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A line $A B$ in three-dimensional space makes angles $45^{\circ}$ and $120^{\circ}$ with the positive $x$-axis and the positive $y$-axis respectively. If $A B$ makes an acute angle $\theta$ with the positive $z$-axis, then $\theta$ equals
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Verified Answer
The correct answer is:
$60^{\circ}$
$60^{\circ}$
$$
\begin{aligned}
& \ell=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\
& m=\cos 120^{\circ}=-\frac{1}{2} \\
& n=\cos \theta
\end{aligned}
$$
where $\theta$ is the angle which line makes with positive z-axis.
Now $\ell^2+m^2+n^2=1$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2}+\frac{1}{4}+\cos ^2 \theta=1 \\
& \cos ^2 \theta=\frac{1}{4} \\
& \Rightarrow \cos \theta=\frac{1}{2} ( $\theta$ Being acute)\\
& \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
\begin{aligned}
& \ell=\cos 45^{\circ}=\frac{1}{\sqrt{2}} \\
& m=\cos 120^{\circ}=-\frac{1}{2} \\
& n=\cos \theta
\end{aligned}
$$
where $\theta$ is the angle which line makes with positive z-axis.
Now $\ell^2+m^2+n^2=1$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2}+\frac{1}{4}+\cos ^2 \theta=1 \\
& \cos ^2 \theta=\frac{1}{4} \\
& \Rightarrow \cos \theta=\frac{1}{2} ( $\theta$ Being acute)\\
& \Rightarrow \theta=\frac{\pi}{3}
\end{aligned}
$$
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