If the vector equation or the plane $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}})+\lambda…
If the vector equation or the plane $\overline{\mathrm{r}}=(2 \hat{\mathrm{i}}+\hat{\mathrm{k}})+\lambda \hat{\mathrm{i}}+\mu(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{i}+2 \hat{k})=\alpha$ then $\alpha=$
2
3
1
0
Solution
Here given plane passes through the point $(2,0,1)$ and let $\overline{\mathrm{b}}=\hat{\mathrm{i}}$ and $\bar{c}=\hat{i}+2 \hat{j}-3 \hat{k}$.
Normal vector $\overline{\mathrm{n}}=\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 0 & 0 \\ 1 & 2 & -3\end{array}\right|=3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
The equation of plane in scalar product from is $\overline{\mathrm{r}} \cdot \overline{\mathrm{n}}=\overline{\mathrm{a}} \cdot \overline{\mathrm{n}}$
Here $\bar{a} \cdot \bar{n}=(2 \hat{i}+\hat{k}) \cdot(3 \hat{j}+2 \hat{k})=2$
$\therefore \alpha=2$