Question

If the position vectors of the vertices A, B and C of a $\Delta$ABC are respectively $4\hat{i}+7\hat{j}+8\hat{k}$, $2\hat{i}+3\hat{j}+4\hat{k}$ and $2\hat{i}+5\hat{j}+7\hat{k}$, then the position vector of the point, where the bisector of $\angle$A meets BC is?

Answer 1

step 1; if the bisector of $\angle A$ meets BC at D,then D divides BC in the ratio AB:AC,

Now,

$$\begin{array}{c}\overrightarrow{AB}=\left|B-A\right|=\overline{b}=(2i+3j+4k)-(4i+7j+8k)\\ =2i+3j-4k-4i-7j-8k\\ \overline{b}=(-2i-4j-4k)\\ \left|\overline{b}\right|=\sqrt{4+16+16}=\sqrt{36}=6\\ \left|\overline{b}\right|=6\\ \overrightarrow{AC}=\left|C-A\right|=\overline{c}=(2i+5j+7k)-(4i+7j+8k)\\ \overline{c}=-2i-2j-k\\ \left|\overline{c}\right|=AC=\sqrt{4+4+1}=\sqrt{9}=3\\ \left|\overline{c}\right|=3\end{array}$$

then,

the Position vector of D by sector formula,

$$\begin{array}{c}\overline{AD}=k\left(\frac{\overline{b}}{\left|\overline{b}\right|}+\frac{\overline{c}}{\left|\overline{c}\right|}\right)whereas,k=\frac{\overline{b}\times \overline{c}}{\overline{b+\overline{c}}}=\frac{6\times 3}{6+3}=\frac{18}{9}=2\\ \end{array}$$$$\begin{array}{c}=2\left(\frac{-2i-4j-4k}{6}+\frac{-2i-2j-k}{3}\right)\\ =2\left(\frac{-2i-4j-4k-4i-4j-2k}{6}\right)\\ =2\left(\frac{-6i-8j-6k}{6}\right)\\ =\frac{2}{3}(-3i-4j-3k)\end{array}$$

$$\begin{array}{c}\overrightarrow{BC}=(c-b)=(2i+5j+7k)-(2i+3j+4k)\\ \overrightarrow{BC}=0+2j+3k\\ then\\ \therefore \frac{h-4}{0}=\frac{k-7}{2}=\frac{z-8}{3}=\frac{-2(14+24)}{13}=\frac{-2\left(38\right)}{13}\\ so,\\ h=4,\\ k=\frac{-4\left(38\right)}{13}+7=\frac{-61}{13}\\ z=\frac{-6\left(38\right)}{13}+8=\frac{-124}{13}\end{array}$$