The unit vector perpendicular to the plane $4 x-3 y+12 z=15$ is

The unit vector perpendicular to the plane $4 x-3 y+12 z=15$ is
  1. $\frac{4 \hat{\imath}+3 \hat{\jmath}+12 \hat{k}}{13}$
  2. $\frac{4 \hat{\imath}-3 \hat{\jmath}+12 \hat{k}}{13}$
  3. $\frac{-4 \hat{\imath}+3 \hat{\jmath}+12 \hat{k}}{13}$
  4. $\frac{-4 \hat{\imath}-3 \hat{\jmath}+12 \hat{k}}{13}$

Solution

The unit vector perpendicular to plane $4 x-3 y+12 z=15$ is $\frac{4 \hat{i}-3 \hat{j}+12 \hat{k}}{\sqrt{16+9+144}}=\frac{4 \hat{i}-3 \hat{j}+12 \hat{k}}{13}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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