Let \( \mathrm{a}=\hat{\mathrm{i}}-\hat{\mathrm{j}} \) and \( \mathrm{b}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}} \) and c be a vector such \( (\mathrm{a} \times \mathrm{c})+\mathrm{b}=0 \) and \( \mathrm{a} \cdot \mathrm{c}=4 \) , then \( |\mathrm{c}|^2 \) equal to:
Step-by-step Solution:
\[ \overrightarrow{a} \times (\overrightarrow{a} \times \overrightarrow{c}) + \overrightarrow{a} \times \overrightarrow{b} = 0 \] \[ (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{a} - (\overrightarrow{a} \cdot \overrightarrow{a}) \overrightarrow{c} + (\overrightarrow{a} \times \overrightarrow{b}) = 0 \] \[ 4\overrightarrow{a} - 2\overrightarrow{c} + \overrightarrow{a} \times \overrightarrow{b} = 0 \] \[ \overrightarrow{a} \times \overrightarrow{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1 \end{vmatrix} \] \[ = (-\hat{i} - \hat{j} + 2\hat{k}) \] \[ 2\overrightarrow{c} = 4(\hat{i} - \hat{j}) + (-\hat{i} - \hat{j} + 2\hat{k}) \] \[ 2\overrightarrow{c} = 3\hat{i} - 5\hat{j} + 2\hat{k} \] \[ 2 |\overrightarrow{c}| = \sqrt{38} \] \[ 4 |\overrightarrow{c}| = 38 \] \[ |\overrightarrow{c}|^2 = \frac{19}{2} \]