Let $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}+\hat{k}$ be two vectors. Consider a vector $\vec{c}=\alpha\vec{a}+\beta\vec{b},\alpha,\beta\in R$. If the projection of $\vec{c}$ on the vector $(\vec{a}+\vec{b})$ is $3\sqrt{2}$, then the minimum value of $\left(\vec{c}-(\vec{a}\times\vec{b})\right)\cdot\vec{c}$ equals _____.

$\text{Given: } \vec{a}=2\hat{i}+\hat{j}-\hat{k},\qquad \vec{b}=\hat{i}+2\hat{j}+\hat{k}$

$\vec{c}=\alpha\vec{a}+\beta\vec{b},\qquad \alpha,\beta\in\mathbb{R}$

$\text{To find: } \min\left[\left(\vec{c}-(\vec{a}\times\vec{b})\right)\cdot\vec{c}\right]$

$\text{Given: projection of } \vec{c} \text{ on } (\vec{a}+\vec{b}) = 3\sqrt{2}$

$\vec{a}+\vec{b} = (2+1)\hat{i}+(1+2)\hat{j}+(-1+1)\hat{k} = 3\hat{i}+3\hat{j}$

$|\vec{a}+\vec{b}| = \sqrt{3^2+3^2} = \sqrt{18} = 3\sqrt{2}$

$\dfrac{\vec{c}\cdot(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|} = 3\sqrt{2}$

$\Rightarrow \vec{c}\cdot(\vec{a}+\vec{b}) = 3\sqrt{2}\times 3\sqrt{2} = 18$

$\vec{a}\cdot\vec{a} = 4+1+1 = 6,\qquad \vec{b}\cdot\vec{b} = 1+4+1 = 6$

$\vec{a}\cdot\vec{b} = 2+2-1 = 3$

$\vec{c}\cdot(\vec{a}+\vec{b}) = \alpha(\vec{a}\cdot\vec{a}+\vec{a}\cdot\vec{b})+\beta(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b})$

$= \alpha(6+3)+\beta(3+6) = 9(\alpha+\beta)$

$9(\alpha+\beta)=18 \;\Rightarrow\; \alpha+\beta=2$

a⃗×b⃗ = î ĵ k̂ 2 1 -1 1 2 1

$= \hat{i}(1\cdot1-(-1)\cdot2) - \hat{j}(2\cdot1-(-1)\cdot1) + \hat{k}(2\cdot2-1\cdot1)$

$\vec{a}\times\vec{b} = 3\hat{i}-3\hat{j}+3\hat{k}$

$\vec{a}\cdot(\vec{a}\times\vec{b}) = 0,\qquad \vec{b}\cdot(\vec{a}\times\vec{b}) = 0$

$\vec{c}\cdot(\vec{a}\times\vec{b}) = \alpha(0)+\beta(0) = 0$

$\left(\vec{c}-(\vec{a}\times\vec{b})\right)\cdot\vec{c} = |\vec{c}|^2 - (\vec{a}\times\vec{b})\cdot\vec{c}$

$= |\vec{c}|^2 - 0 = |\vec{c}|^2$

$|\vec{c}|^2 = \alpha^2|\vec{a}|^2 + 2\alpha\beta(\vec{a}\cdot\vec{b}) + \beta^2|\vec{b}|^2$

$= 6\alpha^2 + 6\alpha\beta + 6\beta^2 = 6(\alpha^2+\alpha\beta+\beta^2)$

$\beta = 2-\alpha \;\Rightarrow\; |\vec{c}|^2 = 6\left[\alpha^2+\alpha(2-\alpha)+(2-\alpha)^2\right]$

$= 6(\alpha^2-2\alpha+4)$

$\alpha^2-2\alpha+4 = (\alpha-1)^2+3$

$|\vec{c}|^2 = 6\left[(\alpha-1)^2+3\right]$

$\text{Minimum when } (\alpha-1)^2 = 0 \;\Rightarrow\; \alpha = 1,\ \ \beta = 1$

$\vec{c} = \vec{a}+\vec{b} = (3,\ 3,\ 0)$

$\left(\vec{c}-(\vec{a}\times\vec{b})\right)\cdot\vec{c} \Big|_{\min} = 6(0+3) = 18$