Question

The angle between the vectors \[ \mathbf{A} = 2\hat{i} + \hat{j} + 3\hat{k} \quad \text{and} \quad \mathbf{B} = 3\hat{i} - 2\hat{j} + \hat{k} \] will be

• A. \( 90^\circ \)
• B. \( 60^\circ \)
• C. \( 30^\circ \)
• D. \( \cos^{-1} \left( \frac{1}{14} \right) \)

Hint:

To find the angle between two vectors, use the formula \( \cos \theta = \frac{\mathbf{A} . \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \), and make sure to calculate both the dot product and the magnitudes correctly.

Answer 1

The Correct Option is B

Step 1: Formula for the Angle Between Vectors:
The angle \( \theta \) between two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is given by the formula: \[ \cos \theta = \frac{\mathbf{A} . \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} \] where: - \( \mathbf{A} . \mathbf{B} \) is the dot product of the vectors, - \( |\mathbf{A}| \) and \( |\mathbf{B}| \) are the magnitudes of the vectors.
Step 2: Find the Dot Product \( \mathbf{A} . \mathbf{B} \):
The dot product is calculated as: \[ \mathbf{A} . \mathbf{B} = (2 \times 3) + (1 \times -2) + (3 \times 1) = 6 - 2 + 3 = 7 \]
Step 3: Find the Magnitudes of the Vectors:
The magnitude of \( \mathbf{A} \) is: \[ |\mathbf{A}| = \sqrt{(2^2 + 1^2 + 3^2)} = \sqrt{4 + 1 + 9} = \sqrt{14} \] The magnitude of \( \mathbf{B} \) is: \[ |\mathbf{B}| = \sqrt{(3^2 + (-2)^2 + 1^2)} = \sqrt{9 + 4 + 1} = \sqrt{14} \]
Step 4: Calculate \( \cos \theta \):
Now, we substitute the values into the formula: \[ \cos \theta = \frac{7}{\sqrt{14} \times \sqrt{14}} = \frac{7}{14} = \frac{1}{2} \]
Step 5: Find the Angle \( \theta \):
Thus, the angle \( \theta \) is: \[ \theta = \cos^{-1} \left( \frac{1}{2} \right) = 60^\circ \]
Step 6: Final Answer:
The angle between the vectors is \( 60^\circ \).