Question

If \( \hat{i} + \hat{j}, \, \hat{j} + \hat{k}, \, \hat{i} + \hat{k} \) are the position vectors of the vertices of a triangle ABC taken in order, then \( \angle A \) is equal to:

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{5} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{\pi}{4} \)
• E. \( \frac{\pi}{3} \)

Hint:

Note that \( \vec{BC} = \vec{OC} - \vec{OB} = \hat{i} - \hat{j} \). Its magnitude is also \( \sqrt{2} \). Since all sides are equal, it's an equilateral triangle, so all angles are \( \pi/3 \).

Answer 1

Answer: (E)

Concept: The angle \( A \) is the angle between the vectors \( \vec{AB} \) and \( \vec{AC} \). If position vectors are \( \vec{OA}, \vec{OB}, \vec{OC} \), then: \( \vec{AB} = \vec{OB} - \vec{OA} \) and \( \vec{AC} = \vec{OC} - \vec{OA} \).

Step 1: Calculate the vectors forming angle A.
\( \vec{OA} = \hat{i} + \hat{j} \), \( \vec{OB} = \hat{j} + \hat{k} \), \( \vec{OC} = \hat{i} + \hat{k} \). \[ \vec{AB} = (\hat{j} + \hat{k}) - (\hat{i} + \hat{j}) = -\hat{i} + \hat{k} \] \[ \vec{AC} = (\hat{i} + \hat{k}) - (\hat{i} + \hat{j}) = -\hat{j} + \hat{k} \]

Step 2: Find magnitudes and dot product.
\[ |\vec{AB}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \] \[ |\vec{AC}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \] \[ \vec{AB} \cdot \vec{AC} = (-1)(0) + (0)(-1) + (1)(1) = 1 \]

Step 3: Calculate the angle.
\[ \cos A = \frac{\vec{AB} \cdot \vec{AC}}{|\vec{AB}| |\vec{AC}|} = \frac{1}{\sqrt{2} \sqrt{2}} = \frac{1}{2} \] Since \( \cos A = 1/2 \), \( A = 60^\circ = \frac{\pi}{3} \).