LinAlg4.1.12

Show that the points represented by $\bar{i}+2\bar{j}+3\bar{k},3\bar{i}+4\bar{j}+7\bar{k},-3\bar{i}-2\bar{j}-5\bar{k}\,$ are collinear.

Let $A=\bar{i}+2\bar{j}+3\bar{k},B=3\bar{i}+4\bar{j}+7\bar{k},C=-3\bar{i}-2\bar{j}-5\bar{k}\,$ Let O be the origin,then

$\bar{OA}=\bar{i}+2\bar{j}+3\bar{k},\bar{OB}=3\bar{i}+4\bar{j}+7\bar{k},\bar{OC}=-3\bar{i}-2\bar{j}-5\bar{k}\,$

Now $\bar{AB}=\bar{OB}-\bar{OA}=(3\bar{i}+4\bar{j}+7\bar{k})-(\bar{i}+2\bar{j}+3\bar{k})=2\bar{i}+2\bar{j}+4\bar{k}\,$

Similarly $\bar{BC}=\bar{OC}-\bar{OB}=(-3\bar{i}-2\bar{j}-5\bar{k})-(3\bar{i}+4\bar{j}+7\bar{k})=-6\bar{i}-6\bar{j}-12\bar{k}\,$

Therefore $\bar{BC}=-3(2\bar{i}+2\bar{j}+4\bar{k})=-3\bar{AB}\,$

$|BC|=|-3|\cdot|AB|=3|AB|\,$

AB and BC are two lines with common point B,hence A,B,C are collinear.