Question

If the vertices $A,B,C$ of a $△ABC$ are $(1,2,3)$, $(-1,0,0),(0,1,2)$ respectively, then $\angle ABC$ is equal to $\angle ABC$ is the angle between the vectors $BA$ 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 $\angle ABC$ i.e., angle between $BA$ 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 $\angle ABC$ is the angle between the vectors $BA$ and $BC$.
Here, $BA=PV$ of $A-PV$ of $B$
$=(\hat{i}+2\hat{j}+3\hat{k})-(-\hat{i}+0\hat{j}+0\hat{k})$
$=[\hat{i}-(-\hat{i})+(2\hat{j}-0)+(3\hat{k}-0)]=2\hat{i}+2\hat{j}+3\hat{k}$
$|BA|=\sqrt{(2{)}^2+(2{)}^2+(3{)}^2}=\sqrt{4+4+9}=\sqrt{17}$
$BC=PV\text{ of }C-PV\text{ of }B$
$=(0\hat{i}+1\hat{j}+2\hat{k})-(-\hat{i}+0\hat{j}+0\hat{k})$
$=[0-(-\hat{i})+(1\hat{j}-0)+(2\hat{k}-0)]=\hat{i}+\hat{j}+2\hat{k}$
$|BC|=\sqrt{(1{)}^2+(1{)}^2+(2{)}^2}=\sqrt{1+1+4}=\sqrt{6}$
Now, $BA\cdot BC=(2\hat{i}+2\hat{j}+3\hat{k})\cdot (\hat{i}+\hat{j}+2\hat{k})$
$=2\times 1+2\times 1+3\times 2=10$
$\cos \theta =\frac{BA\cdot BC}{|BA||BC|}$
$\Rightarrow \cos (\angle ABC)=\frac{10}{\sqrt{17}\sqrt{6}}$
$\Rightarrow \angle ABC={\cos }^{-1}\left(\frac{10}{\sqrt{102}}\right)$