Question

If \( \vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} \), \( |\vec{b}| = 5 \) and the angle between \( \vec{a} \) and \( \vec{b} \) is \( \frac{\pi}{6} \), then the area of the triangle formed by these two vectors as two sides is:

• A. \( \frac{15}{4} \)
• B. \( \frac{15}{2} \)
• C. \( 15 \)
• D. \( \frac{15\sqrt{3}}{2} \)
• E. \( 15\sqrt{3} \)

Hint:

Remember that \( \frac{1}{2}|\vec{a} \times \vec{b}| \) is for a triangle, whereas \( |\vec{a} \times \vec{b}| \) gives the area of a parallelogram.

Answer 1

The Correct Option is A

Concept: The area of a triangle with two sides represented by vectors \( \vec{a} \) and \( \vec{b} \) is given by: \[ \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{1}{2} |\vec{a}| |\vec{b}| \sin \theta \] where \( \theta \) is the angle between the two vectors.

Step 1: Calculate the magnitude of vector \( \vec{a} \).
Given \( \vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} \): \[ |\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

Step 2: Substitute all values into the area formula.
We have \( |\vec{a}| = 3 \), \( |\vec{b}| = 5 \), and \( \theta = \frac{\pi}{6} \) (which is \( 30^\circ \)). \[ \text{Area} = \frac{1}{2} \times 3 \times 5 \times \sin\left(\frac{\pi}{6}\right) \] \[ \text{Area} = \frac{1}{2} \times 15 \times \frac{1}{2} \] \[ \text{Area} = \frac{15}{4} \]