Question

The angle between vectors A=2i-j+2k and B= 6i-3j+6k is

• A. Zero
• B. \(30^\circ\)
• C. \(45^\circ\)
• D. \(60^\circ\)

Hint:

Always check for proportionality between the coefficients of \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\).
Since \(\frac{2}{6} = \frac{-1}{-3} = \frac{2}{6} = \frac{1}{3}\), the vectors are parallel, which immediately yields an angle of zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The direction of a vector is determined by its components.
If one vector is a positive scalar multiple of another vector, they are collinear and point in the exact same direction.

Step 2: Key Formula or Approach:
The angle \(\theta\) between two vectors \(\vec{A}\) and \(\vec{B}\) can be computed using:
\[ \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \] Alternatively, if \(\vec{B} = k\vec{A}\) where \(k \gt 0\), the vectors are parallel and collinear, indicating \(\theta = 0\).

Step 3: Detailed Explanation:
Given the vectors:
\[ \vec{A} = 2\hat{i} - \hat{j} + 2\hat{k} \] \[ \vec{B} = 6\hat{i} - 3\hat{j} + 6\hat{k} \] Factoring out a common term of 3 from \(\vec{B}\):
\[ \vec{B} = 3(2\hat{i} - \hat{j} + 2\hat{k}) \] This shows that:
\[ \vec{B} = 3\vec{A} \] Since \(\vec{B}\) is a positive scalar multiple of \(\vec{A}\) (\(k = 3 \gt 0\)), the two vectors point in the identical direction.
The angle \(\theta\) between two parallel and co-directional vectors is \(0^\circ\) (Zero).

Step 4: Final Answer:
The correct option is 1, which corresponds to Zero.