Question

If the vertices A, B, C of a triangle ABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively then find $\angle ABC$ ($\angle ABC$ is the angle between the vectors BA and BC).

Answer 1

We are given the points A(1,2,3), B(-1,0,0) and C(0,1,2)
Also, it is given that $\angle ABC$ is the angle between the vectors BA and BC.
Here, BA = PV of A - PV of B $=(\hat{i}+2\hat{j}+3\hat{k})-(-\hat{i}+0\hat{j}+0\hat{k})$
$=[\hat{i}-(-\hat{i})+(2\hat{j}-0)+(3\hat{k}-0\left)\right]=2\hat{i}+2\hat{j}+3\hat{k}$
$|BA|=\sqrt{(2{)}^2+(2{)}^2+(3{)}^2}=\sqrt{4+4+9}=\sqrt{17}$
BC = PV of C - PV of B
$=(0\hat{i}+1\hat{j}+2\hat{k})-(-\hat{i}+0\hat{j}+0\hat{k})=[0-(-\hat{i})+(1\hat{j}-0)+(2\hat{k}-0\left)\right]$
$=\hat{i}+\hat{j}+2\hat{k}$
$|BC|=\sqrt{(1{)}^2+(1{)}^2+(2{)}^2}=\sqrt{1+1+4}=\sqrt{6}$
Now, BA.BC $=(2\hat{i}+2\hat{j}+3\hat{k}).(\hat{i}+\hat{j}+2\hat{k})=2\times 1+2\times 1+3\times 2=10$
$cos\theta =\frac{BA.BC}{|BA||BC|}\Rightarrow cos\left(\angle ABC\right)=\frac{10}{\sqrt{17}\sqrt{6}}\Rightarrow \angle ABC=co{s}^{-1}\left(\frac{10}{\sqrt{102}}\right)$