The Cartesian equation of the plane…

The Cartesian equation of the plane $\bar{r}=(\hat{i}-\hat{j})+\lambda(\hat{i}+\hat{j}+\hat{k})+\mu(\hat{i}-2 \hat{j}+3 \hat{k})$ is
  1. $\mathrm{x}+\mathrm{y}+\mathrm{z}=0$
  2. $5 \mathrm{x}+2 \mathrm{y}+3 \mathrm{z}=0$
  3. $2 x+y+z=0$
  4. $5 \mathrm{x}-2 \mathrm{y}-3 \mathrm{z}-7=0$

Solution

$\begin{aligned} & \overline{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}, \overline{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}, \overline{\mathrm{B}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \end{aligned}$ $\overline{\mathrm{n}}$ is $\perp$ er $\overline{\mathrm{A}}$ and $\overline{\mathrm{B}}$ $\begin{aligned} & \therefore \overline{\mathrm{n}}=\overline{\mathrm{A}} \times \overline{\mathrm{B}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{array}\right| \\ & =\hat{\mathrm{i}}(3+2)-\hat{\mathrm{j}}(3-1)+\hat{\mathrm{k}}(-2-1)=5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\ & \overline{\mathrm{a}} \cdot \overline{\mathrm{n}}=\mathrm{d}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}) \cdot(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=5+2=7 \\ & \therefore \overline{\mathrm{r}} \cdot(5 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})=7 \\ & \Rightarrow \text { Cartesian equation is } 5 \mathrm{x}-2 \mathrm{y}-3 \mathrm{z}-7=0 \end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 2)

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