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

The Correct Option is C

∵ \(\vec a\)+\(\vec b\)+\(\vec c\)=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 (S1) is not correct
For (S2) : 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.