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. $\overline{\mathrm{r}} \cdot(2 \hat{\imath}-3 \hat{\jmath}-4 \hat{\mathrm{k}})=-1$
  2. $\overline{\mathrm{r}} \cdot(\hat{\imath}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=3$
  3. $\overline{\mathrm{r}} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}})=-5$
  4. $\overline{\mathrm{r}} \cdot(2 \hat{\imath}+3 \hat{\jmath}-4 \hat{\mathrm{k}})=5$

Solution

Here $\bar{a}=\hat{i}-\hat{j}+\hat{k}$ $\& \quad \bar{n}=2 \hat{i}+3 \hat{j}-4 \hat{k}$ Now $\bar{a} \cdot \bar{n} \quad=2-3-4=-5$ Vector equation of plane passing through $\mathrm{A}(\bar{a})$ is $\begin{aligned} & \overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}} \\ \Rightarrow & \overline{\mathrm{r}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})=-5 \end{aligned}$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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