# Show that the vectors

Question:

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

Solution:

Let vectors $2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k}$ and $3 \hat{i}-4 \hat{j}-4 \hat{k}$ be position vectors of points $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ respectively.

i.e., $\overrightarrow{\mathrm{OA}}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{OB}}=\hat{i}-3 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{OC}}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

Now, vectors $\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}}$, and $\overrightarrow{\mathrm{AC}}$ represent the sides of $\triangle \mathrm{ABC}$.

i.e., $\overrightarrow{\mathrm{OA}}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{\mathrm{OB}}=\hat{i}-3 \hat{j}-5 \hat{k}$, and $\overrightarrow{\mathrm{OC}}=3 \hat{i}-4 \hat{j}-4 \hat{k}$

$\therefore \overrightarrow{\mathrm{AB}}=(1-2) \hat{i}+(-3+1) \hat{j}+(-5-1) \hat{k}=-\hat{i}-2 \hat{j}-6 \hat{k}$

$\overrightarrow{\mathrm{BC}}=(3-1) \hat{i}+(-4+3) \hat{j}+(-4+5) \hat{k}=2 \hat{i}-\hat{j}+\hat{k}$

$\overrightarrow{\mathrm{AC}}=(2-3) \hat{i}+(-1+4) \hat{j}+(1+4) \hat{k}=-\hat{i}+3 \hat{j}+5 \hat{k}$

$|\overrightarrow{\mathrm{AB}}|=\sqrt{(-1)^{2}+(-2)^{2}+(-6)^{2}}=\sqrt{1+4+36}=\sqrt{41}$

$|\overrightarrow{\mathrm{BC}}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{4+1+1}=\sqrt{6}$

$|\overrightarrow{\mathrm{AC}}|=\sqrt{(-1)^{2}+3^{2}+5^{2}}=\sqrt{1+9+25}=\sqrt{35}$

$\therefore|\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{AC}}|^{2}=6+35=41=|\overrightarrow{\mathrm{AB}}|^{2}$

Hence, ΔABC is a right-angled triangle.