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JEE Advanced 2023 Paper 1 · Mathematics · Vector Algebra · medium

Let vec a=hat i+2hat j+2hat k, vec b=2hat i+hat j-2hat k and vec c=hat i+hat j+hat k. A vector vec r satisfies vec r×vec a=vec b and vec r·vec c=3.…

Numerical answer

Let a⃗=i^+2j^+2k^\vec a=\hat i+2\hat j+2\hat k, b⃗=2i^+j^−2k^\vec b=2\hat i+\hat j-2\hat k and c⃗=i^+j^+k^\vec c=\hat i+\hat j+\hat k. A vector r⃗\vec r satisfies r⃗×a⃗=b⃗\vec r\times\vec a=\vec b and r⃗⋅c⃗=3\vec r\cdot\vec c=3. Find ∣r⃗∣2|\vec r|^{2}.
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Answer: 5

Since a⃗⋅b⃗=2+2−4=0\vec a\cdot\vec b=2+2-4=0, a solution exists. Take a⃗×\vec a\times both sides: a⃗×(r⃗×a⃗)=r⃗∣a⃗∣2−a⃗(a⃗⋅r⃗)=a⃗×b⃗\vec a\times(\vec r\times\vec a)=\vec r|\vec a|^2-\vec a(\vec a\cdot\vec r)=\vec a\times\vec b. With ∣a⃗∣2=9|\vec a|^2=9 and a⃗×b⃗=(−6,6,−3)\vec a\times\vec b=(-6,6,-3),
r⃗=(−6,6,−3)9+λa⃗=(−23,23,−13)+λ(1,2,2).\vec r=\frac{(-6,6,-3)}{9}+\lambda\vec a=\left(-\tfrac23,\tfrac23,-\tfrac13\right)+\lambda(1,2,2).
Then r⃗⋅c⃗=−13+5λ=3\vec r\cdot\vec c=-\frac13+5\lambda=3, so λ=23\lambda=\frac23 and r⃗=(0,2,1)\vec r=(0,2,1).
Check: (0,2,1)×(1,2,2)=(2,1,−2)=b⃗(0,2,1)\times(1,2,2)=(2,1,-2)=\vec b and r⃗⋅c⃗=3\vec r\cdot\vec c=3.
Therefore ∣r⃗∣2=0+4+1=5|\vec r|^2=0+4+1=5. The condition a⃗⋅b⃗=0\vec a\cdot\vec b=0 is necessary for r⃗×a⃗=b⃗\vec r\times\vec a=\vec b to have any solution, because a cross product is perpendicular to both of its factors.
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