Question

If $ \vec{a} $ is collinear with $ \vec{b} = 3\hat{i} + 6\hat{j} + 6\hat{k} $ and $ \vec{a} \cdot \vec{b} = 27 $, then find $ |\vec{a}| $.

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When vectors are collinear, one is a scalar multiple of the other. Use this in dot product identities to find unknowns.

Answer 1

The Correct Option is C

Since \( \vec{a} \) is collinear with \( \vec{b} \), we can write: \[ \vec{a} = \lambda \vec{b} \] Then, \[ \vec{a} \cdot \vec{b} = (\lambda \vec{b}) \cdot \vec{b} = \lambda |\vec{b}|^2 \] Given: \[ \vec{a} \cdot \vec{b} = 27,\quad \vec{b} = 3\hat{i} + 6\hat{j} + 6\hat{k} \Rightarrow |\vec{b}|^2 = 3^2 + 6^2 + 6^2 = 9 + 36 + 36 = 81 \] So, \[ \lambda \cdot 81 = 27 \Rightarrow \lambda = \frac{1}{3} \Rightarrow \vec{a} = \frac{1}{3} \vec{b} \Rightarrow |\vec{a}| = \frac{1}{3} |\vec{b}| = \frac{1}{3} \cdot \sqrt{81} = \frac{1}{3} \cdot 9 = \boxed{3} \]