proving that $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is the normal to the plane.

Question

$P$, $Q$, and $R$ are points in $ \mathbb{R}^3 $ which are not on the same line. if $\vec{a} = \vec{OP}$, $\vec{b} = \vec{OQ}$, and $\vec{c} = \vec{OR}$, show that $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$ is perpendicular to the plane containing $P$, $Q$, and $R$.

So far, I have defined:

$\vec{a} = <a_1, a_2, a_3> , \vec{b} = <b_1, b_2, b_3> , \vec{c} = <c_1, c_2, c_3> $

The lines spanning between the points a, b, and c.

$ \vec{ab} = <b_1 - a_1, b_2 - a_2, b_3 - a_3> $

$ \vec{bc} = <c_1 - b_1, c_2 - b_2, c_3 - b_3> $

$ \vec{ca} = <a_1 - c_1, a_2 - c_2, a_3 - c_3> $

The perpendicular lines to the planes ab, bc, and ca.

$ \vec{a \times b} = <a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1> $

$ \vec{b \times c} = <b_2c_3 - b_3c_2, b_3c_1 - b_1c_3, b_1c_2 - b_2c_1> $

$ \vec{c \times a} = <c_2a_3 - c_3a_2, c_3a_1 - c_1a_3, c_1a_2 - c_2a_1> $

Adding them I get:

$ <a_2b_3 + b_2c_3 + c_2a_3 - a_3b_2 - b_3c_2 - c_3a_2, a_3b_1 + b_3c_1 + c_3a_1 - a_1b_3 - b_1c_3 - c_1a_3, a_1b_2 + b_1c_2 + c_1a_2 - a_2b_1 - b_2c_1 - c_2a_1> $

Dot-producting the vectors $\vec{ab}$, $\vec{bc}$, and$\vec{ca}$ with the vectors $\vec{a \times b}$, $\vec{b \times c}$, and $\vec{c \times a}$ expands like crazy.

What I have been going for so far is that since vectors $\vec{ab}$, $\vec{bc}$, and$\vec{ca}$ make up the plane, if I dot product each one with the cross-product, $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$, it should result in zero, or at least everything cancelling out since this is supposed to be the perpendicular to the plane. Maybe I'm just tired, but it doesn't seem to be resulting in that.

I feel like I'm getting close, but not quite getting the result that I am looking for. What exactly am I missing? Am I going in the right direction even or is there something completely obvious that I am missing?

Answer 1

Since $P,Q,R$ are non-collinear, $\vec{PQ}$ and $\vec{PR}$ are two linearly independent vectors in the plane of $PQR$. Using the notation provided in the question,

$$\vec{PQ}=\vec{OQ}-\vec{OP}=\vec b - \vec a,\\ \space \vec{PR}=\vec{OR}-\vec{OP}=\vec c - \vec a.$$

The cross product of $\vec{PQ}$ and $\vec{PR}$ is normal to $PQR$. Repeated application of the linearity property and anti-symmetry property of the cross product,

$$\vec{PQ}\times\vec{PR}=\left(\vec b - \vec a\right)\times\left(\vec c - \vec a\right)\\ =\vec b \times \left(\vec c - \vec a\right) - \vec a \times \left(\vec c - \vec a\right)\\ =\vec b \times \vec c - \vec b \times \vec a - \vec a \times \vec c + \vec a \times \vec a\\ =\vec b \times \vec c + \vec a \times \vec b + \vec c \times \vec a + \vec0\\ =\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a.$$

Answer 2

Note: Nothing changes under a cyclic permutation ($a \rightarrow b \rightarrow c$)

The scalar product $$ ((a \times b)+(b\times c)+(c \times a)) \cdot (b - a) =\\ $$ reduces to $$ (b \times c)\cdot a - (c \times a) \cdot b =0 $$

Answer 3

You are right so far. The sum of cross products seems to be correct, and when I compute the dot products with it I get zero. Make sure you are using $\vec{ab} = \vec{b} - \vec{a}$.

For another way to do this, just use algebra:

$$\left(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \circ \left(\vec{b} - \vec{a}\right)$$

Distribute property of dot product:

$$\left(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \circ \vec{b} - \left(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \right) \circ \vec{a}$$

And again:

$$\left(\vec{a} \times \vec{b}\right) \circ \vec{b} + \left( \vec{b} \times \vec{c}\right) \circ \vec{b} + \left(\vec{c} \times \vec{a} \right) \circ \vec{b} - \left(\vec{a} \times \vec{b}\right) \circ \vec{a} - \left(\vec{b} \times \vec{c}\right) \circ \vec{a} - \left(\vec{c} \times \vec{a} \right) \circ \vec{a}$$

Now since we know that $\left(\vec{x} \times \vec{y}\right) \circ \vec{x} = 0$, most of the terms above are zero:

$$\left(\vec{c} \times \vec{a} \right) \circ \vec{b} - \left(\vec{b} \times \vec{c}\right) \circ \vec{a}$$

This actually leads to a nice property of cross products: $\vec{x} \circ \left (\vec{y} \times \vec{z}\right) = \vec{z} \circ \left (\vec{x} \times \vec{y}\right) = \vec{y} \circ \left (\vec{z} \times \vec{x}\right)$. I don't know if you would actually be expect to prove this or not in your homework since it is well known.

Either way, your work that you've already done is on the right track, you are probably just making some $\pm$ error or something in the part you haven't shown us.

Answer 4

The easy way to do this is to use the following identity obeyed by the vector cross and dot product, which may be found in the Wikipedia page http://en.m.wikipedia.org/wiki/Cross_product: for any three vectors $\vec a, \vec b, \vec c \in \Bbb R^3$,

$\vec a \cdot (\vec b \times \vec c) = \vec b \cdot (\vec c \times \vec a) = \vec c \cdot (\vec a \times \vec b). \tag{1}$

Now observe that $\vec a - \vec b, \vec a - \vec c$ lie in the plane containing $P, Q, R$, and they are linearly independent since these points are not collinear. Then

$(\vec a - \vec b) \cdot (\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a) = \vec a \cdot (\vec b \times \vec c) - \vec b \cdot (\vec c \times \vec a = 0 \tag{2}$

by virtue of (1). Likewise the similar product with $(\vec a - \vec c)$ replacing $(\vec a - \vec b)$ on the left hand side of (2) vanishes as well. Thus $\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a$ is normal to the plane containing $P, Q, R$ since $\vec a - \vec b, \vec a - \vec c$ span it.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!