The two adjacent sides of a parallelogram are represented by vectors 2 i -4 j + 5 k and i -2 j -3 k. Find the unit vector parallel to one of its…

9.(a)One diagonal of the parallelogram\[= \left(2\hat{i} - 4\hat{j} + 5\hat{k}\right) + \left(\hat{i} - 2\hat{j} - 3\hat{k}\right)\]\[= 3\hat{i} - 6\hat{j} + 2\hat{k}\]Unit vector parallel to the diagonal\[= \frac{3\hat{i} - 6\hat{j} + 2\hat{k}}{\sqrt{9 + 36 + 4}}\]\[= \frac{3}{7}\hat{i} - \frac{6}{7}\hat{j} + \frac{2}{7}\hat{k}\]Vector area of parallelogram $\displaystyle = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix}$\[= \hat{i}(22) - \hat{j}(-11) + \hat{k}(0)\]\[= 22\hat{i} + 11\hat{j}\]Area $\displaystyle = \sqrt{484 + 121} = \sqrt{605} = 11\sqrt{5}$OR9.(b)\[(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0 \Rightarrow \left[(2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}\right] \cdot (3\hat{i} + \hat{j}) = 0\]\[\Rightarrow 3(2 - \lambda) + (2 + 2\lambda) \cdot 1 = 0\]\[\Rightarrow -3\lambda + 2\lambda + 6 + 2 = 0\]\[\Rightarrow \lambda = 8\]