The vectors a , b and c are of the same length. If taken pairwise,...

Given that, $\vec{a}=\hat{i}+\hat{j}, \vec{b}=\hat{j}+\hat{k}$ Since, $\vec{a}, \vec{b}, \vec{c}$ has same length. $\Rightarrow|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=\sqrt{1+1}=\sqrt{2}$ Let $\theta$ be the angle between the vectors. $\therefore \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{0+1+0}{\sqrt{2} \times \sqrt{2}}=\frac{1}{2}\because$ All the conditions are satisfied, so it can be vector $\overrightarrow{\mathrm{c}}$ (I) Take $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}+\hat{\mathrm{k}} \Rightarrow|\overrightarrow{\mathrm{c}}|=\sqrt{1+1}=\sqrt{2}$ and $\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}|}=\frac{1+0+0}{\sqrt{2} \sqrt{2}}=\frac{1}{2}$ All $\overrightarrow{\mathrm{c}}$ the conditions are satisfied. So, it can be vector. (II) Take $\overrightarrow{\mathrm{c}}=\frac{\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{3} \Rightarrow|\overrightarrow{\mathrm{c}}|=\frac{1}{3} \sqrt{1+16+1}=\frac{3 \sqrt{2}}{3}=\sqrt{2}$ and $\cos \theta=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}|}=\frac{\frac{-1}{3}+\frac{4}{3}}{\sqrt{2} \sqrt{2}}=\frac{1}{2}\because$ All the conditions are satisfied, so it can be vector $\overrightarrow{\mathrm{c}}$