I need help with finding the cartesian equation for the plane for the following conditions:
$\pi_2$ passed B(4,2,-1) and C(2,2,3) and is also parallel to x-axis.
Usually if the question gives me three points, I will be able to obtain two vector from which I can find the normal vector of the plane (Conducting a cross product of these two vectors).
So where I can find the third point in this question?
The answer key says the right equation is $y=2$ but I am not sure how they obtained this?
Best Answer
General equation of a plane in cartesian form : $$ ax + by + cz = d$$ Given that the plane is parallel to $x$-axis $\Rightarrow a = 0 \Rightarrow$ , the equation becomes $ by + cz = d$, On putting the values $ (4 ,2 ,-1), (2,2,3)$ in the equation, and subtracting one equation from other, you would find $z = 0 \Rightarrow$ The equation is of form $by =d$, On putting the values in the equation you would find $ \dfrac{d}{b} = 2 \Rightarrow$ Equation of the plane is $y =2$