Question

Let ABC be a triangle such that $\vec{BC}$=$\vec a$,$\vec{CA}$=$\vec b$,$\vec AB$=$\vec c$,|$\vec a$|=6$\sqrt2$,|$\vec b$|=2$\sqrt3$ and $\vec b$⋅$\vec c$=12 Consider the statements:
(S1):|($\vec a$×$\vec b$)+($\vec c$×$\vec d$)|−|$\vec c$|=6(2$\sqrt2$−1)
(S2):$\angle$ACB=cos−1⁡($\sqrt{\frac{2}{3}}$)
Then

• A. Both (S1) and (S2) are true
• B. Only (S1) is true
• C. Only (S2) is true
• D. Both (S1) and (S2) are false

Answer 1

Answer: (C)

Only (S2) is true

Answer 2

∵ $\vec a$+$\vec b$+$\vec c$=0⋯(i)
then
$\vec a$+$\vec c$=−$\vec b$
then
($\vec a$+$\vec c$)×$b$=−$\vec b$×$\vec b$
∴ $\vec a$×$\vec b$+$\vec c$×$\vec b$=$\vec 0$⋯(ii)
For
(S1):|$\vec a$×$\vec b$+$\vec c$×$\vec b$|−|$\vec c$|=6(2$\sqrt2$−1)
|($\vec a$+$\vec c$)×$\vec b$|−|$\vec c$|=6(2$\sqrt2$−1)
|$\vec c$|=6−12$\sqrt2$ (not possible)
Hence (S 1) is not correct
For (S 2) : from (i)
$\vec b$+$\vec c$=−$\vec a$
⇒ $\vec b$⋅$\vec b$+$\vec c$⋅$\vec b$=−$\vec a$⋅$\vec b$
⇒ 12+12=−6$\sqrt2$⋅2$\sqrt3$cos⁡(π−$\angle$ACB)
∴ cos⁡($\angle$ACB)=$\sqrt{\frac{2}{3}}$
∴ ∠ACB=cos−1$\sqrt{\frac{2}{3}}$
∴ S(2) is correct.