question_answer

Show that the three points

Answer:

The position vectors of the points A B and C are \[\overrightarrow{OA}=\hat{i}-2\hat{j}-8\hat{k},\] \[\overrightarrow{OB}=5\hat{i}-2\hat{k}\] and \[\overrightarrow{OC}=11\hat{i}+3\hat{j}+7\hat{k}\] respectively. \[\therefore \]      \[\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(5\hat{i}-2\hat{k})-(\hat{i}-2\hat{j}-8\hat{k})\]             \[=4\hat{i}+2\hat{j}+6\hat{k}\] \[\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(11\hat{i}+3\hat{j}+7\hat{k})-(5\hat{i}-2\hat{k})\]            \[=6\hat{i}+3\hat{j}+9\hat{k}\] and \[\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(11\hat{i}+3\hat{j}+7\hat{k})-(\hat{i}\,-2\hat{j}-8\hat{k})\]                         \[=10\hat{i}+5\hat{j}+15\hat{k}\] Here, we see that \[\overrightarrow{AB}=2(2\hat{i}+\hat{j}+3\hat{k})=\frac{2}{5}\overrightarrow{AC}\] Thus, the vectors \[\overrightarrow{AB}\] and \[\overrightarrow{AC}\] are parallel. But AB and AC have a common point. So, the points A, B and C are collinear. Now,     \[\overrightarrow{AB}=2(2\hat{i}+\hat{j}+3\hat{k})=\frac{2}{3}\overrightarrow{BC}\] \[\Rightarrow \]   \[|\overrightarrow{AB}|\,\,=\frac{2}{3}|\overrightarrow{BC}|\,\,\Rightarrow \,\,AB=\frac{2}{3}BC\Rightarrow \frac{AB}{BC}=\frac{2}{3}\] Hence, B divides the line segment AC internally in the ratio 2 : 3.    \[A(1,\,\,\,-2,\,\,-8),\]  \[B(5,\,\,0,\,\,-2),\] and C(11, 3, 7) are collinear, and also find the ratio in which B divides AC.