Search any question & find its solution
Question:
Answered & Verified by Expert
If a line makes angles $\tan ^{-1} \sqrt{7}, \tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}}$ with $X$-axis, $Y$-axis respectively, then the angle made by it with $Z$-axis is
Options:
Solution:
2378 Upvotes
Verified Answer
The correct answer is:
$\frac{\pi}{4}$ or $\frac{3 \pi}{4}$
A line makes angle $\tan ^{-1} \sqrt{7}$ and $\tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}}$ with $X$-axis and $Y$-axis respectively.
So,
Sc
$$
\begin{aligned}
& \alpha=\tan ^{-1} \sqrt{7} \\
& \tan \alpha=\sqrt{7} \\
& \Rightarrow \quad \cos \alpha=\frac{1}{\sqrt{8}} \\
& \text { and } \\
& \beta=\tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}} \\
& \tan \beta=\frac{\sqrt{5}}{\sqrt{3}} \\
& \Rightarrow \quad \cos \beta=\frac{\sqrt{3}}{\sqrt{8}} \\
&
\end{aligned}
$$

Let angle make with $Z$-axis is $\gamma$. So,
$$
\begin{aligned}
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\
& \Rightarrow\left(\frac{1}{\sqrt{8}}\right)^2+\left(\frac{\sqrt{3}}{\sqrt{8}}\right)+\cos ^2 \gamma=1 \\
& \Rightarrow \quad \frac{4}{8}+\cos ^2 \gamma=1 \quad \Rightarrow \quad \frac{1}{2}+\cos ^2 \gamma=1
\end{aligned}
$$
$$
\begin{array}{lr}
\Rightarrow & \cos ^2 \gamma=\frac{1}{2} \\
\Rightarrow & \cos \gamma= \pm \frac{1}{\sqrt{2}} \\
\Rightarrow & \cos \gamma=\frac{1}{\sqrt{2}} \text { or } \frac{-1}{\sqrt{2}} \\
\Rightarrow & \gamma=\frac{\pi}{4} \text { or } \pi-\frac{\pi}{4} \\
\Rightarrow & \gamma=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}
\end{array}
$$
So, angle made by line with $Z$-axis is $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$.
So,
Sc
$$
\begin{aligned}
& \alpha=\tan ^{-1} \sqrt{7} \\
& \tan \alpha=\sqrt{7} \\
& \Rightarrow \quad \cos \alpha=\frac{1}{\sqrt{8}} \\
& \text { and } \\
& \beta=\tan ^{-1} \frac{\sqrt{5}}{\sqrt{3}} \\
& \tan \beta=\frac{\sqrt{5}}{\sqrt{3}} \\
& \Rightarrow \quad \cos \beta=\frac{\sqrt{3}}{\sqrt{8}} \\
&
\end{aligned}
$$

Let angle make with $Z$-axis is $\gamma$. So,
$$
\begin{aligned}
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\
& \Rightarrow\left(\frac{1}{\sqrt{8}}\right)^2+\left(\frac{\sqrt{3}}{\sqrt{8}}\right)+\cos ^2 \gamma=1 \\
& \Rightarrow \quad \frac{4}{8}+\cos ^2 \gamma=1 \quad \Rightarrow \quad \frac{1}{2}+\cos ^2 \gamma=1
\end{aligned}
$$
$$
\begin{array}{lr}
\Rightarrow & \cos ^2 \gamma=\frac{1}{2} \\
\Rightarrow & \cos \gamma= \pm \frac{1}{\sqrt{2}} \\
\Rightarrow & \cos \gamma=\frac{1}{\sqrt{2}} \text { or } \frac{-1}{\sqrt{2}} \\
\Rightarrow & \gamma=\frac{\pi}{4} \text { or } \pi-\frac{\pi}{4} \\
\Rightarrow & \gamma=\frac{\pi}{4} \text { or } \frac{3 \pi}{4}
\end{array}
$$
So, angle made by line with $Z$-axis is $\frac{\pi}{4}$ and $\frac{3 \pi}{4}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.