how do I find the general equation for the plane with vector equation:

Question

Solving: Need a normal vector $n = [a,b,c]$.

We already know two directional vectors: $d_1= [-2,0,3]$, $d_2= [3,-1,2]$ $$ \left\lbrace\begin{array}{} n \times d_1 = 0 \\ n \times d_2 = 0 \\ \end{array}\right. \;\Rightarrow\; \left\lbrace\begin{array}{} -2a + 3c = 0 \\ 3a - b + 2c = 0\\ \end{array}\right. $$ let $c = 1$:

$a=\frac{3}{2}, b=-\frac{13}{2}$

$n= \left[\frac{3}{2}, -\frac{13}{2}, 1\right]$ then I get stuck here.

I'm not really sure if i'm doing this right, anything to help finish this question will be greatly appreciated.

Answer 1

We have given $$-2a+3c=0$$ and $$3a-b+2c=0$$ form the first equation we get

$$c=\frac{2}{3}a$$ so we get with the seond equation $$3a-b+\frac{4}{3}a=0$$ so $$a=\frac{3}{13}b$$ and $$c=\frac{2}{13}b$$ and we get $$a=\frac{3}{13}b,b=b,c=\frac{2}{13}b$$

Answer 2

Let $c=1 \implies a=\frac32$ and $3a-b+2=0\implies b=\frac{13}2$.

So $n=(\frac32,\frac{13}2,1)$.

Now the equation of the plane is: $$3x+13y+2z=d$$. To get $d$ plug in a point, say $(4,5,-2)$: $3\cdot4+13\cdot 5+2\cdot (-2)=73$.

Thus $$3x+13y+2z=73$$ is an equation of the plane.