Understanding Triple Scalar Product and Its Properties: Explained Simply

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The discussion centers on the confusion surrounding the triple scalar product property, specifically why a.(bxc) = b.(cxa) = c.(axb) holds true for any three vectors a, b, and c. The participant understands the property that a.(ax(anything)) = 0 but struggles with the first property due to a lack of depth in their high school textbook. A suggestion is made to explore the interchangeability of the dot and cross products to gain clarity on the relationship between the vectors. The property a.(bxc) can also be expressed as (axb).c, which may help in understanding the equality. Overall, the thread emphasizes the need for clearer explanations of vector operations.
phospho
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Im having trouble understanding this property

my book states that: a.(bxc) = b.(cxa) = c.(axb)

it also states that a.(ax(anything)) = 0

I understand the second point and why that's true, what I don't understand is why a.(bxc) = b.(cxa) = c.(axb) is true

If I name any 3 vectors a b and c would this be true? I'm just really confused to why it works, and my book doesn't really go into depth as it's a high school book, and Wikipedia seems to be vague or too complex for me.
 
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phospho said:
Im having trouble understanding this property

my book states that: a.(bxc) = b.(cxa) = c.(axb)

it also states that a.(ax(anything)) = 0

I understand the second point and why that's true, what I don't understand is why a.(bxc) = b.(cxa) = c.(axb) is true

If I name any 3 vectors a b and c would this be true? I'm just really confused to why it works, and my book doesn't really go into depth as it's a high school book, and Wikipedia seems to be vague or too complex for me.

Do you have the property that you can interchange the dot and cross? If not, you should show that first. Then try it on a.(bxc) and see if you can get one of the other forms remembering that the dot product is commutative.
 
LCKurtz said:
Do you have the property that you can interchange the dot and cross? If not, you should show that first. Then try it on a.(bxc) and see if you can get one of the other forms remembering that the dot product is commutative.

no, I don't - what is this property?
 
LCKurtz said:
Do you have the property that you can interchange the dot and cross? If not, you should show that first. Then try it on a.(bxc) and see if you can get one of the other forms remembering that the dot product is commutative.

phospho said:
no, I don't - what is this property?

a.(bxc) = (axb).c
 

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