The position vectors of vertices A, B and C of triangle ABC are respectively \hat{j}+\hat{k}, 3\hat{i}+\hat{j}+5\hat{k} and 3\hat{j}+3\hat{k}. What is angle C equal to ?
- A. \frac{\pi}{6}
- B. \frac{\pi}{4}
- C. \frac{\pi}{3}
- D. \frac{\pi}{2} ✓
Correct Answer: D. \frac{\pi}{2}
Explanation
We define the vectors radiating from C: \vec{CA} = A - C = (0, 1, 1) - (0, 3, 3) = (0, -2, -2). \vec{CB} = B - C = (3, 1, 5) - (0, 3, 3) = (3, -2, 2). Their dot product is \vec{CA} \cdot \vec{CB} = (0)(3) + (-2)(-2) + (-2)(2) = 0 + 4 - 4 = 0. Since their dot product is zero, \vec{CA} and \vec{CB} are perpendicular, meaning the angle C is \frac{\pi}{2}.
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