The length of the perpendicular to the plane $\bar{r} \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=14$ from…

The length of the perpendicular to the plane $\bar{r} \cdot(\hat{\imath}-2 \hat{\jmath}+3 \hat{k})=14$ from the origin is
  1. $\sqrt{7}$ units
  2. 7 units
  3. 14 units
  4. $\sqrt{14}$ units

Solution

Length of $\perp$ er from the point $A(\bar{a})$ to the plane $\bar{r} \cdot \bar{n}=p$ is $\frac{|\bar{a} \cdot \bar{n}-p|}{|\bar{n}|}$ Here $\overline{\mathrm{a}}=0 \hat{\mathrm{i}}+0 \hat{\mathrm{j}}+0 \hat{\mathrm{k}}$ and $\overline{\mathrm{n}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ $\therefore \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=0 \quad$ and $|\overline{\mathrm{n}}|=\sqrt{1+4+9}=\sqrt{14}$ Hence required distance is $\frac{|0-14|}{\sqrt{14}}=\sqrt{14}$

Asked in: MHT CET 2020 (12 Oct Shift 2)

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