Finding the Equation of a Plane provided with normal vector and point

Question

Okay, so I am trying to find an equation of a plane that passes through the point (-1,3,5) and has a normal vector of <2,4,-3>.

The format of the answer must be such that

_____x + _____y + ______z +1 = 0

My work so far

Okay, so I attempted to start using point-normal form for the plane. Given that we already know the attitude numbers to be [2,4,-3] from the normal vector.

Working with this I get:

2(x+1) + 4(y-3) -3(z-5) = 0 Converting to standard form I get 2x +4y -3z +5=0

My problem is that I can not figure out how to get 1 as the constant rather than 5.

Answer 1

I think you've made a mistake with your expansion: $$2(x+1)+4(y-3)-3(z-5)=0$$ $$2x+4y-3z\color{#bb0000}{+5}=0$$

To get it in the form required, just simply divide the whole equation by $5$: $$2x+4y-3z+5=0$$ $$\frac{2x+4y-3z+5}{5}=\frac{0}{5}$$ $$\frac{2}{5}x+\frac{4}{5}y-\frac{3}{5}z+1=0$$ Which is in the form $ax+by+cz+1=0$.

Answer 2

Why don't you just divide what you already have by $-5$?