Physics - Vectors Question with Solution | TestHub

$AB=3i+j+\hat{k}$ and $AC=i+2j+i$ $8A=-(3i+j+k)$ $\angle ABC$ is angle between $BA$ and $BC$ $BC=AC+BA$ $=AC-AB$ $=\hat{i}+2\hat{j}+\hat{k}-(3\hat{i}+\hat{j}+\hat{k})=-2\hat{i}+\hat{j}$ $BA\cdot BC=|BA\Vert BC|\cos \theta$ $-(3\hat{i}+\hat{j}+\hat{k})\cdot (-2\hat{i}+\hat{j})$ $=\left(\sqrt{3^2+1^2+1^2}\right)\cdot \left(\sqrt{2^2+1^2}\right)\cos \theta$ $\Rightarrow 6-1=\sqrt{11\times 5}\cdot \cos \theta$ $\cos \theta =\frac{5}{\sqrt{55}}=\sqrt{\frac{5}{11}}=\theta ={\cos }^{-1}=\sqrt{\frac{5}{11}}$