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JEE Advanced 2024 Paper 2 · Mathematics · Vector Algebra · easy

Let vec a,vec b be unit vectors with angle (π)/(3) between them, and let vec c=2(vec a×vec b)+3vec a-vec b. The value of 4 [vec a vec b vec c]^2 is

Numerical answer

Let a⃗,b⃗\vec a,\vec b be unit vectors with angle π3\dfrac\pi3 between them, and let c⃗=2(a⃗×b⃗)+3a⃗−b⃗\vec c=2(\vec a\times\vec b)+3\vec a-\vec b. The value of 4 [a⃗ b⃗ c⃗]24\,[\vec a\ \vec b\ \vec c]^2 is
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Answer: 9

The scalar triple product [a⃗ b⃗ c⃗]=c⃗⋅(a⃗×b⃗)[\vec a\ \vec b\ \vec c]=\vec c\cdot(\vec a\times\vec b). The components 3a⃗3\vec a and −b⃗-\vec b of c⃗\vec c lie in the plane of a⃗,b⃗\vec a,\vec b and are perpendicular to a⃗×b⃗\vec a\times\vec b, so they contribute 00. Thus [a⃗ b⃗ c⃗]=2∣a⃗×b⃗∣2=2sin⁡2π3=2⋅34=32[\vec a\ \vec b\ \vec c]=2|\vec a\times\vec b|^2=2\sin^2\dfrac\pi3=2\cdot\dfrac34=\dfrac32. Then 4[a⃗ b⃗ c⃗]2=4⋅94=94[\vec a\ \vec b\ \vec c]^2=4\cdot\dfrac94=9. A direct numerical computation with a⃗=(1,0,0)\vec a=(1,0,0), b⃗=(12,32,0)\vec b=(\tfrac12,\tfrac{\sqrt3}2,0) gives the same value.
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