Let $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{i} - \hat{j}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{a} \cdot \hat{n} = 0$ and $\vec{b} \cdot \hat{n} = 0$, then find $|\vec{c} \cdot \hat{n}|$.

1

Given, $\vec{a} = \hat{i} + \hat{j}, \vec{b} = \hat{i} - \hat{j}$

and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$

Also, given $\vec{a} \cdot \hat{n} = 0$

and $\vec{b} \cdot \hat{n} = 0$

Here, $\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

Here, $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{vmatrix}$

$= \hat{i}(0 - 0) - \hat{j}(0 - 0) + \hat{k}(-1 - 1) = -2\hat{k}$

$∴\hat{n} = \frac{-2\hat{k}}{\sqrt{(-2)^2}} = -\hat{k}$

Therefore, $|\vec{c} \cdot \hat{n}| = |(\hat{i} + \hat{j} + \hat{k}) \cdot (-\hat{k})| = |-1| = 1$