Question

If A, B, C, D are the points with position vectors $\hat{i}-\hat{j}+\hat{k},2\hat{i}-\hat{j}+3\hat{k},2\hat{i}-3\hat{k},3\hat{i}-2hatj+\hat{k}$, respectively, find the projection of $\overline{AB}$ along $\overline{CD}$

Answer 1

Here, $\overline{OA}=\hat{i}+\hat{j}-\hat{k},\overline{OB}=2\hat{i}-\hat{j}+3\hat{k},\overline{OC}=2\hat{i}-3\hat{k}$ and $\overline{OD}=3i-2\hat{i}+\hat{k}$

$\therefore \overline{AB}=\overline{OB}-\overline{OA}=(2-1)\hat{i}+(-1-1)\hat{j}+(3+1)\hat{k}$

$=\hat{i}-2\hat{j}+4\hat{k}$

And $\overline{CD}=\overline{OD}-\overline{OC}=(3-2)\hat{i}+(-2-0)\hat{j}+(1+3)\hat{k}$$=\hat{i}-2\hat{j}+4\hat{k}$

So, the projection of $\overline{AB}$ along $\overline{CD}=\overline{A}B.\frac{\overline{CD}}{|\overline{CD}|}$

$=\frac{(\hat{i}-2\hat{j}+4\hat{k})(\hat{i}-2\hat{j}+4\hat{k})}{\sqrt{1^2+2^2+4^2}}$

$=\frac{1+4+16}{\sqrt{21}}=\frac{21}{\sqrt{21}}$

$=\sqrt{21}$ units.