Question

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

Answer 1

$\vec{c}=\left(2\alpha+\beta\right)\hat{i}+\hat{j}\left(\alpha+2\beta\right)+\hat{k}\left(\beta-\alpha\right)$
$\frac{\vec{c}.\left(\vec{a}+\vec{b}\right)}{\left|\vec{a}+\vec{b}\right|}=3\sqrt{2}$
$\Rightarrow \alpha+\beta=2 .......... \left(1\right)$
$\left(\vec{c}-\left(\vec{a}\times\vec{b}\right)\right).\left(\alpha\vec{a}+\beta\vec{b}\right)$
$=\left|\vec{c}\right|^{2}=\alpha^{2}\left|\vec{a}\right|^{2}+\beta^{2}\left|b\right|^{2}+2\alpha\beta\left(\vec{a}.\vec{b}\right)$
$=6\left(\alpha^{2}+\left(2-\alpha\right)^{2}+\alpha\left(2-\alpha\right)\right)$
$=6\left(\alpha^{2}+\beta^{2}+\alpha\beta\right)$
$=6\left(\left(\alpha-1\right)^{2}+3\right)$
$\Rightarrow $ Min. value $= 18$