I have the following question here:
Let $$\vec{u}=(u_1,u_2,u_3),\vec{v}=(v_1,v_2,v_3),\vec{w}=(w_1,w_2,w_3)$$
be $3$ vectors in $\mathbb{R}^3$. Define a matrix whose rows are these three vectors:
$$\begin{bmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{bmatrix}$$
a) Show that $$\det(A)=\vec{u} \cdot (\vec{v} \times \vec{w})$$
This proof was trivial. Simply compute $\vec{v} \times \vec{w} = \begin{vmatrix} i & j & k \\ v_1 & v_2 & v_3\\ w_1 & w_2 & w_3\\
\end{vmatrix}$ and dot the result with $\vec{u}=(u_1,u_2,u_3)$. Compute the determinant (I used cofactor expansion) and you'll get both sides to match.
b) Suppose that $\vec{v},\vec{w}$ are not colinear. Show that $\vec{v} \times \vec{w}$ is normal to the plane given by $\text{Span}\{\vec{v},\vec{w}\}$.
Since $\vec{v}$ and $\vec{w}$ are not collinear, this implies any vecotor can be written as the span of the two vectors. Let $\vec{u}=\alpha\vec{v}+\beta\vec{w}$ where $\alpha, \beta$ are real numbers. If $\vec{u}$ is normal to $\vec{v} \times \vec{w}$ then $\vec{u}\cdot(\vec{v} \times \vec{w})$ implies:
$\vec{u}\cdot(\vec{v} \times \vec{w})$
$(\alpha\vec{v}+\beta\vec{w})\cdot(\vec{v} \times \vec{w})$
$(\alpha\vec{v})\cdot(\vec{v} \times \vec{w})+(\beta\vec{w})\cdot(\vec{v} \times \vec{w})=0$
It's $0$ since two of the rows are the same so the determinant (Since the is the same as part a) evaluates to $0$ as well.
c) Use parts (a) and (b) to argue that $\vec{u},\vec{v},\vec{w}$ all lie on the same plane if and only if $\det(A)=0$.
I can do this geometrically quite easily since this is just the volume of the parallelipiped spanned by the three vectors $\vec{u}$,$\vec{v}$ and $\vec{w}$.
Is there a way way to argue this algebraically though? I keep reading various links about how this relates to the normal vector but I can't seem to put it together.
The most common answer I see is:
"If $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are coplanar, their triple product $\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})$ is $0$ since $\mathbf{u}$ must be perpendicular to $\mathbf{v} \times \mathbf{w}$."
Why though? I feel silly asking this but why does $\vec{u}$ have the be orthogonal to $\vec{v}$ and $\vec{w}$? I feel like I'm missing a very simple and obvious step here.
Can someone draw out some kind of visual for this?
Best Answer
This could also follow by the point-normal definition of a plane: $\mathbf{\vec{n}}\cdot \vec{x} = 0$.
Given two points $x_1,x_2 \in \mathcal{P}$ then the vector $\vec{x} = \mathbf{x_2} - \mathbf{x_1 }\in \mathcal{P}$, and the normal has the property that is perpendicular to all vectors in $\mathcal{P}$:
If $\vec{x} \in \mathcal{P}$ (the plane), then $\mathbf{\vec{n}}\cdot \vec{x} = 0$.
The normal of the plane $\mathcal{P}_{\vec{v},\vec{w}}$ made by noncolinear vectors $\vec{v},\vec{w}$ is $\mathbf{\vec{n}} = \vec{v} \times \vec{w}\not= \vec{0}$, and if $\vec{u} \in \mathcal{P}_{\vec{v},\vec{w}}$ then $\det(A)=(\vec{v} \times \vec{w})\cdot \vec{u}= \mathbf{\vec{n}}\cdot \vec{u} =0$.
If $\vec{v},\vec{w}$ are colinear then $\mathbf{\vec{n}}=\vec{0}$ and certainly $\vec{0}\cdot\vec{u} =0$.
There's an image of the normal vector to a plane in this question.