Find a vector of magnitude 4 units perpendicular to each of the vectors 2 i - j + k and i + j - k and hence verify your answer.

Let $\displaystyle \overrightarrow{\mathbf{a}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\displaystyle \overrightarrow{\mathbf{c}}$ be the vector perpendicular to both $\displaystyle \overrightarrow{\mathbf{a}} \& \overrightarrow{\mathbf{b}}$ then, $\displaystyle \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right|=3 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ Let $\displaystyle \overrightarrow{\mathbf{d}}$ the vector perpendicular to both the vectors $\displaystyle \overrightarrow{\mathbf{a}} \& \overrightarrow{\mathbf{b}}$ and having magnitude $\displaystyle 4$, $\displaystyle \overrightarrow{\mathbf{d}}=4 \hat{\mathbf{c}}=2 \sqrt{2} \hat{\mathbf{j}}+2 \sqrt{2} \hat{\mathbf{k}}($ or $\displaystyle -2 \sqrt{2} \hat{\mathbf{j}}-2 \sqrt{2} \hat{\mathbf{k}})$ Verification: $\displaystyle |\overrightarrow{\mathbf{d}}|=\sqrt{\mathbf{8 + 8}}=\mathbf{4}, \overrightarrow{\mathbf{d}} \cdot \overrightarrow{\mathbf{a}}=\overrightarrow{\mathbf{d}} \cdot \overrightarrow{\mathbf{b}}=\mathbf{0} \Rightarrow \overrightarrow{\mathbf{d}} \perp \overrightarrow{\mathbf{a}}$ and $\displaystyle \overrightarrow{\mathbf{d}} \perp \overrightarrow{\mathbf{b}}$