Discussion
Let \(\vec{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}\) and \(\vec{AC} = -\hat{i} + 2\hat{j} + 2\hat{k}\) .If θ is the angle between \(\vec{AB}\) and \(\vec{AC}\) then sinθ=
Answer: D
\(\sin\theta = \frac{\sqrt{17}}{9}\).
Find the cosine from the dot product:
\[ \begin{aligned} \vec{AB}\cdot\vec{AC} &= -2 + 20 + 22 = 40 \\ |\vec{AB}| &= \sqrt{4 + 100 + 121} = 15, \ |\vec{AC}| = 3 \\ \cos\theta &= \frac{40}{45} = \frac{8}{9} \end{aligned} \]
Then \(\sin\theta = \sqrt{1 - \frac{64}{81}} = \frac{\sqrt{17}}{9}\). The other square roots come from arithmetic slips in the magnitudes or the dot product, and \(\frac{4}{9}\) is half the cosine.
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