let $$\mathbb{R}^{3}\supseteq S = span_{\mathbb{R}}{\left(\begin{array}{c}1\\2\\3\end{array}\right)} , \left(\begin{array}{c}4\\5\\6\end{array}\right)$$
Is a Plane going through Cartesian coordinate system - $(0, 0, 0)$.
I need to prove that S is the Kernel of equation of the form $$ax + by + cz = 0$$ for $a, b ,c \in\mathbb{{R}}$ (given the assumption which not all objects = $0$)
i tried to put a, b , c in a matrix of the form: $$\begin{array}{ccc} a & 2b & 3c\\ 4a & 5b & 6c \end{array}$$
but when i ranked it i could only say what a and b are in terms of c, which i don't think is the answer. what did i do wrong? (it's defiantly not because of a mistake in the ranking of the matrix)
When you have a plane in our 3 dimensional world, the polynomial equation $ax+by+cz=d$ can be found as follow: Find a normal vector that is normal to both directional vectors, i.e. ${\left(\begin{array}{c}1\\2\\3\end{array}\right)} , \left(\begin{array}{c}4\\5\\6\end{array}\right)$. This can be done with the dot product (that's handy when the problem is expanded into a higher dimension) or the cross product, which works here. Since you indicated that you know how to do cross product, I'll leave it up to you to verify that a normal vector to both directional vectors is $(-3,6,-3)$ and thus the equation of the plane would become $-3x+6y-3z=d$ with $d$ in this case being zero. You may simplify if needed. Verify by substitution that both directional vectors are a solution to the equation (and therefore any linear combination of them will also be a solution)