Find the angle between the vectors \overrightarrow{\mathrm{a}}=\mathrm{i}+\mathrm{j}-\mathrm{k} and \overrightarrow{\mathrm{b}}=\mathrm{i}-\mathrm{j}+\overrightarrow{\mathrm{k}}
Find the angle between the vectors \overrightarrow{\mathrm{a}}=\mathrm{i}+\mathrm{j}-\mathrm{k} and \overrightarrow{\mathrm{b}}=\mathrm{i}-\mathrm{j}+\overrightarrow{\mathrm{k}}

    \[\begin{array}{l} \overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \quad \text { and } \quad \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}} \\ |\overrightarrow{\mathrm{a}}|=\sqrt{1^{2}+1^{2}+(-1)^{2}}=\sqrt{3} \\ |\overrightarrow{\mathrm{b}}|=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{3} \\ \overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}=1.1+1(-1)+(-1) .1=-1 \\ \overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{b}}| \cos \theta \\ -1=\sqrt{3} * \sqrt{3} \cos \theta \\ \Longrightarrow \cos \theta=-\frac{1}{3} \\ \Longrightarrow \theta=\cos ^{-1}\left(-\frac{1}{3}\right) \end{array}\]