Question

The angle between the two vectors \( \hat{i} + \hat{j} + \hat{k} \) and \( 2\hat{i} - 2\hat{j} + 2\hat{k} \) is equal to

• A. \( \cos^{-1}\left(\frac{2}{3}\right) \)
• B. \( \cos^{-1}\left(\frac{1}{6}\right) \)
• C. \( \cos^{-1}\left(\frac{5}{6}\right) \)
• D. \( \cos^{-1}\left(\frac{1}{18}\right) \)
• E. \( \cos^{-1}\left(\frac{1}{3}\right) \)

Hint:

Always simplify magnitudes before substituting to avoid mistakes.

Answer 1

Answer: (E)

Concept:
• Angle between vectors: \[ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|} \]

Step 1: Write vectors.
\[ \vec{a} = (1,1,1), \quad \vec{b} = (2,-2,2) \]

Step 2: Find dot product.
\[ \vec{a}\cdot\vec{b} = 1\cdot2 + 1\cdot(-2) + 1\cdot2 \] \[ = 2 - 2 + 2 = 2 \]

Step 3: Find magnitudes.
\[ |\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt{3} \] \[ |\vec{b}| = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3} \]

Step 4: Compute cosine.
\[ \cos\theta = \frac{2}{\sqrt{3}\cdot2\sqrt{3}} = \frac{2}{6} = \frac{1}{3} \]

Step 5: Final Answer.
\[ \boxed{\cos^{-1}\left(\frac{1}{3}\right)} \]