Let $\bar{u}=\hat{i}+\hat{j}, \bar{v}=\hat{i}-\hat{j}$ and $\bar{w}=\hat{i}+2 \hat{j}+3 \hat{k}$. If…

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

Solution

We have, $\overline{\mathrm{u}} \cdot \hat{\mathrm{n}}=0$ and $\overline{\mathrm{v}} \cdot \hat{\mathrm{n}}=0$ $\therefore \quad \hat{\mathrm{n}} \perp \overline{\mathrm{u}}$ and $\hat{\mathrm{n}} \perp \stackrel{\rightharpoonup}{\mathrm{v}}$ $\Rightarrow \hat{\mathrm{n}}= \pm \frac{\overline{\mathrm{u}} \times \overline{\mathrm{v}}}{|\overline{\mathrm{u}} \times \overline{\mathrm{v}}|}$
Now, $\bar{u} \times \bar{v}=(\hat{i}+\hat{j}) \times(\hat{i}-\hat{j})=-2 \hat{k}$ $\therefore \quad \hat{\mathrm{n}}= \pm \hat{\mathrm{k}}$
Hence, $|\overline{\mathrm{w}} \cdot \hat{\mathrm{n}}|=|(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot( \pm \hat{\mathrm{k}})|=3$

Asked in: MHT CET 2024 (15 May Shift 1)

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