CBSE 2022 · Region 1 · Set 1 · Q7 · 3 marks
If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ and $\displaystyle \overrightarrow{\mathrm{d}}$ are four non-zero vectors such that $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ and $\displaystyle \vec{\mathrm{a}} \times \vec{\mathrm{c}}=4 \vec{\mathrm{b}} \times \vec{\mathrm{d}}$, then show that $\displaystyle (\vec{\mathrm{a}}-2 \vec{\mathrm{d}})$ is parallel to $\displaystyle (2 \vec{\mathrm{b}}-\vec{\mathrm{c}})$ where $\displaystyle \vec{\mathrm{a}} \neq 2 \vec{\mathrm{d}}, \vec{\mathrm{c}} \neq 2 \vec{\mathrm{b}}$.The two adjacent sides of a parallelogram are represented by $\displaystyle 2 \hat{i}-4 \hat{j}-5 \hat{k}$ and $\displaystyle 2 \hat{i}+2 \hat{j}+3 \hat{k}$. Find the unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram also.
If $\displaystyle \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}$ and $\displaystyle \overrightarrow{\mathrm{d}}$ are four non-zero vectors such that $\displaystyle \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{d}}$ and $\displaystyle \vec{\mathrm{a}} \times \vec{\mathrm{c}}=4 \vec{\mathrm{b}} \times \vec{\mathrm{d}}$, then show that $\displaystyle (\vec{\mathrm{a}}-2 \vec{\mathrm{d}})$ is parallel to $\displaystyle (2 \vec{\mathrm{b}}-\vec{\mathrm{c}})$ where $\displaystyle \vec{\mathrm{a}} \neq 2 \vec{\mathrm{d}}, \vec{\mathrm{c}} \neq 2 \vec{\mathrm{b}}$.
The two adjacent sides of a parallelogram are represented by $\displaystyle 2 \hat{i}-4 \hat{j}-5 \hat{k}$ and $\displaystyle 2 \hat{i}+2 \hat{j}+3 \hat{k}$. Find the unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram also.
Marking-scheme solution
(a)
Consider $\displaystyle (\vec{a} - 2\vec{d}) \times (2\vec{b} - \vec{c})$
$\displaystyle = \vec{a} \times 2\vec{b} - \vec{a} \times \vec{c} - 4\vec{d} \times \vec{b} + 2\vec{d} \times \vec{c}$
$\displaystyle = 0$
$\displaystyle \therefore (\vec{a} - 2\vec{d}) \parallel (2\vec{b} - \vec{c})$
Or(b) Let ABCD be a parallelogram with
\[\overrightarrow{AB} = \overrightarrow{DC} = 2\hat{i} - 4\hat{j} - 5\hat{k}\]
and
\[\overrightarrow{BC} = \overrightarrow{AD} = 2\hat{i} + 2\hat{j} + 3\hat{k}\]
\[\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = 4\hat{i} - 2\hat{j} - 2\hat{k}\]
and $\displaystyle \overrightarrow{BD} = 6\hat{j} + 8\hat{k}$
$\displaystyle \therefore |\overrightarrow{AC}| = 2\sqrt{6}$ and $\displaystyle |\overrightarrow{BD}| = 10$
$\displaystyle \therefore$ Required unit vectors $\displaystyle \hat{d}_1$ and $\displaystyle \hat{d}_2$ are
\[\hat{d}_1 = \frac{2}{\sqrt{6}}\hat{i} - \frac{1}{\sqrt{6}}\hat{j} - \frac{\hat{k}}{\sqrt{6}} \quad \text{and} \quad \hat{d}_2 = \frac{3}{5}\hat{j} + \frac{4}{5}\hat{k}\]
Now,
\[\text{Area of } \parallel ABCD = \frac{1}{2}|\vec{d}_1 \times \vec{d}_2|\]
\[= \frac{1}{2}\left\| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & -2 \\ 0 & 6 & 8 \end{array} \right\|\]
\[= \frac{1}{2}|-4\hat{i} - 32\hat{j} + 24\hat{k}|\]
\[= \frac{1}{2}\sqrt{1616} = 2\sqrt{101}\]
Vector AlgebraProduct of Two VectorsApplyshort_answerhard
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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.