If two sides of a triangle are represented by the vectors √3(^a×→b) and →b−(^a⋅→b)^a where →b is a non-zero vector and ^a is a unit vector in the direction of →a, then the angles of triangle are
A
tan−1(1√3);tan−1(12);tan−1(√3+21−2√3)
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B
tan−1(√3);tan−1(1√3);cot−1(0)
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C
tan−1(√3);tan−1(2);tan−1(√3+22√3−1)
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D
tan−1(1);tan−1(1);cot−1(0)
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Solution
The correct option is Btan−1(√3);tan−1(1√3);cot−1(0) Let →V=√3(^a×→b)
and →V2=→b−(^a⋅→b)^a=(^a×→b)×^a →V1⋅→V2=√3(^a×→b)⋅[(^a×→b)×^a] ⇒→V1⋅→V2=0
So, the angle between →V1 and →V2 is ∠A=90∘
Now, using sine law in ΔABC, ∣∣∣→b−(^a⋅→b)^a∣∣∣sinθ=√3∣∣∣^a×→b∣∣∣cosθ ⇒tanθ=1√3|→b−(^a⋅→b)^a||→a×→b| ⇒tanθ=1√3|(^a×→b)×^a||→a×→b| ⇒tanθ=1√3|^a×→b||→a|sin90∘|→a×→b| ⇒tanθ=1√3 ∴θ=π6
The angles of ΔABC are π3,π6,π2 or tan−1(√3);tan−1(1√3);cot−1(0)