# If the vertices A, B, C of a triangle ABC are

Question:

If the vertices $A, B, C$ of a triangle $A B C$ are $(1,2,3),(-1,0,0),(0,1,2)$, respectively, then find $\angle A B C$. [ $\angle A B C$ is the angle between the vectors $\overrightarrow{B A}$ and $\overrightarrow{B C}$ ]

Solution:

The vertices of ΔABC are given as A (1, 2, 3), B (–1, 0, 0), and C (0, 1, 2).

Also, it is given that $\angle \mathrm{ABC}$ is the angle between the vectors $\overrightarrow{\mathrm{BA}}$ and $\overrightarrow{\mathrm{BC}}$.

$\overrightarrow{\mathrm{BA}}=\{1-(-1)\} \hat{i}+(2-0) \hat{j}+(3-0) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}$

$\overrightarrow{\mathrm{BC}}=\{0-(-1)\} \hat{i}+(1-0) \hat{j}+(2-0) \hat{k}=\hat{i}+\hat{j}+2 \hat{k}$

$\therefore \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})=2 \times 1+2 \times 1+3 \times 2=2+2+6=10$

$|\overrightarrow{\mathrm{BA}}|=\sqrt{2^{2}+2^{2}+3^{2}}=\sqrt{4+4+9}=\sqrt{17}$

$|\overrightarrow{\mathrm{BC}}|=\sqrt{1+1+2^{2}}=\sqrt{6}$

Now, it is known that:

$\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}| \cos (\angle \mathrm{ABC})$

$\therefore 10=\sqrt{17} \times \sqrt{6} \cos (\angle \mathrm{ABC})$

$\Rightarrow \cos (\angle \mathrm{ABC})=\frac{10}{\sqrt{17} \times \sqrt{6}}$

$\Rightarrow \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)$