Question

In a $\Delta$ ABC, $\overrightarrow{AB}=6\overrightarrow{i}+3\overrightarrow{j}+3\overrightarrow{k}$ ; $\overrightarrow{AC}=3\overrightarrow{i}-3\overrightarrow{j}+6\overrightarrow{k}$. D and D' are points of tri-sections of sides BC. Find $\overrightarrow{AD}$ and $\overrightarrow{A{D}^{\prime }}$.

Answer 1

We have

$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$

or, $\overrightarrow{BC}=(3\hat{i}-3\hat{j}+6\hat{k})$$-(6\hat{i}+3\hat{j}+3\hat{k})=-3\hat{i}-6\hat{j}+3\hat{k}$.

Now, according to the problem,

$\overrightarrow{BD}=\overrightarrow{D^{\prime }C}=\frac{1}{3}\overrightarrow{BC}=(-\hat{i}-2\hat{j}+\hat{k})$.

$\therefore \overrightarrow{AD}=\overrightarrow{BD}-\overrightarrow{BA}=\overrightarrow{BD}+\overrightarrow{AB}=5\hat{i}+\hat{j}+4\hat{k}$.

And $\overrightarrow{A{D}^{\prime }}=\overrightarrow{AC}-\overrightarrow{D^{\prime }C}=4\hat{i}-\hat{j}+5\hat{k}$.