I have a question here which says:
"The solution set of:
$$ax+by+cz=0$$ $$dx+ey+fz=0$$
is a line in $\mathbb{R^3}$"
I am supposed to show that this is true or false. If it's true, I have to explain why and if it's false, I have to explain why it's false.
I don't think it's true. We COULD have a line if the planes intersected but if $a,b,c$ are scalar multiples of $d,e,f$, then we would just have the same plane since there are infinitely many solutions. Is that the right should process or did I mess up my reasoning?
When absolutely nothing is provided about $a,b,c,d,e,f$ then for one you may even just take say all of them to be equal to zero, and then the system of solutions is the whole of $\mathbb R^3$, which is not a line.
Of course, a plane is also obtainable : it is obtained exactly when the ratios $d:a,e:b,f:c$ are the same (else, the planes would not be parallel and hence would intersect at a line passing through zero). Then, the equations $ax+by+cz = 0$ and $dx+ey+fz = 0$ describe the same plane. This is clearly seen for example with $a,b,c,d,e,f$ all equal to $1$, or say $a,b,c=1$, $d,e,f = 2$. So, obviously the statement given can be false.
Note , however, that a solution must exist. This is because of the rank-nullity theorem, applied on the linear transformation $(x,y,z) \to (ax+by+cz,dx+ey+fz)$. This is a map from $\mathbb R^3$ to $\mathbb R^2$. It has rank atmost $2$, therefore nullity at least one i.e. the null space has at least one line.