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: Use the condition of collinearity.
Since \(\vec a\) is collinear with \(\vec b\), \[ \vec a=\lambda \vec b \] where \(\lambda\) is a scalar.
Given, \[ \vec b=3\hat i+6\hat j+6\hat k \] Therefore, \[ \vec a=\lambda(3\hat i+6\hat j+6\hat k) \]

Step 2: Compute \(\vec b\cdot \vec b\).
\[ \vec b\cdot \vec b = 3^2+6^2+6^2 \] \[ =9+36+36 \] \[ =81 \] Thus, \[ |\vec b|=\sqrt{81}=9 \]

Step 3: Use the dot product condition.
Given, \[ \vec a\cdot \vec b=27 \] Substitute \[ \vec a=\lambda \vec b \] \[ (\lambda \vec b)\cdot \vec b=27 \] \[ \lambda(\vec b\cdot \vec b)=27 \] \[ 81\lambda=27 \] \[ \lambda=\frac13 \]

Step 4: Find \(|\vec a|\).
Since \[ \vec a=\lambda \vec b, \] we get \[ |\vec a|=|\lambda||\vec b| \] \[ =\frac13 \times 9 \] \[ =3 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{3} \]