Show that if $v_1$ and $v_2$ are any two vectors in this plane, then for any two scalars, $c_1v_1 + c_2v_2$ is also a vector in the plane

Question

Let $a,\,b$ and $c$ be constants (not all zero) and consider the equation $ax + by + cz = 0$, which has a graph that is a plane that passes through the origin in $\mathbb{R}^3$. Show that if $v_1$ and $v_2$ are any two vectors in this plane, then for any two scalars, $c_1$ and $c_2$, $c_1v_1 + c_2v_2$ is also a vector in the plane (hint: use the parametric form of the plane).

I'm not even sure I understand what this question is asking me to show. (and I'm not sure what the parametric form is referring to). How do I show $c_1v_1 + c_2v_2$ is a vector in this plane? How do I even figure out what the actual plane is in relation to this vector?

Answer 1

The hint makes ittle sense indeed. If $v_1=(x_1,y_1,z_1)$ is in the plane, then this means that $$\tag1ax_1+by_1+cz_1=0.$$ And if $v_2=(x_2,y_2,z_2)$ i sin the plane, this mean sthat $$\tag2ax_2+by_2+cz_2=0.$$ Then $c_1c_1+c_2v_2$ is the vector $(c_1x_1+c_2v_2,c_1y_1+c_2y_2,c_1z_1+c_2z_2)$. Use $(1)$ and $(2)$ to show that the expression $$a(c_1x_1+c_2v_2)+b(c_1y_1+c_2y_2)+c(c_1z_1+c_2z_2) $$ evaluates to $0$, thus showing that that $c_1c_1+c_2v_2$ is in the plane

Answer 2

If $\vec{v}_1$ and $\vec{v}_2$ are vectors in the plane then you know that $\vec{N}\circ\vec{v} = 0$ where $\vec{N}$ is the normal vector defining the plane. This is all you need, because then you would get:

$$ \vec{N}\circ\left(c_1\vec{v}_1 + c_2\vec{v}_2\right) = c_1\left(\vec{N}\circ\vec{v}_1\right) + c_2\left(\vec{N}\circ\vec{v}_2\right) = 0 $$

It's zero by the fact that $\vec{v}_1$ and $\vec{v}_2$ are originally in the plane and thus $\vec{N}\circ\vec{v}_1 = 0$ and $\vec{N}\circ\vec{v}_2 = 0$.

A second interpretation of the question:

If you are not talking about vectors, but rather coordinates (which are not vectors), then the question is completely different.

If we have a planar equation:

$$ \vec{N}\circ \vec{x} = \vec{N}\circ \vec{x}_0 $$

Then we are given two values $\vec{v}_1$ and $\vec{v}_2$ such that the above equation is satisfied:

$$ \vec{N}\circ \vec{v}_1 = \vec{N}\circ \vec{x}_0\\ \vec{N}\circ \vec{v}_2 = \vec{N}\circ \vec{x}_0 $$

The question is what is the following value:

\begin{align} \vec{N}\circ\left(c_1\vec{v}_1 + c_2\vec{v}_2\right) \stackrel{?}{=} & \vec{N}\circ \vec{x}_0 \\ =&c_1\vec{N}\circ\vec{v}_1 + c_2\vec{N}\circ\vec{v}_2 \\ =& c_1\vec{N}\circ\vec{x}_0 + c_2\vec{N}\circ\vec{x}_0 \\ =& \left(c_1 + c_2\right)\vec{N}\circ\vec{x}_0 \\ \neq & \vec{N}\circ \vec{x}_0 \end{align}

It's clear that this is not true in general, unless $c_1 + c_2 = 1$ and/or $\vec{N}\circ\vec{x}_0 = 0$ (as explained below).

Particular planar equation: $\vec{N}\circ\vec{x} = 0$:

In this particular case, the statement is true but that is because it's equivalent to the the first thing I said--the coordinate becomes the vector. Algebraically it's true because $\vec{N}\circ\vec{x}_0 = 0$ and thus $\left(c_1 + c_2\right)\cdot 0 = 0 = \vec{N}\cdot\vec{x}_0$ for any $c_1$ or $c_2$.