When dealing with direction cosines, remember that the sum of the squares of the direction cosines is always equal to 1. This property can help you solve for unknown angles when given the other angles.
The correct answer is: (C): \(\frac{\pi}{4}\)
We are given that the line makes an angle of \( \frac{\pi}{3} \) with both the X-axis and the Y-axis, and we are asked to find the acute angle it makes with the Z-axis.
Step 1: Understand the direction cosines
The direction cosines of a line are the cosines of the angles the line makes with the coordinate axes. If the line makes angles \( \alpha \), \( \beta \), and \( \gamma \) with the X, Y, and Z axes respectively, then the direction cosines are given by:
\( \cos \alpha, \cos \beta, \cos \gamma \)
Step 2: Use the known angles
We are told that the line makes an angle of \( \frac{\pi}{3} \) with both the X-axis and the Y-axis. Therefore, the direction cosines for the X and Y axes are:
\( \cos \alpha = \cos \beta = \cos \frac{\pi}{3} = \frac{1}{2} \)
Step 3: Use the property of direction cosines
The sum of the squares of the direction cosines of a line is always equal to 1. Therefore, we have:
\( \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \)
Substitute \( \cos \alpha = \cos \beta = \frac{1}{2} \) into the equation:
\( \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 + \cos^2 \gamma = 1 \)
Simplifying:
\( \frac{1}{4} + \frac{1}{4} + \cos^2 \gamma = 1 \)
\( \frac{2}{4} + \cos^2 \gamma = 1 \)
\( \cos^2 \gamma = \frac{2}{4} = \frac{1}{2} \)
Step 4: Solve for \( \gamma \)
Now, take the square root of both sides:
\( \cos \gamma = \frac{1}{\sqrt{2}} \)
Therefore, the acute angle \( \gamma \) is:
\( \gamma = \frac{\pi}{4} \)
Conclusion:
The acute angle made by the Z-axis is \( \frac{\pi}{4} \), so the correct answer is (C): \(\frac{\pi}{4}\).
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}
Match the following:
In the following, \( [x] \) denotes the greatest integer less than or equal to \( x \). 
Choose the correct answer from the options given below:
For x < 0:
f(x) = ex + ax
For x ≥ 0:
f(x) = b(x - 1)2