Let ABC be a triangle with BC= a, CA= b, AB= c, | a|=6√2, | b|=2√3, b c=12.

Question

Let $ABC$ be a triangle with $\vec{BC}=\vec a$, $\vec{CA}=\vec b$, $\vec{AB}=\vec c$, $|\vec a|=6\sqrt2$, $|\vec b|=2\sqrt3$, $\vec b\cdot\vec c=12$. Consider:
(S1): $|\vec a	imes\vec b+\vec c	imes\vec b-\vec c| = 6(2\sqrt2-1)$
(S2): $\angle ACB = \cos^{-1}\left(\dfrac23ight)$

Accepted Answer

Since $\vec a+\vec b+\vec c=\vec 0$ (triangle law), $\vec a+\vec c = -\vec b$, and taking cross product with $\vec b$: $\vec a\times\vec b+\vec c\times\vec b = \vec 0$ \ldots  (i)

For (S1): using (i), the claimed quantity reduces to a value requiring $|\vec c|=6-12\sqrt2 < 0$, which is impossible --- so (S1) is false.

For (S2): from $\vec b+\vec c=-\vec a$: $|\vec b|^2+\vec c\cdot\vec b = -\vec a\cdot\vec b$, i.e. $12+12 = -(6\sqrt2)(2\sqrt3)\cos(\pi-\angle ACB)$, giving
$$\cos(\angle ACB) = \frac23 \implies \angle ACB=\cos^{-1}\left(\frac23\right)$$

So (S2) is true. Only (S2) holds.

*(Note: the source's block header shows "Answer: D", but its own worked solution explicitly concludes "(S2) is correct" --- i.e. option C. The derivation is retained as shown; option C is used here since it matches the shown work.)*

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