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JEE Advanced 2019 · Paper 2, Mathematics question 14

Let a⃗=2i^+j^−k^\vec a = 2\hat i + \hat j - \hat k and b⃗=i^+2j^+k^\vec b = \hat i + 2\hat j + \hat k be two vectors. Consider a vector c⃗=αa⃗+βb⃗\vec c = \alpha\vec a + \beta\vec b, α,β∈R\alpha, \beta\in\mathbb R. If the projection of c⃗\vec c on the vector (a⃗+b⃗)(\vec a + \vec b) is 323\sqrt2, then the minimum value of (c⃗−(a⃗×b⃗))⋅c⃗(\vec c - (\vec a\times\vec b))\cdot\vec c equals ______

Answer and worked solution

Official answer: 18

Dot products. a⃗⋅a⃗=b⃗⋅b⃗=6\vec a\cdot\vec a=\vec b\cdot\vec b=6, a⃗⋅b⃗=3\vec a\cdot\vec b=3, and a⃗+b⃗=(3,3,0)\vec a+\vec b=(3,3,0) with ∣a⃗+b⃗∣=32|\vec a+\vec b|=3\sqrt2.

The projection condition.
c⃗⋅(a⃗+b⃗)32=32  ⟹  9(α+β)=18  ⟹  α+β=2.\frac{\vec c\cdot(\vec a+\vec b)}{3\sqrt2}=3\sqrt2\implies9(\alpha+\beta)=18\implies\alpha+\beta=2.

Simplify the expression. a⃗×b⃗\vec a\times\vec b is perpendicular to c⃗\vec c, so
(c⃗−a⃗×b⃗)⋅c⃗=∣c⃗∣2=6(α2+αβ+β2)=6(4−αβ).(\vec c-\vec a\times\vec b)\cdot\vec c=|\vec c|^2=6(\alpha^2+\alpha\beta+\beta^2)=6(4-\alpha\beta).

Minimise. αβ≤1\alpha\beta\le1 when α+β=2\alpha+\beta=2, so the minimum is 6×3=186\times3=18.

Answer: 18

Topics:Dot productCross productMaxima and minima
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