One vertex of a rectangular parallelopiped is at the origin O and the lengths of its edges along x,y and z axes are 3,4 and 5 units respectively. Let…

Given,

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

And $P$P be the vertex $(3,4,5)$(3,4,5),

Now plotting the diagram we get,

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

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

Now finding, $\vec{{n}_{1}}\times \vec{{n}_{2}}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 0 & 0 & 1\end{matrix}|$n1​​×n2​​=∣i^30​j^​40​k^51​∣

$$\Rightarrow \vec{{n}_{1}}\times \vec{{n}_{2}}=\hat{i}(4)-\hat{j}(3)+\hat{k}(0)$$⇒n1​​×n2​​=i^(4)−j^​(3)+k^(0)

$$\Rightarrow \vec{{n}_{1}}\times \vec{{n}_{2}}=4\hat{i}-3\hat{j}$$⇒n1​​×n2​​=4i^−3j^​

Now we know that,

Shortest Distance, $S.D=\frac{(\vec{{a}_{2}}-\vec{{a}_{1}})\cdot (\vec{{n}_{1}}\times \vec{{n}_{2}})}{|\vec{{n}_{1}}\times \vec{{n}_{2}}|}$S.D=∣n1​​×n2​​∣(a2​​−a1​​)⋅(n1​​×n2​​)​

$$\Rightarrow S.D=\frac{(3\hat{i}+0\hat{j}+0\hat{k}-0\hat{i}+0\hat{j}+0\hat{k})\cdot (4\hat{i}-3\hat{j})}{5}$$⇒S.D=5(3i^+0j^​+0k^−0i^+0j^​+0k^)⋅(4i^−3j^​)​

$$\Rightarrow S.D=\frac{(3\hat{i})\cdot (4\hat{i}-3\hat{j})}{5}$$⇒S.D=5(3i^)⋅(4i^−3j^​)​

$$\Rightarrow S.D=\frac{12}{5}$$⇒S.D=512​