Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $\alpha ,\beta ,\gamma$ be the angles made by the directed line with the X, Y, and Z axes respectively.

Given $\alpha =45^{\circ }$ and $\beta =60^{\circ }$.

The direction cosines satisfy the relation ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$.

Substituting the given values, we get ${\cos }^2(45^{\circ })+{\cos }^2(60^{\circ })+{\cos }^2\theta =1$.

${\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\cos }^2\theta =1$

$\frac{1}{2}+\frac{1}{4}+{\cos }^2\theta =1$

$\frac{3}{4}+{\cos }^2\theta =1$

${\cos }^2\theta =\frac{1}{4}$

$\cos \theta =\pm \frac{1}{2}$

Since $\theta$ is an obtuse angle, $\cos \theta <0$.

$\cos \theta =-\frac{1}{2}$

$\theta =120^{\circ }$

Answer: $120^{\circ }$