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Let $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ and $\overrightarrow{\mathbf{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$. Find a vector $\overrightarrow{\mathrm{d}}$ which is perpendicula to both $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$, and $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=15$.
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Verified Answer
Let the vector $\overrightarrow{\mathrm{d}}$ be $\mathrm{d}_1 \hat{\mathrm{i}}+\mathrm{d}_2 \hat{\mathrm{j}}+\mathrm{d}_3 \hat{\mathrm{k}}$, vector $\overrightarrow{\mathrm{d}}$ is perpendicular to vector
$$
\begin{aligned}
&\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{a}}=0 \\
&\Rightarrow \mathrm{d}_1+4 \mathrm{~d}_2+2 \mathrm{~d}_3=0
\end{aligned}
$$
Again vector $\mathrm{d}$ is perpendicular to vector $\overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{b}}=0$
$\therefore \quad 3 \mathrm{~d}_1-2 \mathrm{~d}_2+7 \mathrm{~d}_3=0$
From (i) \& (ii) $\frac{\mathrm{d}_1}{28+4}=\frac{\mathrm{d}_2}{6-7}=\frac{\mathrm{d}_3}{-2-12}$
or $\frac{\mathrm{d}_1}{32}=\frac{\mathrm{d}_2}{-1}=\frac{\mathrm{d}_3}{14}=\mathrm{p}$
$\therefore \quad \mathrm{d}_1=32 \mathrm{p}, \mathrm{d}_2=-\mathrm{p}, \quad \mathrm{d}_3=-14 \mathrm{p}$
$\therefore \quad \overrightarrow{\mathrm{d}}=32 \mathrm{p} \hat{\mathrm{i}}-\mathrm{p} \hat{\mathrm{j}}-14 \mathrm{p} \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=15$
or $(2 \hat{i}-\hat{j}+4 \hat{k})(32 p \hat{i}-p \hat{j}-14 p \hat{k})=15$
$64 \mathrm{p}+\mathrm{p}-56 \mathrm{p}=15 \quad \therefore \mathrm{p}=\frac{15}{9}=\frac{5}{3}$
$\therefore$ vector $\overrightarrow{\mathrm{d}}=\frac{160}{3} \hat{\mathrm{i}}-\frac{5}{3} \hat{\mathrm{j}}-\frac{70}{3} \hat{\mathrm{k}}$
$$
\begin{aligned}
&\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \Rightarrow \overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{a}}=0 \\
&\Rightarrow \mathrm{d}_1+4 \mathrm{~d}_2+2 \mathrm{~d}_3=0
\end{aligned}
$$
Again vector $\mathrm{d}$ is perpendicular to vector $\overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{d}} \cdot \overrightarrow{\mathrm{b}}=0$
$\therefore \quad 3 \mathrm{~d}_1-2 \mathrm{~d}_2+7 \mathrm{~d}_3=0$
From (i) \& (ii) $\frac{\mathrm{d}_1}{28+4}=\frac{\mathrm{d}_2}{6-7}=\frac{\mathrm{d}_3}{-2-12}$
or $\frac{\mathrm{d}_1}{32}=\frac{\mathrm{d}_2}{-1}=\frac{\mathrm{d}_3}{14}=\mathrm{p}$
$\therefore \quad \mathrm{d}_1=32 \mathrm{p}, \mathrm{d}_2=-\mathrm{p}, \quad \mathrm{d}_3=-14 \mathrm{p}$
$\therefore \quad \overrightarrow{\mathrm{d}}=32 \mathrm{p} \hat{\mathrm{i}}-\mathrm{p} \hat{\mathrm{j}}-14 \mathrm{p} \hat{\mathrm{k}}, \quad \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=15$
or $(2 \hat{i}-\hat{j}+4 \hat{k})(32 p \hat{i}-p \hat{j}-14 p \hat{k})=15$
$64 \mathrm{p}+\mathrm{p}-56 \mathrm{p}=15 \quad \therefore \mathrm{p}=\frac{15}{9}=\frac{5}{3}$
$\therefore$ vector $\overrightarrow{\mathrm{d}}=\frac{160}{3} \hat{\mathrm{i}}-\frac{5}{3} \hat{\mathrm{j}}-\frac{70}{3} \hat{\mathrm{k}}$
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