The equation of a plane containing the point $(1,-1,1)$ and parallel to the plane $2 x+3 y-4 z=17$ is

The equation of a plane containing the point $(1,-1,1)$ and parallel to the plane $2 x+3 y-4 z=17$ is
  1. $\bar{r}_{\cdot}(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{k})=-5$
  2. $\bar{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{k})=-15$
  3. $\bar{r} \cdot(4 \hat{\imath}+3 \hat{\jmath}-4 \hat{k})=-3$
  4. $\bar{r} \cdot(3 \hat{\imath}+4 \hat{\jmath}-2 \hat{k})=-3$

Solution

Equation of plane passing through the point having position vector $\bar{a}$ and normal to $\overline{\mathrm{n}}$ is $\therefore \overline{\mathrm{r}} \cdot \overline{\mathrm{n}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=2-3-4=-5$

Asked in: MHT CET 2020 (12 Oct Shift 1)

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