# Scalar divided by a Vector

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How do you simplify a scalar divided by a vector?

For example if we have

an unknown vector (say x y z) = a known scalar (say lambda) / a known vector (say i j k), how do you find out the values of x, y and z in terms of lamda, i, j and k?

For example if we have

an unknown vector (say x y z) = a known scalar (say lambda) / a known vector (say i j k), how do you find out the values of x, y and z in terms of lamda, i, j and k?

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#2

Do you have a specific question that you could perhaps screenshot/ share a picture of?

Straight division/ multiplication by vectors aren't defined operations.

There are two main operations on vectors which can be compared to the ordinary multiplication of numbers but are slightly different for vectors, namely the dot product and the cross product.

We also have normal addition of vectors that work as you would expect, i.e add component wise, e.g (1,2,3)+(4,5,6)=(5,7,9).

And we can multiply a vector by a scalar, so for example 2*(1,2,3)=(2,4,6).

So something like (lambda)/(a vector) doesn't really make any sense, or at least isn't defined mathemtically. Unless there is an function applied to the vector in the denominator, for example the norm function which takes a vector (x,y,z) and returns the magnitude of the vector sqrt(x

Straight division/ multiplication by vectors aren't defined operations.

There are two main operations on vectors which can be compared to the ordinary multiplication of numbers but are slightly different for vectors, namely the dot product and the cross product.

We also have normal addition of vectors that work as you would expect, i.e add component wise, e.g (1,2,3)+(4,5,6)=(5,7,9).

And we can multiply a vector by a scalar, so for example 2*(1,2,3)=(2,4,6).

So something like (lambda)/(a vector) doesn't really make any sense, or at least isn't defined mathemtically. Unless there is an function applied to the vector in the denominator, for example the norm function which takes a vector (x,y,z) and returns the magnitude of the vector sqrt(x

^{2}+y^{2}+z^{2}).
Last edited by Tayls102; 3 months ago

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(Original post by

Do you have a specific question that you could perhaps screenshot/ share a picture of?

Straight division/ multiplication by vectors aren't defined operations.

There are two main operations on vectors which can be compared to the ordinary multiplication of numbers but are slightly different for vectors, namely the dot product and the cross product.

We also have normal addition of vectors that work as you would expect, i.e add component wise, e.g (1,2,3)+(4,5,6)=(5,7,9).

And we can multiply a vector by a scalar, so for example 2*(1,2,3)=(2,4,6).

So something like (lambda)/(a vector) doesn't really make any sense, or at least isn't defined mathemtically. Unless there is an function applied to the vector in the denominator, for example the norm function which takes a vector (x,y,z) and returns the magnitude of the vector sqrt(x

**Tayls102**)Do you have a specific question that you could perhaps screenshot/ share a picture of?

Straight division/ multiplication by vectors aren't defined operations.

There are two main operations on vectors which can be compared to the ordinary multiplication of numbers but are slightly different for vectors, namely the dot product and the cross product.

We also have normal addition of vectors that work as you would expect, i.e add component wise, e.g (1,2,3)+(4,5,6)=(5,7,9).

And we can multiply a vector by a scalar, so for example 2*(1,2,3)=(2,4,6).

So something like (lambda)/(a vector) doesn't really make any sense, or at least isn't defined mathemtically. Unless there is an function applied to the vector in the denominator, for example the norm function which takes a vector (x,y,z) and returns the magnitude of the vector sqrt(x

^{2}+y^{2}+z^{2}).
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#4

Okay so I have a feeling that here w

Does this help?

_{3}is a vector rather than a scalar. There are two reasons why I think this, the first is that the use of 'x' in the context of vectors really means cross product, which is only defined on vectors (i.e you can't take the cross product between a scalar and a vector). The second reason is that if you do treat it as a scalar, and interpret 'x' as normal multiplication by a scalar, what you get doesn't really make sense. Of course, I can't be sure without some context of the question.Does this help?

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(Original post by

Okay so I have a feeling that here w

Does this help?

**Tayls102**)Okay so I have a feeling that here w

_{3}is a vector rather than a scalar. There are two reasons why I think this, the first is that the use of 'x' in the context of vectors really means cross product, which is only defined on vectors (i.e you can't take the cross product between a scalar and a vector). The second reason is that if you do treat it as a scalar, and interpret 'x' as normal multiplication by a scalar, what you get doesn't really make sense. Of course, I can't be sure without some context of the question.Does this help?

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#6

Okay, lets assume that w

Are you familiar with the cross product of two vectors?

_{3}is a vector which I'm fairly sure it would be.Are you familiar with the cross product of two vectors?

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(Original post by

Okay, lets assume that w

Are you familiar with the cross product of two vectors?

**Tayls102**)Okay, lets assume that w

_{3}is a vector which I'm fairly sure it would be.Are you familiar with the cross product of two vectors?

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#8

Great! that sounds right, what system of equations* did you get?

What methods do you know to solve a system of equations?

Edit: *

What methods do you know to solve a system of equations?

Edit: *

Spoiler:

Show

Hint: There may or may not be a

*unique*solution
Last edited by Tayls102; 3 months ago

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(Original post by

Great! that sounds right, what system of equations* did you get?

What methods do you know to solve a system of equations?

Edit: *

**Tayls102**)Great! that sounds right, what system of equations* did you get?

What methods do you know to solve a system of equations?

Edit: *

Spoiler:

Show

Hint: There may or may not be a

*unique*solution.
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#10

That looks good, is there perhaps a way you could write this in the form A

Also, remember that w

**w**=_{3}**v**, where**v**is a vector, and A is a 3x3 matrix, both to be found?Also, remember that w

_{x},w_{y},w_{z}are components of**w**so don't need to be in bold_{3}
Last edited by Tayls102; 3 months ago

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