swaha Posted April 7, 2010 Share Posted April 7, 2010 what is A.B.C? where A,B,C are vectors. Link to comment Share on other sites More sharing options...

D H Posted April 7, 2010 Share Posted April 7, 2010 Do you mean the scalar triple product [math]\vec a \cdot (\vec b \times \vec c)[/math], or the vector triple product [math]\vec a \times (\vec b \times \vec c)[/math]? Link to comment Share on other sites More sharing options...

swaha Posted April 7, 2010 Author Share Posted April 7, 2010 i just mean A.B.C Link to comment Share on other sites More sharing options...

D H Posted April 7, 2010 Share Posted April 7, 2010 That doesn't make sense. Link to comment Share on other sites More sharing options...

swaha Posted April 7, 2010 Author Share Posted April 7, 2010 ctually i saw it in a book, may be a misprint then. shall wait fr a few more negative answers before requesting the moderators to close. Link to comment Share on other sites More sharing options...

D H Posted April 7, 2010 Share Posted April 7, 2010 First off, swaha, do you understand the difference between the inner product between two vectors [math]\vec a \cdot \vec b[/math] and the scalar product [math]\vec a \times \vec b[/math]? There are two products of three vectors in three-space. I named both in post #2, perhaps a bit to tersely. The first is the scalar triple product [math]\vec a \cdot (\vec b \times \vec c)[/math]. Since the inner product is a commutative operation, this is the same as [math](\vec b \times \vec c)\cdot \vec a[/math]. One could eliminate the parentheses in these forms because [math]\vec a \cdot \vec b \times \vec c[/math] has only one viable interpretation. One geometric interpretation of this product is the volume of a parallelepiped with sides specified by the vectors [math]\vec a[/math], [math]\vec b[/math], and [math]\vec c[/math]. Rearrangements (permutations) of the vectors [math]\vec a[/math], [math]\vec b[/math], and [math]\vec c[/math] might change the sign of the result, but never the absolute value. The second triple product is the vector triple product [math]\vec a \times (\vec b \times \vec c)[/math]. Unlike the scalar triple product, those parentheses are essential here. Specifying things in the right order is also essential. In other words, [math]\vec a \times (\vec b \times \vec c)\ne(\vec a \times \vec b) \times \vec c\ne \vec b \times (\vec a \times \vec c)[/math], and so on. One use of the vector triple product is to compute the component of a vector normal to vector. Suppose [math]\hat a[/math] is a unit vector in [math]\mathbb R^3[/math] and [math]\vec b[/math] is some other vector in [math]\mathbb R^3[/math]. The component of [math]\vec b[/math] normal to [math]\hat a[/math] is [math]\hat a \times (\vec b \times \hat a)[/math]. Link to comment Share on other sites More sharing options...

triclino Posted April 7, 2010 Share Posted April 7, 2010 i just mean A.B.C If B and C are on the same currier ,or parallel to A ,then : A.B.C = A.(B.C) =(A.B).C = C.(B.A) =(C.B).A = B.(A.C) ........e.t.c e.t.c Otherwise : A.(B.C)[math]\neq (A.B).C[/math] Link to comment Share on other sites More sharing options...

D H Posted April 7, 2010 Share Posted April 7, 2010 Use the right nomenclature, please. There are two products defined for vectors in [math]\mathbb R^3[/math], the inner product and the cross product. Neither is denoted with a period. Link to comment Share on other sites More sharing options...

swaha Posted April 8, 2010 Author Share Posted April 8, 2010 If B and C are on the same currier ,or parallel to A ,then : A.B.C = A.(B.C) =(A.B).C = C.(B.A) =(C.B).A = B.(A.C) ........e.t.c e.t.c Otherwise : A.(B.C)[math]\neq (A.B).C[/math] why? pls explain. i think its so when they are perpendicular not parallel. Link to comment Share on other sites More sharing options...

D H Posted April 8, 2010 Share Posted April 8, 2010 Ignore triclino. What he wrote doesn't make sense. Please, people. Learn to use the correct nomenclature. There are two well-defined products for 3-vectors, the scalar product denoted by a center dot, and the cross product denoted by [math]\times[/math]. This doesn't make a lick of sense: [math]\vec a \cdot \vec b \cdot \vec c[/math]. That can only mean triclino was talking about the cross product, and what he wrote isn't correct for that either. The correct condition under which [math]\vec a \times (\vec b \times \vec c) = (\vec a \times \vec b)\times \vec c[/math] is that [math]\vec c[/math] is parallel to [math]\vec a[/math], i.e., [math]\vec c = \alpha \vec a[/math] where [math]\alpha[/math] is some scalar. There is no constraint on [math]\vec b[/math]. If all three are parallel to one another the vector triple product is identically zero for all arrangements of the factors in the product. Link to comment Share on other sites More sharing options...

swaha Posted April 9, 2010 Author Share Posted April 9, 2010 oh thanks. Link to comment Share on other sites More sharing options...

triclino Posted April 10, 2010 Share Posted April 10, 2010 . This doesn't make a lick of sense: [math]\vec a \cdot \vec b \cdot \vec c[/math]. That can only mean triclino was talking about the cross product, and what he wrote isn't correct for that either. Why you did not ask me what i meant ,but make such a fuss over minor details ?? This is a physics forum and people know what a dot product is , and very easily can understand that: A.(B.C) is really A(B.C) since the dot product is always a scalar. Now is not true that if the vectors are on the same currier or parallel then : A(B.C) =(A.B)C ??? Link to comment Share on other sites More sharing options...

D H Posted April 10, 2010 Share Posted April 10, 2010 You left out your "otherwise" in post #7 from the above, triclino. Furthermore, using A.B for the cross product is very, very bad form. That period looks a lot more like a dot than a cross. This is not a minor detail since there are many products for vectors. For example, the inner or dot product, the cross product for vectors in 3- and 7- space, the outer product, the exterior or wedge product, etc. Each has its own symbol and none of them is denoted with a period. Link to comment Share on other sites More sharing options...

Cap'n Refsmmat Posted April 10, 2010 Share Posted April 10, 2010 This is much easier: [math]\vec{a} (\vec{b} \cdot \vec{c})[/math] Use LaTeX to get the point across. Link to comment Share on other sites More sharing options...

darkenlighten Posted April 11, 2010 Share Posted April 11, 2010 [math] \mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c} [/math] is not equal to [math] a(\mathbf{b} \cdot \mathbf{c})[/math] Because you cannot dot a vector and a scalar, you can multiply them, but not dot them. This is kinda trivial, but the reason I say it is if there was maybe some type of proof or equation that had a similar form, you would not be able to an operation like this. Or whatever you are trying to say but either way [math] \mathbf{a} \cdot \mathbf{b} \cdot \mathbf{c} [/math] this cannot work due to the reason above. Link to comment Share on other sites More sharing options...

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