The equation of the plane through $(-1,1,2)$ whose normal makes equal acute angles with co-ordinate axes is

The equation of the plane through $(-1,1,2)$ whose normal makes equal acute angles with co-ordinate axes is
  1. $x+y+z-3=0$
  2. $x+y+z-2=0$
  3. $x+y-z-2=0$
  4. $x-y+z-3=0$

Solution

Note that $(-1,1,2)$ is satisfied by only option (B) Alternate Method: Let $A \equiv(-1,1,2)$ $\begin{aligned} \therefore \quad \bar{a} & =-\hat{i}+\hat{j}+2 \hat{k} \\ & \bar{n}=\hat{i}+\hat{j}+\hat{k} \end{aligned}$ $\therefore \quad$ equation of plane is $\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}$ $\begin{aligned} & \Rightarrow \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\ & \Rightarrow \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2 \\ & \Rightarrow(x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=2 \\ & \Rightarrow x+y+\mathrm{z}-2=0 \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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