Q. If the position vectors of points A, B,C and D are given by $\overline{a} = 2\hat{i}+3\hat{j}+4\hat{k},$  $\overline{b}=5\hat{i}+4\hat{j}-\hat{k},$ $\overline{c}=3\hat{i}+6\hat{j}+2\hat{k} and \overline{d}=-2\hat{i}+6\hat{j}-\hat{k}$ respectively, show that $\overline{AB} \perp \overline{CD}$. 

Dear Student,
Here is the solution of your asked query:

With reference to origin O, we have
$$\begin{gathered} \mathrm{OA}=2\mathrm{i}+3\mathrm{j}+4\mathrm{k} \\ \mathrm{OB}=5\mathrm{i}+4\mathrm{j}-\mathrm{k} \\ \mathrm{OC}=3\mathrm{i}+6\mathrm{j}+2\mathrm{k} \\ \mathrm{OD}=-2\mathrm{i}+6\mathrm{j}-\mathrm{k} \\ \mathrm{Now} \mathrm{we} \mathrm{have}; \\ \mathrm{AB}=\mathrm{OB}-\mathrm{OA} =\left(5\mathrm{i}+4\mathrm{j}-\mathrm{k}\right) -\left(2\mathrm{i}+3\mathrm{j}+4\mathrm{k}\right)=5\mathrm{i}+4\mathrm{j}-\mathrm{k}-2\mathrm{i}-3\mathrm{j}-4\mathrm{k}=3\mathrm{i}+\mathrm{j}-5\mathrm{k} \\ \mathrm{CD}=\mathrm{OD}-\mathrm{OC}=\left(-2\mathrm{i}+6\mathrm{j}-\mathrm{k}\right)-\left(3\mathrm{i}+6\mathrm{j}+2\mathrm{k}\right)=-2\mathrm{i}+6\mathrm{j}-\mathrm{k}-3\mathrm{i}-6\mathrm{j}-2\mathrm{k}=-5\mathrm{i}-3\mathrm{k} \\ \mathrm{cos\theta }=\frac{\overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{CD}}}{\left|\overrightarrow{\mathrm{AB}}\right|\left|\overrightarrow{\mathrm{CD}}\right|} \\ =\frac{\left(3\mathrm{i}+\mathrm{j}-5\mathrm{k}\right).\left(-5\mathrm{i}-3\mathrm{k}\right)}{\sqrt{3^2+1^2+{\left(-5\right)}^2}\times \sqrt{{\left(-5\right)}^2+{\left(-3\right)}^2}} \\ =\frac{3\times \left(-5\right)+15}{\sqrt{35}\times \sqrt{34}} \\ =0 \\ \Rightarrow \mathrm{cos\theta }=\cos 90° \\ \Rightarrow \mathrm{\theta }=90° \end{gathered}$$

Therefore vectors AB and CD are perpendicular.
Regards