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JEE Main 2024 · Mathematics · Vector Algebra · hard

Let vec a=hat i+hat j+hat k and let vec b be a vector with vec a×vec b=hat j-hat k and vec a·vec b=3. Then vec b·(hat i-hat j) equals

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Let a⃗=i^+j^+k^\vec a=\hat i+\hat j+\hat k and let b⃗\vec b be a vector with a⃗×b⃗=j^−k^\vec a\times\vec b=\hat j-\hat k and a⃗⋅b⃗=3\vec a\cdot\vec b=3. Then b⃗⋅(i^−j^)\vec b\cdot(\hat i-\hat j) equals
  1. A. 11
  2. B. 33
  3. C. 13\dfrac13
  4. D. 22
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Answer: (A) 11

Write c⃗=a⃗×b⃗=j^−k^\vec c=\vec a\times\vec b=\hat j-\hat k. Take the cross product with a⃗\vec a: a⃗×(a⃗×b⃗)=(a⃗⋅b⃗)a⃗−∣a⃗∣2b⃗\vec a\times(\vec a\times\vec b)=(\vec a\cdot\vec b)\vec a-|\vec a|^2\vec b, so a⃗×c⃗=3a⃗−3b⃗\vec a\times\vec c=3\vec a-3\vec b and b⃗=a⃗−13(a⃗×c⃗)\vec b=\vec a-\dfrac13(\vec a\times\vec c). Now a⃗×c⃗=(1⋅(−1)−1⋅1,  1⋅0−1⋅(−1),  1⋅1−0)=(−2,1,1)\vec a\times\vec c=(1\cdot(-1)-1\cdot1,\;1\cdot0-1\cdot(-1),\;1\cdot1-0)=(-2,1,1). Hence b⃗=(53,23,23)\vec b=\left(\dfrac53,\dfrac23,\dfrac23\right), which indeed gives a⃗⋅b⃗=3\vec a\cdot\vec b=3. Therefore b⃗⋅(i^−j^)=53−23=1\vec b\cdot(\hat i-\hat j)=\dfrac53-\dfrac23=1. The data are consistent because a⃗⋅(a⃗×b⃗)=a⃗⋅(j^−k^)=0\vec a\cdot(\vec a\times\vec b)=\vec a\cdot(\hat j-\hat k)=0, as required for any cross product.
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