Question

If \( \vec{a} \) and \( \vec{b}=3\hat{i}+6\hat{j}+6\hat{k} \) are collinear and \( \vec{a}\cdot\vec{b}=27 \), then \( \vec{a} \) is

• A. \( 3(\hat{i}+\hat{j}+\hat{k}) \)
• B. \( \hat{i}+2\hat{j}+2\hat{k} \)
• C. \( 2\hat{i}+2\hat{j}+2\hat{k} \)
• D. \( \hat{i}+3\hat{j}+3\hat{k} \)
• E. \( \hat{i}-3\hat{j}+2\hat{k} \)

Hint:

Collinear vectors are scalar multiples.

Answer 1

The Correct Option is C

Concept: Collinear ⇒ \( \vec{a} = \lambda \vec{b} \)

Step 1: Assume: \[ \vec{a} = \lambda(3,6,6) \]

Step 2: Compute dot product.
\[ \vec{a}\cdot\vec{b} = \lambda (3^2+6^2+6^2) \]

Step 3: Simplify.
\[ = \lambda (9+36+36) = 81\lambda \]

Step 4: Given: \[ 81\lambda = 27 ⇒ \lambda = \frac{1}{3} \]

Step 5: So: \[ \vec{a} = (1,2,2) \]