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Vector Algebra question

2022 · 25 Jul · Shift 1 · Q34
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  5. /2022 · 25 Jul · Shift 1 · Q34

Vector Algebra question

2022 · 25 Jul · Shift 1 · Q34

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let ABC\mathrm{ABC}ABC be a triangle such that BC→=a→,CA→=b→,AB→=c→,∣a→∣=62,∣b→∣=23\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{CA}}=\overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{c}},|\overrightarrow{\mathrm{a}}|=6 \sqrt{2},|\overrightarrow{\mathrm{b}}|=2 \sqrt{3}BC=a,CA=b,AB=c,∣a∣=62​,∣b∣=23​ and b⃗⋅c⃗=12\vec{b} \cdot \vec{c}=12b⋅c=12. Consider the statements : (S1):∣(a→×b→)+(c→×b→)∣−∣c⃗∣=6(22−1)(S2):∠ACB=cos⁡−1(23)(\mathrm{S} 1):|(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})|-|\vec{c}|=6(2 \sqrt{2}-1)(\mathrm{S} 2): \angle \mathrm{ACB}=\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)(S1):∣(a×b)+(c×b)∣−∣c∣=6(22​−1)(S2):∠ACB=cos−1(32​​) Then
  1. A
    both (S1) and (S2) are true
  2. B
    only (S1) is true
  3. C
    only (S2) is true
  4. D
    both (S1) and (S2) are false
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