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JEE Advanced 2016 Paper 1 · Mathematics · Vector Algebra · hard

Let vec a,vec b,vec c be unit vectors such that the angle between each pair is (π)/(3). If vec d=vec a×vec b+vec b×vec c+vec c×vec a, then the value…

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Let a⃗,b⃗,c⃗\vec a,\vec b,\vec c be unit vectors such that the angle between each pair is π3\dfrac{\pi}{3}. If d⃗=a⃗×b⃗+b⃗×c⃗+c⃗×a⃗\vec d=\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a, then the value of 4 ∣d⃗∣24\,|\vec d|^2 is
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Answer: 3

Use the Gram matrix: a⃗⋅a⃗=1\vec a\cdot\vec a=1 and every cross dot product is 12\tfrac12. For two cross products, (a⃗×b⃗)⋅(c⃗×d⃗)=(a⃗⋅c⃗)(b⃗⋅d⃗)−(a⃗⋅d⃗)(b⃗⋅c⃗)(\vec a\times\vec b)\cdot(\vec c\times\vec d)=(\vec a\cdot\vec c)(\vec b\cdot\vec d)-(\vec a\cdot\vec d)(\vec b\cdot\vec c). Squares: ∣a⃗×b⃗∣2=1−14=34|\vec a\times\vec b|^2=1-\tfrac14=\tfrac34, and likewise for the other two, giving 94\tfrac94 in total. Cross terms: (a⃗×b⃗)⋅(b⃗×c⃗)=(a⃗⋅b⃗)(b⃗⋅c⃗)−(a⃗⋅c⃗)(b⃗⋅b⃗)=14−12=−14(\vec a\times\vec b)\cdot(\vec b\times\vec c)=(\vec a\cdot\vec b)(\vec b\cdot\vec c)-(\vec a\cdot\vec c)(\vec b\cdot\vec b)=\tfrac14-\tfrac12=-\tfrac14, and each of the three pairs gives the same value. So ∣d⃗∣2=94+2(3⋅(−14))=94−32=34|\vec d|^2=\tfrac94+2\left(3\cdot\left(-\tfrac14\right)\right)=\tfrac94-\tfrac32=\tfrac34. Hence 4∣d⃗∣2=34|\vec d|^2=3. A numerical construction of the vectors with this Gram matrix reproduced 0.750.75.
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