If $V_1$ and $V_2$ are any two vectors with the same initial point and $k_1, k_2$ are any two scalars, prove that $k_1V_1+k_2V_2$ lies in the same plane as $V_1$ and $V_2$.
I'm not sure if I'm going about this correctly, but I'll give it a shot.
Proof. Since $V_1$ and $V_2$ share the same initial point, say $O$, we have $k_1V_1$ and $k_2V_2$ share the same initial point $O$. Let $k_1V1 = \vec{OQ}$ and $k_2V_2 = \vec{OP}$. Then
$$
k_1V_1+k_2V_2 = \vec{OP}+\vec{OQ} \tag1
$$
This sum is defined by constructing a vector $\vec{PR}=\vec{OQ}$ which uniquely determines the point $R$, and thus the resultant vector of $(1)$ is $\vec{OR}$.
If $\vec{OP}$ and $\vec{OQ}$ do not lie in the same straight line, then we have parallelogram $OQRP$, with $\vec{OR}$ lying in the diagonal of $OQRP$. And so
$$
\vec{OR} = \vec{OP}+\vec{OQ} = k_1V_1+k_2V_2
$$
lies in parallelogram $OQRP$ as do vectors $V_1$ and $V_2$.
Similar conclusion if $\vec{OP}$ and $\vec{OQ}$ do lie in the same straight line.
Have I done anything here? Please help.
Best Answer
Consider a normal vector Ñ defining the plane of interest(P) such that: Ñ•V¹ = O and Ñ•V² = O. For any vector V to lie in plane P, V•Ñ = O. Considering V=k¹V¹+k²V², (k¹V¹+k²V²)•Ñ = k¹(V¹•Ñ)+k²(V²•Ñ) = k¹(0) + k²(0) = 0, Implying that our desired vector lies in P