Question

The incentre of the triangle formed by the points $\hat{i}+\hat{j}+\hat{k},$ $4\hat{i}+\hat{j}+\hat{k}$, and $4\hat{i}+5\hat{j}+\hat{k}$ is

• A. $\frac{\hat{i}+\hat{j}+\hat{k}}{3}$
• B. $\hat{i}+2\hat{j}+3\hat{k}$
• C. $3\hat{i}+2\hat{j}+\hat{k}$
• D. $\hat{i}+\hat{j}+\hat{k}$

Answer 1

Answer: (C)

$3\hat{i}+2\hat{j}+\hat{k}$

Position vector of $A$ is $\hat{i}+\hat{j}+\hat{k}$, gives coordinates as $(1,1,1)$.

Position vector of $B$ is $4\hat{i}+\hat{j}+\hat{k}$, gives coordinates as $(4,1,1)$.

Position vector of $C$ is $4\hat{i}+5\hat{j}+\hat{k}$, gives coordinates as $(4,5,1)$.

$\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=3\hat{i}$

$\overrightarrow{BC}=\overrightarrow{C}-\overrightarrow{B}=4\hat{j}$

$\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A}=3\hat{i}+4\hat{j}$

Here, we can see that $△ABC$ is a right angled triangle with right angle at $B$.

$|\overrightarrow{AB}|=\sqrt{3^2}=3$

$|\overrightarrow{BC}|=\sqrt{4^2}=4$

$|\overrightarrow{AC}|=\sqrt{3^2+4^2}=5$

Point of Intersection of angle bisectors can be found using formula:

$I=\frac{|\overrightarrow{BC}|\times \overrightarrow{A}+|\overrightarrow{CA}|\times \overrightarrow{B}+|\overrightarrow{AB}|\times \overrightarrow{C}}{|\overrightarrow{AB}|+|\overrightarrow{BC}|+|\overrightarrow{CA}|}$

$I=\frac{4\times (\hat{i}+\hat{j}+\hat{k})+5\times (4\hat{i}+\hat{j}+\hat{k})+3\times (4\hat{i}+5\hat{j}+\hat{k})}{3+4+5}$

$\Rightarrow I=\frac{1}{12}(36\hat{i}+24\hat{j}+12\hat{j})=3\hat{i}+2\hat{j}+\hat{k}$

Hence, incenter will be $3\hat{i}+2\hat{j}+\hat{k}$