Question

The angle between the vectors \[ \vec{a} = 2\hat{i} - 3\hat{j} + 2\hat{k} \quad \text{and} \quad \vec{b} = \hat{i} + 4\hat{j} + 5\hat{k} \] is:

• A. \( 30^\circ \)
• B. \( 90^\circ \)
• C. \( 45^\circ \)
• D. \( 60^\circ \)

Hint:

The angle between two vectors can be found using the dot product formula: \( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \). If the dot product is zero, the vectors are perpendicular (i.e., \( 90^\circ \)).

Answer 1

The Correct Option is D

The angle \( \theta \) between two vectors \( \vec{a} \) and \( \vec{b} \) is given by the formula: \[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \] First, calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (2)(1) + (-3)(4) + (2)(5) \] \[ = 2 - 12 + 10 = 0 \] Next, calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \): \[ |\vec{a}| = \sqrt{(2)^2 + (-3)^2 + (2)^2} = \sqrt{4 + 9 + 4} = \sqrt{17} \] \[ |\vec{b}| = \sqrt{(1)^2 + (4)^2 + (5)^2} = \sqrt{1 + 16 + 25} = \sqrt{42} \] Now, use the formula for the cosine of the angle: \[ \cos \theta = \frac{0}{\sqrt{17} \times \sqrt{42}} = 0 \] Since \( \cos \theta = 0 \), the angle \( \theta \) is: \[ \theta = 90^\circ \] Thus, the correct answer is: \[ \boxed{90^\circ} \]