Question

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

Answer 1

Intercepts are $a=3,b=-4,c=2$

The intercept form of a plane is as follows:

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

$\therefore \frac{x}{3}+\frac{y}{-4}+\frac{z}{2}=1$

The euqation of the given plane is

$4x-3y+6z=12$.

$(x\hat{i}+y\hat{j}+z\hat{k}).(4\hat{i}-3\hat{j}+6\hat{k})=12$

$\overrightarrow{r}.(4\hat{i}-3\hat{j}+6\hat{k})=12$