\circ) + \cos^2(\gamma) = 1$.$(\frac{1}{\sqrt{2}})^2 + (-\frac{1}{2})^2 + \cos^2(\gamma) = 1$.$\frac{1}{2} + \frac{1}{4} + \cos^2(\gamma) = 1$.$\frac{3}{4} + \cos^2(\gamma) = 1$.$\cos^2(\gamma) = 1 - \frac{3}{4} = \frac{1}{4}$.$\cos(\gamma) = \pm \frac{1}{2}$.Since $	heta$ is an acute angle,$\cos(	heta) = \frac{1}{2}$.Therefore,$	heta = 60^\circ$.

In three-dimensional space,a line $AB$ makes angles of $45^\circ$ and $120^\circ$ with the positive $x$-axis and $y$-axis,respectively. If $AB$ makes an acute angle $\theta$ with the positive $z$-axis,then $\theta = \dots^\circ$.

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