The angle between the vectors $2\hat{k}-3\hat{j}$ and $\hat{i}-2\hat{k}$ is

Given Vectors are $2\hat{k}-3\hat{j} \& \hat{i}-2\hat{k}$

Now, For $\overrightarrow{a} \& \overrightarrow{b}$ their dot product is $\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}=\left|\overrightarrow{\mathrm{a}}\right|\left|\overrightarrow{\mathrm{b}}\right|\mathrm{cos\theta }$

$\Rightarrow \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}$

$\therefore \cos \theta =\frac{\left(2\hat{k}-3\hat{j}\right).\left(\hat{i}-2\hat{k}\right)}{\left|2\hat{k}-3\hat{j}\right|\left|\hat{i}-2\hat{k}\right|}$

                  $=\frac{-4}{\sqrt{{\left(2\right)}^2+{\left(-3\right)}^2}\times \sqrt{{\left(1\right)}^2+{\left(-2\right)}^2}}$

     $\Rightarrow \cos \theta =\frac{-4}{\sqrt{65}}$

$\Rightarrow \mathrm{\theta }={\cos }^{-1}\left(\frac{-4}{\sqrt{65}}\right)$