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Mathematics · 2022 · 3 marks
CBSE 2022 · Region 5 · Set 2 · Q9
(a) Let a = i + j, b = i - j and c = i + j + k. If n is a unit vector such that a. n =0 and b. n =0, then find | c. n |.
Let $\displaystyle \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\displaystyle \overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$. If $\displaystyle \hat{\mathrm{n}}$ is a unit vector such that $\displaystyle \overrightarrow{\mathrm{a}} . \hat{\mathrm{n}}=0$ and $\displaystyle \overrightarrow{\mathrm{b}} . \hat{\mathrm{n}}=0$, then find $\displaystyle |\overrightarrow{\mathrm{c}} . \hat{\mathrm{n}}|$.
If $\displaystyle \overrightarrow{\mathrm{a}}$ and $\displaystyle \overrightarrow{\mathrm{b}}$ are unit vectors inclined at an angle $\displaystyle 30$° to each other, then find the area of the parallelogram with $\displaystyle (\overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{~b}})$ and $\displaystyle (3 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$ as adjacent sides.
Marking-scheme solution
\[\begin{aligned}
& \vec{a} \cdot \hat{n}=0, \vec{b} \cdot \hat{n}=0 \Rightarrow \hat{n} \text { is } \perp \text { to both } \vec{a} \text { and } \vec{b} \\
& \text { Now, } \hat{n}=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \\
& \text { Here, } \vec{n}=\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 1 & 0 \\
1 & -1 & 0
\end{array}\right|=-2 \hat{k} \\
& \Rightarrow \hat{n}=-\hat{k} \quad(\text { or } \hat{n}=\hat{k}) \\
& \therefore|\vec{c} \cdot \hat{n}|=|\vec{c} \cdot(-\hat{k})|=1
\end{aligned}
\]
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