Let ABC be a triangle such that $\overrightarrow{BC}=\vec{a}$, $\overrightarrow{CA}=\vec{b}$, $\overrightarrow{AB}=\vec{c}$, $|\vec{a}|=6\sqrt{2}$, $|\vec{b}|=2\sqrt{3}$ and $\vec{b} \cdot \vec{c}=12$. Consider the statements: (S1): $|(\vec{a} \times \vec{b})+(\vec{c} \times \vec{b})|-|\vec{c}|=6(2\sqrt{2}-1)$ (S2): $\angle ABC=\cos^{-1}\left(\sqrt{\frac{2}{3}}\right)$. Then

(D) both (S1) and (S2) are false

(A) both (S1) and (S2) are true

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Vector Algebra

Let ABC be a triangle such that

B C = a

C A = b

A B = c

∣ a ∣ = 62

∣ b ∣ = 23

and

b \cdot c = 12

.
Consider the statements:
(S1):

∣ (a \times b) + (c \times b) ∣ - ∣ c ∣ = 6 (22 - 1)

(S2):

∠ A B C = cos^{- 1} (\frac{2}{3})

.
Then

Answer: (D)

Since

a + b + c = 0

(triangle closure), we have

b + c = - a

. Taking magnitude squared:

∣ b ∣^{2} + ∣ c ∣^{2} + 2 b \cdot c = ∣ a ∣^{2}

, so

12 + ∣ c ∣^{2} + 24 = 72

, giving

∣ c ∣ = 6

. For S1:

(a \times b) + (c \times b) = (a + c) \times b = (- b) \times b = 0

, so the expression equals

0 - 6 = - 6 \neq = 6 (22 - 1)

. S1 is false. For S2: Using the cosine rule in triangle ABC at vertex C,

∣ c ∣^{2} = ∣ a ∣^{2} + ∣ b ∣^{2} - 2∣ a ∣∣ b ∣ cos (∠ A C B)

, we get

cos (∠ A C B) = 2/3

. But the statement claims this is

∠ A B C

, not

∠ A C B

. S2 is false. Both statements are false.

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Single correct option