Let vec a = hat i - hat j, vec b = hat i + ha | vedclass

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(A) Given $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.From the equation $\vec

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$\frac{19}{2}$

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(A) Given $\vec{a} = \hat{i} - \hat{j}$,$\vec{b} = \hat{i} + \hat{j} + \hat{k}$,and $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$.From the equation $\vec{a} 	imes \vec{c} + \vec{b} = 0$,we have $\vec{a} 	imes \vec{c} = -\vec{b}$.Calculating the cross product $\vec{a} 	imes \vec{c}$:$\vec{a} 	imes \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ x & y & z \end{vmatrix} = \hat{i}(-z) - \hat{j}(z) + \hat{k}(y + x) = -z\hat{i} - z\hat{j} + (x + y)\hat{k}$.Equating this to $-\vec{b} = -(\hat{i} + \hat{j} + \hat{k}) = -\hat{i} - \hat{j} - \hat{k}$:$-z = -1 \Rightarrow z = 1$.$-z = -1 \Rightarrow z = 1$ (consistent).$x + y = -1$.Given $\vec{a} \cdot \vec{c} = 4$,we have $(\hat{i} - \hat{j}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = x - y = 4$.Solving the system of equations:$x + y = -1$$x - y = 4$Adding the equations: $2x = 3 \Rightarrow x = \frac{3}{2}$.Subtracting the equations: $2y = -5 \Rightarrow y = -\frac{5}{2}$.Thus,$\vec{c} = \frac{3}{2}\hat{i} - \frac{5}{2}\hat{j} + 1\hat{k}$.$|\vec{c}|^2 = x^2 + y^2 + z^2 = \left(\frac{3}{2}ight)^2 + \left(-\frac{5}{2}ight)^2 + (1)^2 = \frac{9}{4} + \frac{25}{4} + 1 = \frac{34}{4} + 1 = \frac{17}{2} + 1 = \frac{19}{2}$.

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

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