Question

The angle made by a vector \( \vec{B} = 3\hat{i} + 2\hat{j} + 4\hat{k} \) with the y-axis is

• A. \( \cos^{-1} \left( \frac{5}{\sqrt{23}} \right) \)
• B. \( \cos^{-1} \left( \frac{4}{\sqrt{11}} \right) \)
• C. \( \cos^{-1} \left( \frac{3}{\sqrt{17}} \right) \)
• D. \( \cos^{-1} \left( \frac{2}{\sqrt{29}} \right) \)

Hint:

In vector problems, to find the angle with any axis, use the dot product formula and the magnitude of the vectors. The result will give the cosine of the angle, from which you can find the angle using the inverse cosine.

Answer 1

The Correct Option is D

Step 1: Vector Components.
The vector \( \vec{B} = 3\hat{i} + 2\hat{j} + 4\hat{k} \) represents a 3-dimensional vector with components along the \( \hat{i} \), \( \hat{j} \), and \( \hat{k} \) directions. The angle made by the vector with the y-axis is the angle between the vector and the y-axis unit vector \( \hat{j} \).
Step 2: Using the dot product formula.
The cosine of the angle \( \theta \) between the vector \( \vec{B} \) and the y-axis is given by: \[ \cos \theta = \frac{\vec{B} \cdot \hat{j}}{|\vec{B}| |\hat{j}|} \] Since \( \hat{j} \) is a unit vector, \( |\hat{j}| = 1 \). The dot product \( \vec{B} \cdot \hat{j} \) is simply the coefficient of \( \hat{j} \), which is 2. The magnitude of \( \vec{B} \) is: \[ |\vec{B}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{29} \] Thus, the cosine of the angle is: \[ \cos \theta = \frac{2}{\sqrt{29}} \] So, the angle is \( \cos^{-1} \left( \frac{2}{\sqrt{29}} \right) \).
Step 3: Conclusion.
The correct answer is \( \cos^{-1} \left( \frac{2}{\sqrt{29}} \right) \), which is option (D).