A line AB in three-dimensional space 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

$\text{Given: line AB makes } 45^\circ \text{ with } +x\text{-axis}$

$120^\circ \text{ with } +y\text{-axis}$

$\text{To find: acute } \theta \text{, where AB makes } \theta \text{ with } +z\text{-axis}$

$(1)\ 45^\circ \qquad (2)\ 60^\circ \qquad (3)\ 75^\circ \qquad (4)\ 30^\circ$

$1.\ l=\cos\alpha,\ m=\cos\beta,\ n=\cos\gamma$ $2.\ l^2+m^2+n^2 = 1$ $3.\ \text{find } n \Rightarrow \gamma,\ \text{then acute } \theta$ $\text{Direction angles: } \alpha,\ \beta,\ \gamma \text{ with } x,\ y,\ z \text{ axes}$ $l=\cos\alpha,\qquad m=\cos\beta,\qquad n=\cos\gamma$ $l^2 + m^2 + n^2 = 1$ $(l,\ m,\ n) \text{ is a unit vector}$ $l = \cos 45^\circ = \dfrac{1}{\sqrt{2}}$ $m = \cos 120^\circ = -\dfrac{1}{2}$ $\dfrac{1}{2} + \dfrac{1}{4} + n^2 = 1$ $n^2 = 1 - \dfrac{3}{4} = \dfrac{1}{4}$ $n = \cos\gamma = \pm\dfrac{1}{2}$ $\cos\gamma = \dfrac12 \Rightarrow \gamma = 60^\circ$ $\cos\gamma = -\dfrac12 \Rightarrow \gamma = 120^\circ$ $\vec{d} = \left(\dfrac{1}{\sqrt{2}},\ -\dfrac{1}{2},\ \dfrac{1}{2}\right),\qquad |\vec{d}| = 1$ $60^\circ < 90^\circ\ \text{(acute)},\qquad 120^\circ > 90^\circ\ \text{(obtuse)}$ $\theta = 60^\circ$ $\boxed{\text{Answer: (2)}\quad 60^\circ}$