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): \(\frac{\pi}{4}\).
List - I | List - II | ||
(P) | γ equals | (1) | \(-\hat{i}-\hat{j}+\hat{k}\) |
(Q) | A possible choice for \(\hat{n}\) is | (2) | \(\sqrt{\frac{3}{2}}\) |
(R) | \(\overrightarrow{OR_1}\) equals | (3) | 1 |
(S) | A possible value of \(\overrightarrow{OR_1}.\hat{n}\) is | (4) | \(\frac{1}{\sqrt6}\hat{i}-\frac{2}{\sqrt6}\hat{j}+\frac{1}{\sqrt6}\hat{k}\) |
(5) | \(\sqrt{\frac{2}{3}}\) |
A block of certain mass is placed on a rough floor. The coefficients of static and kinetic friction between the block and the floor are 0.4 and 0.25 respectively. A constant horizontal force \( F = 20 \, \text{N} \) acts on it so that the velocity of the block varies with time according to the following graph. The mass of the block is nearly (Take \( g = 10 \, \text{m/s}^2 \)):
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is
The circuit shown in the figure contains two ideal diodes \( D_1 \) and \( D_2 \). If a cell of emf 3V and negligible internal resistance is connected as shown, then the current through \( 70 \, \Omega \) resistance (in amperes) is: