A line AB in three dimensions makes angles 45 | vedclass

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(C) Let the direction angles of the line $AB$ be $\alpha = 45^{\circ}$,$\beta = 120^{\circ}$,and $\gamma = 	heta$.The sum of the squares of the direc

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the direction angles of the line $AB$ be $\alpha = 45^{\circ}$,$\beta = 120^{\circ}$,and $\gamma = 	heta$.The sum of the squares of the direction cosines of a line is always $1$,i.e.,$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.Substituting the given values,we have $\cos^2 45^{\circ} + \cos^2 120^{\circ} + \cos^2 	heta = 1$.Since $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ and $\cos 120^{\circ} = -\frac{1}{2}$,we get:$\left(\frac{1}{\sqrt{2}}ight)^2 + \left(-\frac{1}{2}ight)^2 + \cos^2 	heta = 1$.$\frac{1}{2} + \frac{1}{4} + \cos^2 	heta = 1$.$\frac{3}{4} + \cos^2 	heta = 1$.$\cos^2 	heta = 1 - \frac{3}{4} = \frac{1}{4}$.Since $	heta$ is an acute angle,$\cos 	heta = \frac{1}{2}$.Therefore,$	heta = 60^{\circ}$.

$A$ line $AB$ in three dimensions makes angles $45^{\circ}$ and $120^{\circ}$ with the positive $X$-axis and the positive $Y$-axis respectively. If $AB$ makes an acute angle $\theta$ with the positive $Z$-axis,then $\theta$ equals (in $^{\circ}$)

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