Question

Write the sum of intercepts cut off by the plane $\overrightarrow{r}.(2\hat{i}+\hat{j}-\hat{k})-5=0$ on the three axes.

Answer 1

The equation of plane is $r.(2i+j-k)-5=0$ , here $r$ is position vector of a point on the given plane.
Let the position vector $r$ be $xi+yj+zk$ , we get $(xi+yj+zk).(2i+j-k)-5=0$
which implies the equation is $2x+y-z-5=0$
The plane cuts $x$-axis at $5/2$ , $Y$-axis at $5$ and $Z$-axis at $-5.$
So the $x$-intercept is $(\frac{5}{2},0,0)$ , $Y$-intercept is $(0,5,0)$ and $Z$-intercept is $(0,0,-5).$