Minimum Value of Vector Expression with Projection Constraint

To find the minimum possible value of the given expression, we proceed step-by-step:

Step 1: Simplify the expression to be minimized We are given: $E = (\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c} = |\vec{c}|^2 - (\vec{a} \times \vec{b}) \cdot \vec{c}$

Since $\vec{c} = \alpha\vec{a} + \beta\vec{b}$, the vector $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. Therefore, the scalar triple product is: $(\vec{a} \times \vec{b}) \cdot \vec{c} = [\vec{a} \quad \vec{b} \quad \vec{c}] = 0$

Thus, the expression simplifies to: $E = |\vec{c}|^2$

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Step 2: Express $|\vec{c}|^2$ in terms of $\alpha$ and $\beta$ Given the vectors: $\vec{a} = 2\hat{i} + \hat{j} - \hat{k} \implies |\vec{a}|^2 = 2^2 + 1^2 + (-1)^2 = 6$ $\vec{b} = \hat{i} + 2\hat{j} + \hat{k} \implies |\vec{b}|^2 = 1^2 + 2^2 + 1^2 = 6$ $\vec{a} \cdot \vec{b} = (2)(1) + (1)(2) + (-1)(1) = 2 + 2 - 1 = 3$

Now, calculating the magnitude squared of $\vec{c}$: $|\vec{c}|^2 = |\alpha\vec{a} + \beta\vec{b}|^2 = \alpha^2 |\vec{a}|^2 + \beta^2 |\vec{b}|^2 + 2\alpha\beta (\vec{a} \cdot \vec{b})$ $|\vec{c}|^2 = 6\alpha^2 + 6\beta^2 + 2\alpha\beta(3) = 6(\alpha^2 + \alpha\beta + \beta^2)$

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Step 3: Use the projection condition to relate $\alpha$ and $\beta$ The vector $\vec{a} + \vec{b}$ is: $\vec{a} + \vec{b} = (2+1)\hat{i} + (1+2)\hat{j} + (-1+1)\hat{k} = 3\hat{i} + 3\hat{j}$ $|\vec{a} + \vec{b}| = \sqrt{3^2 + 3^2 + 0^2} = 3\sqrt{2}$

The scalar projection of $\vec{c}$ on $(\vec{a} + \vec{b})$ is given by: $\text{Projection} = \frac{\vec{c} \cdot (\vec{a} + \vec{b})}{|\vec{a} + \vec{b}|} = 3\sqrt{2}$

Substitute $\vec{c} = \alpha\vec{a} + \beta\vec{b}$: $\vec{c} \cdot (\vec{a} + \vec{b}) = (\alpha\vec{a} + \beta\vec{b}) \cdot (\vec{a} + \vec{b})$ $\vec{c} \cdot (\vec{a} + \vec{b}) = \alpha(|\vec{a}|^2 + \vec{a}\cdot\vec{b}) + \beta(\vec{a}\cdot\vec{b} + |\vec{b}|^2)$ $\vec{c} \cdot (\vec{a} + \vec{b}) = \alpha(6 + 3) + \beta(3 + 6) = 9(\alpha + \beta)$

Using the projection equation: $\frac{9(\alpha + \beta)}{3\sqrt{2}} = 3\sqrt{2} \implies \frac{3(\alpha + \beta)}{\sqrt{2}} = 3\sqrt{2} \implies \alpha + \beta = 2$

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Step 4: Minimize $|\vec{c}|^2$ Substitute $\beta = 2 - \alpha$ into the expression for $|\vec{c}|^2$: $|\vec{c}|^2 = 6\left(\alpha^2 + \alpha(2 - \alpha) + (2 - \alpha)^2\right)$ $|\vec{c}|^2 = 6\left(\alpha^2 + 2\alpha - \alpha^2 + 4 - 4\alpha + \alpha^2\right)$ $|\vec{c}|^2 = 6\left(\alpha^2 - 2\alpha + 4\right)$ $|\vec{c}|^2 = 6\left((\alpha - 1)^2 + 3\right)$

Since $(\alpha - 1)^2 \ge 0$, the minimum value occurs when $\alpha = 1$ (which gives $\beta = 1$): $|\vec{c}|^2_{\text{min}} = 6(0 + 3) = 18$

Thus, the minimum value of $(\vec{c} - (\vec{a} \times \vec{b})) \cdot \vec{c}$ is $18$.