Question

Let $A=(1,2,3)$ , $B=(3,-1,5)$ , $C=(4,0,-3)$, then

• A. $\angle A={90}^0$
• B. $\angle B={90}^0$
• C. $\angle C={90}^0$
• D. $\angle A={45}^0$

Answer 1

The correct option is A $\angle A={90}^0$

Given points are $A=(1,2,3),B=(3,-1,5),C=(4,0,-3)$

Therefore, $\overrightarrow{AB}=2i-3j+2k$
$\overrightarrow{BC}=i+j-8k$
and $\overrightarrow{AC}=3i-2j-6k$
Now $\overrightarrow{AB}.\overrightarrow{AC}=(2i-3j+2k).(3i-2j-6k)$
$=6+6-12$
$=0$
Hence, $\overrightarrow{AB}$ is perpendicular to $\overrightarrow{AC}$.
Therefore, $\angle A={90}^0$