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 $\angle A B C$ is equal to $\angle A B C$ is the angle between the vectors $B A$ and $BC)$

• A. $\cos ^{-1}\left(\frac{10}{\sqrt{114}}\right)$
• B. $\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)$
• C. $\cos ^{-1}\left(\frac{10}{7}\right)$
• D. None of these

Answer 1

Answer: (B)

Here, we have to find $\angle A B C$ i.e., angle between $B A$ and BC, so first of all we have to calculate both these vectors after that we can find the angle between them by $\cos \theta=\frac{( BA ) \cdot( BC )}{| BA | BC \mid}$
We are given the points $A(1,2,3), B(-1,0,0)$ and $C(0,1,2)$.
Also, it is given that $\angle A B C$ is the angle between the vectors $BA$ and $BC$.
Here, $B A=P V$ of $A-P V$ 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)]=2 \hat{i}+2 \hat{j}+3 \hat{k} $
$|B A| =\sqrt{(2)^2+(2)^2+(3)^2}=\sqrt{4+4+9}=\sqrt{17} $
$B C =P V \text { of } C-P V \text { of } B$
$ =(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)]=\hat{i}+\hat{j}+2 \hat{k} $
$|B C| =\sqrt{(1)^2+(1)^2+(2)^2}=\sqrt{1+1+4}=\sqrt{6}$
Now, $ B A \cdot B C=(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})$
$=2 \times 1+2 \times 1+3 \times 2=10$
$\cos \theta=\frac{B A \cdot B C}{|B A||B C|}$
$\Rightarrow \cos (\angle A B C)=\frac{10}{\sqrt{17} \sqrt{6}}$
$\Rightarrow \angle A B C=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right)$