Question

Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.

Answer 1

The equation of the plane in the intercept form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where a, b and c are the intercepts on the x, y and z-axis, respectively.

It is given that the intercepts made by the plane on the x, y and z-axis are 3, –4 and 2, respectively.

∴ a = 3, b = −4, c = 2

Thus, the equation of the plane is

$$\begin{gathered} \frac{x}{3}+\frac{y}{\left(-4\right)}+\frac{z}{2}=1 \\ \Rightarrow 4x-3y+6z=12 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(4\hat{i}-3\hat{j}+6\hat{k}\right)=12 \\ \Rightarrow \overrightarrow{r}.\left(4\hat{i}-3\hat{j}+6\hat{k}\right)=12 \end{gathered}$$
This is the vector form of the equation of the given plane.