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If $\overline{\mathrm{p}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overline{\mathrm{q}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$. Then a vector of magnitude $5 \sqrt{3}$ units perpendicular to the vector $\overline{\mathrm{q}}$ and coplanar with $\overline{\mathrm{p}}$ and $\overline{\mathrm{q}}$ is
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Verified Answer
The correct answer is:
$5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})$
Let $\bar{r}=a \hat{i}+b \hat{j}+c \hat{k}$
As $\overline{\mathrm{r}}$ is perpendicular to $\overline{\mathrm{q}}$.
$\begin{array}{ll}
\therefore \quad \bar{r} \cdot \bar{q}=0 \\
\quad \Rightarrow a-2 b+c=0 ...(i)
\end{array}$
Also, $\bar{r}$ is coplanar with vectors $\bar{p}$ and $\bar{q}$
$\begin{aligned}
\therefore \quad & {\left[\begin{array}{ccc}
\bar{p} & \overline{\mathrm{q}} & -\mathrm{r}
\end{array}\right]=0 } \\
& \Rightarrow\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right|=0 \\
& \Rightarrow 3 \mathrm{a}-3 \mathrm{c}=0 \\
& \Rightarrow \mathrm{a}-\mathrm{c}=0 \\
& \Rightarrow \mathrm{a}=\mathrm{c}...(ii)
\end{aligned}$
From (i) and (ii), we get
$\mathrm{b}=\mathrm{c}$
$\therefore \quad-\overline{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}^{\mathrm{s}}$
Now, the magnitude of required vector is $5 \sqrt{3}$ units.
$\begin{aligned}
\text { Required vector } & =5 \sqrt{3} \times \frac{r}{\sqrt{r}} \\
& =5 \sqrt{3} \times \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}=5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})
\end{aligned}$
As $\overline{\mathrm{r}}$ is perpendicular to $\overline{\mathrm{q}}$.
$\begin{array}{ll}
\therefore \quad \bar{r} \cdot \bar{q}=0 \\
\quad \Rightarrow a-2 b+c=0 ...(i)
\end{array}$
Also, $\bar{r}$ is coplanar with vectors $\bar{p}$ and $\bar{q}$
$\begin{aligned}
\therefore \quad & {\left[\begin{array}{ccc}
\bar{p} & \overline{\mathrm{q}} & -\mathrm{r}
\end{array}\right]=0 } \\
& \Rightarrow\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right|=0 \\
& \Rightarrow 3 \mathrm{a}-3 \mathrm{c}=0 \\
& \Rightarrow \mathrm{a}-\mathrm{c}=0 \\
& \Rightarrow \mathrm{a}=\mathrm{c}...(ii)
\end{aligned}$
From (i) and (ii), we get
$\mathrm{b}=\mathrm{c}$
$\therefore \quad-\overline{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}^{\mathrm{s}}$
Now, the magnitude of required vector is $5 \sqrt{3}$ units.
$\begin{aligned}
\text { Required vector } & =5 \sqrt{3} \times \frac{r}{\sqrt{r}} \\
& =5 \sqrt{3} \times \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}=5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})
\end{aligned}$
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