# Understanding Scalar Triple Product (STP)

1. Jan 26, 2015

### PcumP_Ravenclaw

Dear all,
My question is from the text of Alan F. Beardon, Algebra and Geometry concerning the scalar triple product. I have attached the text in this post.

In order for the STP to be non-zero. The 3 vectors must be distinct and they are not coplanar. 2 vectors can be coplanar and only the 3rd vector has to lie on a different plane right??

Geometrically, STP can be interpreted as the volume of the parallelepiped formed by the 3 vectors.

Now, STP is also used to define a point x with reference to any given set of non-orthogonal coordinates axes along the directions of a, b & c. a, b & c can be considered to be like unit vectors right? Can two of the line segments [0,a], [0,b] or [0,c] lie on the same plane??

The formula in 4.4.3 shows that point x can be written in terms of the 3 axes vectors, a,b & c. The scalar multiple of each vector is

$\dfrac{[x,b,c]}{[a,b,c]}$

I am trying to understand how this came about!! all 3 a,b & c have a common denominator $[a,b,c]$. b cross with c is orthogonal to b and c and it is dot with a. This is like the component of a in the direction of b cross c.
because for any 2 vectors L and M
$L \cdot M = |L| |M| Cos θ$

Is there a better way to understand/interpret this as it is used for other two components b and c as well?

Next, the numerator for a, b and c are as follows $[x,b,c], [a,x,c], [a,b,x]$
$[x,b,c]$ can be understood as the component of b in the direction b cross c. $[a,x,c] & [a,b,x]$ can be understood likewise. But I don't get the general idea as to why we are doing this?? I cannot use the geometrical interpretation of the volume of a parallelepiped.

Also please show how the following is derived so I can apply this to get μ and ν.

$[x,b,c] = [λa + μb + νc, b, c] = λ[a,b,c]$

Danke....

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2. Jan 27, 2015

### UVCatastrophe

Concepts

The notion of linear dependence is helpful here. It might also help to review the axioms/definitions of vector spaces. Long story short, in $D$ dimensions, you need exactly $D$ vectors to provide a basis. Furthermore, all of these basis vectors must be linearly independent. Traditionally, we make basis vectors orthogonal (perpendicular) because that makes calculation a lot simpler, but this isn't necessary (see "Graham-Schmidt Procedure").

Having a basis is great because it allows you to express any vector in your $D$ dimensional space as a linear combination of $D$ basis vectors. A linear combination is something that looks like this:

$\overset{\rightarrow}{x} = c_1 \overset{\rightarrow}{e}_1 + c_2 \overset{\rightarrow}{e}_2 + \ldots + c_D \overset{\rightarrow}{e}_D$

The $c_i$'s are coefficients (just numbers) and each $\overset{\rightarrow}{e}_i$ is a basis vector. What this problem is asking you to do is to express the vector $x$ with respect to the basis $a,b,c$. You are assuming that $a,b,c$ are linearly independent so that this is possible [in 3dimensions the fact that $[a,b,c] \neq 0$ is enough to guarantee linear independence]. We write

$\overset{\rightarrow}{x} = \lambda \overset{\rightarrow}{a} + \mu \overset{\rightarrow}{b} + \nu \overset{\rightarrow}{c}$

and take scalar triple products. You should be able to prove the property

$[c_1 \overset{\rightarrow}{u}_1 + c_2 \overset{\rightarrow}{u}_2, \overset{\rightarrow}{v},\overset{\rightarrow}{w}] = c_1 [\overset{\rightarrow}{u}_1, \overset{\rightarrow}{v},\overset{\rightarrow}{w} ] + c_2 [\overset{\rightarrow}{u}_2, \overset{\rightarrow}{v},\overset{\rightarrow}{w} ]$

In the problem above, this is

$[\overset{\rightarrow}{x},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] = \lambda [\overset{\rightarrow}{a},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] + \mu [\overset{\rightarrow}{b},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] + \nu [\overset{\rightarrow}{c},\overset{\rightarrow}{b},\overset{\rightarrow}{c}]$

Since $\overset{\rightarrow}{b} \times \overset{\rightarrow}{c}$ is perpendicular to both $\overset{\rightarrow}{b}$ and $\overset{\rightarrow}{c}$, two of the terms cancel.

What you actually need to do

Three calculations: one for $\lambda$ (you did that already), one for $\mu$ and one for $\nu$. You can get $\mu$ and $\nu$ by triple-producting $x$ with ($a$ and $c$) and ($a$ and $b$), though I am going to leave it up to you whether I mean respectively or not.

3. Jan 27, 2015

### PcumP_Ravenclaw

Hey UVCatastrophe, Firstly, thanks for your detailed explanation.

From the quote above, we are trying to express the point x in terms of the basis a,b & c. In cartesian orthogonal coordinates what you do is just add up the components in terms of x,y & z but with a, b & c (Non-orthogonal) coordinates you simply cannot do this so you need to use the scalar triple product. Am I right??

why are we doing this, $[\overset{\rightarrow}{x},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] = \lambda [\overset{\rightarrow}{a},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] + \mu [\overset{\rightarrow}{b},\overset{\rightarrow}{b},\overset{\rightarrow}{c}] + \nu [\overset{\rightarrow}{c},\overset{\rightarrow}{b},\overset{\rightarrow}{c}]$ for a and for b ($[a,x,c]$) and c ($[a,b,x]$) as well??

4. Jan 27, 2015

### UVCatastrophe

NP :p

Yes. To test this, you could specialize to a case where a,b, and c are all orthogonal vectors of unit length. What do the triple products work out to be?

You want to write x as a linear combo of a,b,c. So you need to find the coefficients lambda, mu, nu. Appealing to aforementioned properties of stp's allows you to determine these.

5. Jan 27, 2015

### PcumP_Ravenclaw

for $[a,b,c] = a \cdot (b x c)$ b cross c is in the same direction as a and if they are all unit vectors then it has same length as a so it becomes $a \cdot a$ which is equal to 1. Apart from this being the volume of the cube formed from a,b,c, what does this imply?

In the non-orthogonal and non-coplanar coordinates axes, Can't we just say for example, the point x is 5 (λ = 5) units in the direction of $a$, 3 (μ = 3) units in the direction of $b$ and 2 (ν = 2) units in the direction $c$ therefore x is the linear sum of $x = 5a + 3b + 2c$. why does the coeffecients 5,3 and 2 have to be derived from the scalar triple product $[x,b,c] [a,x,c], [a,b,x]$

Danke..