Question

The vertices of a triangle ABC are $A(-1,-2,-3),B(-1,2,3)$ and $C(0,0,0).$ Then the direction ratios of internal bisector of $\angle C$ are :

• A. $0,0,1$
• B. $1,-1,1$
• C. $-1,0,0$
• D. none

Answer 1

The correct option is C $-1,0,0$

Given $A(-1,-2,-3),B(-1,2,3),C(0,0,0)$

$\overrightarrow{BC}=\overrightarrow{b}=\overrightarrow{C}-\overrightarrow{B}$

$\overrightarrow{BC}=\overrightarrow{b}=\hat{i}-2\hat{j}-3\hat{k}$

$\overrightarrow{AC}=\overrightarrow{c}=\overrightarrow{C}-\overrightarrow{A}$

$\overrightarrow{AC}=\overrightarrow{c}=\hat{i}+2\hat{j}+3\hat{k}$

$\overrightarrow{AC}=\overrightarrow{c}=\hat{i}+2\hat{j}+2\hat{k}$

internal angle bisector$=\hat{c}+\hat{b}$

internal angle bisector$=\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{1^2+2^2+3^2}}+\frac{\hat{i}-2\hat{j}-3\hat{k}}{\sqrt{1^2+2^2+3^2}}$

internal angle bisector$=\frac{\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{14}}+\frac{\hat{i}-2\hat{j}-3\hat{k}}{\sqrt{14}}$

internal angle bisector$=\frac{2\hat{i}}{\sqrt{14}}$

$a,b,c$ will be $2,0,0$

Dr's may be $-1,0,0$