Question

Let $\overrightarrow{AB}=3\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{AC}=\hat{i}-\hat{j}+3\hat{k}$ and a point p on the line segment BC is equidistant from AB and AC, then $\overrightarrow{AP}$ is

Answer 1

We have,

A point equidistant from $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is the bisector of $\angle BAC$.

Then,

We know that,

$\frac{\overrightarrow{AB}}{\left|\overrightarrow{AB}\right|}+\frac{\overrightarrow{AC}}{\left|\overrightarrow{AC}\right|}=\frac{3\hat{i}+\hat{j}-\hat{k}}{\sqrt{3^2+1^2+{(-1)}^2}}+\frac{\hat{i}-\hat{j}+3\hat{k}}{\sqrt{1^2+{(-1)}^2+{\left(3\right)}^2}}$

$=\frac{3\hat{i}+\hat{j}-\hat{k}}{\sqrt{11}}+\frac{\hat{i}-\hat{j}+3\hat{k}}{\sqrt{11}}$

$\frac{\overrightarrow{AB}}{\left|\overrightarrow{AB}\right|}+\frac{\overrightarrow{AC}}{\left|\overrightarrow{AC}\right|}=\frac{1}{\sqrt{11}}(4\hat{i}+2\hat{k})$

$So,$

$\overrightarrow{AP}=t(2\hat{i}+\hat{k})where,t=\frac{2}{\sqrt{11}}$

Now, we know that,

$\overrightarrow{BP}=\overrightarrow{AP}-\overrightarrow{AB}=t(2\hat{i}+\hat{k})-(3\hat{i}+\hat{j}-\hat{k})$

$=(2t-3)\hat{i}-\hat{j}+(t+1)\hat{k}$

Also,

$\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=(\hat{i}-\hat{j}+3\hat{k})-(3\hat{i}+\hat{j}-\hat{k})$

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

But

According to given question,

$\overrightarrow{BP}=a\left(\overrightarrow{BC}\right)$

$So,(2t-3)\hat{i}-\hat{j}+(t+1)\hat{k}=a(-2\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k})$

$\text{Now,}\text{compairing}\text{that,}$

$So,(2t-3)=-2a,$

$-1=-2a$

$a=\frac{1}{2}$

$Now,$

$t+1=4a$

$t=1$

$So,$

$\overrightarrow{AP}=2\hat{i}+\hat{k}$

Hence, this is the answer.