CBSE 2022 · Region 3 · Set 1 · Q9 · 3 marks
The two adjacent sides of a parallelogram are represented by vectors $\displaystyle 2 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+$ $\displaystyle 5 \hat{\mathrm{k}}$ and $\displaystyle \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$. Find the unit vector parallel to one of its diagonals. Also, find the area of the parallelogram.
OR If $\displaystyle \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\displaystyle \overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}$ are such that the vector $\displaystyle (\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}})$ is perpendicular to vector $\displaystyle \overrightarrow{\mathrm{c}}$, then find the value of $\displaystyle \lambda$.
Marking-scheme solution
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)\[(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0 \Rightarrow \left[(2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}\right] \cdot (3\hat{i} + \hat{j}) = 0\]\[\Rightarrow 3(2 - \lambda) + (2 + 2\lambda) \cdot 1 = 0\]\[\Rightarrow -3\lambda + 2\lambda + 6 + 2 = 0\]\[\Rightarrow \lambda = 8\]
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.