Let $\vec{a}=\hat{i}-\hat{j}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} \times \vec{c})+\vec{b}=0$ and $\vec{a}.\vec{c}=4$, then $|\vec{c}|^2$ is equal to 

Solution:

Given vectors:

• \(\vec{a} = \hat{i} - \hat{j}\)
• \(\vec{b} = \hat{i} + \hat{j} + \hat{k}\)
• And \((\vec{a} \times \vec{c}) + \vec{b} = 0\)
• \(\vec{a} \cdot \vec{c} = 4\)

From \((\vec{a} \times \vec{c}) + \vec{b} = 0\), we get: \[ \vec{a} \times \vec{c} = -\vec{b} \] Let \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\). The cross product \(\vec{a} \times \vec{c}\) is: \[ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ x & y & z \end{vmatrix} \] Expanding this determinant: \[ \vec{a} \times \vec{c} = (z \hat{i} + z \hat{j} + (x + y) \hat{k}) \] Setting \(\vec{a} \times \vec{c} = -\vec{b}\), we get: \[ z = -1, \quad z = -1, \quad x + y = -1 \] Therefore: \[ x + y = -1 \] Now, from \(\vec{a} \cdot \vec{c} = 4\): \[ \vec{a} \cdot \vec{c} = 1 \cdot x + (-1) \cdot y = 4 \] Simplifying: \[ x - y = 4 \] Solving the system of equations: \[ x + y = -1 \] \[ x - y = 4 \] Adding the two equations: \[ 2x = 3 \quad \Rightarrow \quad x = \frac{3}{2} \] Substituting into \(x + y = -1\): \[ \frac{3}{2} + y = -1 \quad \Rightarrow \quad y = -\frac{5}{2} \] Now, \(\vec{c} = \frac{3}{2} \hat{i} - \frac{5}{2} \hat{j} - \hat{k}\). To find \(|\vec{c}|^2\), we compute: \[ |\vec{c}|^2 = \left( \frac{3}{2} \right)^2 + \left( -\frac{5}{2} \right)^2 + (-1)^2 = \frac{9}{4} + \frac{25}{4} + 1 \] \[ |\vec{c}|^2 = \frac{9 + 25 + 4}{4} = \frac{38}{4} = 9.5 \]
Final Answer:
$$ \boxed{9.5} $$