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If $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ are sides of a parallelogram, then a unit vector is parallel to one of the diagonals of the parallelogram is
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Verified Answer
The correct answer is:
$\frac{\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}}$
Let the position vector
$$
\mathbf{O A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{O B}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}
$$
$\therefore$ Diagonal vector
$$
\begin{aligned}
& \mathrm{r} \mathbf{O C}=\mathbf{O A}+\mathbf{A C} \\
& =\mathbf{O A}+\mathbf{O B} \quad[\because \mathbf{O B} \| \mathbf{A C}] \\
& =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& =4(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})
\end{aligned}
$$
$\therefore$ Unit vector of a diagonal
$$
=\frac{4(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{4^2+4^2+4^2}}=\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{3}}
$$
$$
\mathbf{O A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{O B}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}
$$
$\therefore$ Diagonal vector
$$
\begin{aligned}
& \mathrm{r} \mathbf{O C}=\mathbf{O A}+\mathbf{A C} \\
& =\mathbf{O A}+\mathbf{O B} \quad[\because \mathbf{O B} \| \mathbf{A C}] \\
& =\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}+3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\
& =4(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})
\end{aligned}
$$
$\therefore$ Unit vector of a diagonal
$$
=\frac{4(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{4^2+4^2+4^2}}=\frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{3}}
$$
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