Question

Show that the vectors $\overrightarrow{a}=3\hat{i}-2\hat{j}+\hat{k},\overrightarrow{b}=\hat{i}-3\hat{j}+5\hat{k}and\overrightarrow{c}=2\hat{i}+\hat{j}-4\hat{k}$ form a right angled triangle.

Answer 1

We have $\overrightarrow{b}+\overrightarrow{c}=(\hat{i}-3\hat{j}+5\hat{k})+(2\hat{i}+\hat{j}-4\hat{k})=3\hat{i}-2\hat{j}+\hat{k}=\overrightarrow{a}$

Hence, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar.

Also, we observe that no two of these vectors are parallel, therefore, the given vectors form a triangle.

Further, $\overrightarrow{a}\cdot \overrightarrow{c}=(3\hat{i}-2\hat{j}+\hat{k})\cdot (2\hat{i}+\hat{j}-4\hat{k})=0$

Dot product of two non-zero vectors is zero. Hence, they are perpendicular so they form a right angled triangle.

$\left|\overrightarrow{a}\right|=\sqrt{9+4+1}=\sqrt{14},$

$\left|\overrightarrow{b}\right|=\sqrt{1+9+25}=\sqrt{35}$

and $\left|\overrightarrow{c}\right|=\sqrt{4+1+16}=\sqrt{21}$