Question

If a line makes an angle of \(\frac{\pi}{3}\) with each of x and y-axis, then the acute angle made by z-axis is

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{2}\)

Answer 1

The Correct Option is A

If a line makes an angle of \(\frac{\pi}{3}\) with each of the x and y-axis, then we need to find the acute angle made by the line with the z-axis.

Let \(\alpha, \beta, \gamma\) be the angles the line makes with the x, y, and z-axis respectively. Then \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).

We are given that \(\alpha = \frac{\pi}{3}\) and \(\beta = \frac{\pi}{3}\). Therefore:

\(\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{3}) + \cos^2 \gamma = 1\)

\((\frac{1}{2})^2 + (\frac{1}{2})^2 + \cos^2 \gamma = 1\)

\(\frac{1}{4} + \frac{1}{4} + \cos^2 \gamma = 1\)

\(\frac{1}{2} + \cos^2 \gamma = 1\)

\(\cos^2 \gamma = 1 - \frac{1}{2} = \frac{1}{2}\)

\(\cos \gamma = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}\)

Since we want the acute angle, we take the positive value:

\(\cos \gamma = \frac{1}{\sqrt{2}}\)

\(\gamma = \cos^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}\)

Therefore, the acute angle made by the line with the z-axis is \(\frac{\pi}{4}\).

Thus, the correct option is (A) \(\frac{\pi}{4}\).

Answer 2

Let the direction cosines be $ l, m, n $. Then $ l = \cos \alpha $, $ m = \cos \beta $, $ n = \cos \gamma $. 

Given $ \alpha = \frac{\pi}{3} $, $ \beta = \frac{\pi}{3} $, so:

$$ l = \cos \frac{\pi}{3} = \frac{1}{2}, \quad m = \cos \frac{\pi}{3} = \frac{1}{2}. $$

We know that $ l^2 + m^2 + n^2 = 1 $. Substituting $ l $ and $ m $:

$$ \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \implies \frac{1}{4} + \frac{1}{4} + n^2 = 1 \implies \frac{1}{2} + n^2 = 1 \implies n^2 = \frac{1}{2}. $$ $$ n = \pm \frac{1}{\sqrt{2}}. $$

If $ \cos \gamma = \frac{1}{\sqrt{2}} $, then $ \gamma = \frac{\pi}{4} $. If $ \cos \gamma = -\frac{1}{\sqrt{2}} $, 

then $ \gamma = \frac{3\pi}{4} $. 

Since we need the acute angle, $ \gamma = \frac{\pi}{4} $.