Let $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}},…

Let $\overrightarrow{\mathrm{u}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{v}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{w}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. If $\hat{\mathrm{n}}$ is a unit vector such that $\vec{u} \cdot \hat{n}=0$ and $\overrightarrow{\mathrm{v}} \cdot \hat{\mathrm{n}}=0$, then $|\overrightarrow{\mathrm{w}} \cdot \hat{\mathrm{n}}|$ is equal to
  1. 3
  2. 0
  3. 1
  4. 2

Solution

Since $\overrightarrow{\mathrm{n}}$ is perpendicular $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}, \overrightarrow{\mathrm{n}}=\overrightarrow{\mathrm{u}} \times \overrightarrow{\mathrm{v}}$ $\hat{\mathrm{n}}=\frac{\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 1 & 0 \\ 1 & -1 & 0\end{array}\right|}{\sqrt{2} \times \sqrt{2}}=\frac{-2 \hat{\mathrm{k}}}{2}=-\hat{\mathrm{k}}$ $|\vec{\omega} \cdot \hat{n}|=|(\mathrm{i}+2 \mathrm{j}+3 \mathrm{k}) \cdot(-\mathrm{k})|=|-3|=3$

Asked in: JEE Main 2003

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