<p>Let \(\vec{a}=\hat{i}-\hat{j}\), \(\vec{b}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{c}\) be a vector such that \(\vec{a}\times\vec{c}+\vec{b}=\vec{0}\) and \(\vec{a}\cdot\vec{c}=4\), then \(|\vec{c}|^2\) is equal to:</p>
Step-by-Step Solution
Key Concept: Use the constraint equations systematically: from $\vec{a} \times \vec{c} + \vec{b} = \vec{0}$, express $\vec{a} \times \vec{c} = -\vec{b}$, then use the vector triple product identity and the dot product constraint $\vec{a} \cdot \vec{c} = 4$ to find $|\vec{c}|^2$.
Step 1: From the constraint $\vec{a} \times \vec{c} + \vec{b} = \vec{0}$, we get $\vec{a} \times \vec{c} = -\vec{b}$. Step 2: Calculate $|\vec{a} \times \vec{c}|^2 = |-\vec{b}|^2 = |\vec{b}|^2$. $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, so $|\vec{b}|^2 = 1 + 1 + 1 = 3$ Therefore $|\vec{a} \times \vec{c}|^2 = 3$ Step 3: Apply Lagrange's identity: $|\vec{a} \times \vec{c}|^2 + (\vec{a} \cdot \vec{c})^2 = |\vec{a}|^2|\vec{c}|^2$ $\vec{a} = \hat{i} - \hat{j}$, so $|\vec{a}|^2 = 1 + 1 = 2$ $3 + (4)^2 = 2|\vec{c}|^2$ $3 + 16 = 2|\vec{c}|^2$ $19 = 2|\vec{c}|^2$ Step 4: Solve for $|\vec{c}|^2$: $|\vec{c}|^2 = \frac{19}{2}$ ∴ Answer: A
Correct Answer: A