Understanding Scalar Triple Product (STP)

  • #1

Main Question or Discussion Point

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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Answers and Replies

  • #2
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
Hey UVCatastrophe, Firstly, thanks for your detailed explanation.

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

x⃗ =λa⃗ +μb⃗ +νc⃗ \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

[c1u⃗ 1+c2u⃗ 2,v⃗ ,w⃗ ]=c1[u⃗ 1,v⃗ ,w⃗ ]+c2[u⃗ 2,v⃗ ,w⃗ ][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
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
Hey UVCatastrophe, Firstly, thanks for your detailed explanation.
NP :p

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??
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?

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??
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
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?
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?

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.
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..
 

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