Question
Show that the vectores $\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}},\vec{\text{c}}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$from a right-angled triangle.

Answer

Let ABC be the given triangle and
$\overrightarrow{\text{AC}}=\vec{\text{b}}=\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}}$
$\overrightarrow{\text{CB}}=\vec{\text{a}}=3\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{AB}}=\vec{\text{c}}=2\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}.\vec{\text{b}}=3+6+5=14$
$\vec{\text{b}}.\vec{\text{c}}=2-3-20=-21$
$\vec{\text{c}}.\vec{\text{a}}=6-2-4=0$
So, $\overrightarrow{\text{AB}}$ is perpendicular to $\overrightarrow{\text{CB}}.$
Thus, $\triangle\text{ABC}$ is aright-angled triangle.

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