Question

The angle between the vectors $ a = \hat{i} + 2 \hat{j} + 2 \hat{k} \quad \text{and} \quad b = \hat{i} + 2 \hat{j} - 2 \hat{k} \quad \text{is} $

• A. \( \sin^{-1} \left( \frac{1}{9} \right) \)
• B. \( \sin^{-1} \left( \frac{8}{9} \right) \)
• C. \( \cos^{-1} \left( \frac{8}{9} \right) \)
• D. \( \cos^{-1} \left( \frac{1}{9} \right) \)

Hint:

For the angle between two vectors, use the dot product formula and remember to calculate the magnitudes of both vectors.

Answer 1

The Correct Option is D

The formula for the angle \( \theta \) between two vectors \( a \) and \( b \) is given by the dot product formula: \[ \cos(\theta) = \frac{a \cdot b}{|a| |b|} \] Where: - \( a \cdot b \) is the dot product of the vectors. - \( |a| \) and \( |b| \) are the magnitudes of the vectors. The dot product \( a \cdot b \) is calculated as: \[ a \cdot b = (1)(1) + (2)(2) + (2)(-2) = 1 + 4 - 4 = 1 \] The magnitudes of the vectors are: \[ |a| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] \[ |b| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] Now, applying the formula: \[ \cos(\theta) = \frac{1}{3 \times 3} = \frac{1}{9} \] Thus, the angle \( \theta \) is: \[ \theta = \cos^{-1} \left( \frac{1}{9} \right) \]