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 \[ |\vec a|= \]

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

If two vectors are collinear, one vector is always a scalar multiple of the other: \[ \vec a=\lambda \vec b. \] Then use the dot product condition to determine the scalar \(\lambda\).

Answer 1

The Correct Option is C

Step 1: Write down the given information.
$\vec{b} = 3\hat{i}+6\hat{j}+6\hat{k}$, $\vec{a}$ is collinear with $\vec{b}$, and $\vec{a} \cdot \vec{b} = 27$.
Step 2: Express $\vec{a}$ in terms of $\vec{b}$.
Since $\vec{a}$ is collinear with $\vec{b}$, we write $\vec{a} = \lambda\vec{b} = \lambda(3\hat{i}+6\hat{j}+6\hat{k})$ for some scalar $\lambda$.
Step 3: Compute $\vec{a} \cdot \vec{b}$.
$\vec{a} \cdot \vec{b} = \lambda(\vec{b} \cdot \vec{b}) = \lambda |\vec{b}|^2$.
Step 4: Find $|\vec{b}|^2$.
$|\vec{b}|^2 = 9 + 36 + 36 = 81$, so $|\vec{b}| = 9$.
Step 5: Solve for $\lambda$.
$\lambda \cdot 81 = 27$, so $\lambda = \frac{27}{81} = \frac{1}{3}$.
Step 6: Find $|\vec{a}|$.
$|\vec{a}| = |\lambda| \cdot |\vec{b}| = \frac{1}{3} \times 9 = 3$. This is option (3).
\[ \boxed{|\vec{a}| = 3} \]