Mathematics - Vector Question with Solution | TestHub

Given,

One vertex of a rectangular parallelopiped is at the origin $O$ and the lengths of its edges along $x\mathit{,}\mathit{ }y$ and $z$ axes are $3, 4$ and $5$ units respectively,

And $P$ be the vertex $(3, 4, 5)$,

Now plotting the diagram we get,

Now equation of line $OP:\frac{x}{3}=\frac{y}{4}=\frac{z}{5}$

And equation of line $AB:\frac{x-3}{0}=\frac{y}{0}=\frac{z}{1}$

Now finding, $\overrightarrow{n_1}\mathit{\times }\overrightarrow{n_2}=\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 3& 4& 5\\ 0& 0& 1\end{array}\right|$

$\Rightarrow \overrightarrow{n_1}\mathit{\times }\overrightarrow{n_2}=\hat{\mathrm{i}}\left(4\right)-\hat{\mathrm{j}}\left(3\right)+\hat{\mathrm{k}}\left(0\right)$

$\Rightarrow \overrightarrow{n_1}\mathit{\times }\overrightarrow{n_2}=4\hat{\mathrm{i}}-3\hat{\mathrm{j}}$

Now we know that,

Shortest Distance, $\mathrm{S}.\mathrm{D}=\frac{\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)·\left(\overrightarrow{n_1}\mathit{\times }\overrightarrow{n_2}\right)}{\left|\overrightarrow{n_1}\mathit{\times }\overrightarrow{n_2}\right|}$

$\Rightarrow \mathrm{S}.\mathrm{D}=\frac{\left(3\hat{\mathrm{i}}+0\hat{\mathrm{j}}+0\hat{\mathrm{k}}-0\hat{\mathrm{i}}+0\hat{\mathrm{j}}+0\hat{\mathrm{k}}\right)·\left(4\hat{\mathrm{i}}-3\hat{\mathrm{j}}\right)}{5}$

$\Rightarrow \mathrm{S}.\mathrm{D}=\frac{\left(3\hat{\mathrm{i}}\right)·\left(4\hat{\mathrm{i}}-3\hat{\mathrm{j}}\right)}{5}$

$\Rightarrow \mathrm{S}.\mathrm{D}=\frac{12}{5}$