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Physics Vector manipulation

dwsmith

Well-known member
Feb 1, 2012
1,673
The unknown vector $\mathbf{v}$ satisfies $\mathbf{b}\cdot\mathbf{v} = \lambda$ and $\mathbf{b}\times\mathbf{v} = \mathbf{c}$, where $\lambda$, $\mathbf{b}$, and $\mathbf{c}$ are fixed and known.
Find $\mathbf{v}$ in terms of $\lambda$, $\mathbf{b}$, and $\mathbf{c}$.

I tried writing stuff out in Einstein summation notation but that didn't lead anywhere at least I didn't see it. Is there another method or approach?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,868
Hi dwsmith!

Here's a suggestion.

Start with the known orthogonal basis [TEX]\{\mathbf{b}, \mathbf{c}, \mathbf{c} \times \mathbf{b} \}[/TEX].
Write [TEX]\mathbf{v}[/TEX] as a linear combination of these basis vectors.
Fill in and solve.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
Hi dwsmith!

Here's a suggestion.

Start with the known orthogonal basis [TEX]\{\mathbf{b}, \mathbf{c}, \mathbf{c} \times \mathbf{b} \}[/TEX].
Write [TEX]\mathbf{v}[/TEX] as a linear combination of these basis vectors.
Fill in and solve.
Maybe this is obvious but how do we know that $\{\mathbf{b}, \mathbf{c}, \mathbf{c} \times \mathbf{b} \}$ spans our vector space? What if these are just lin ind but don't span? If it spans, yes v is a lin combination.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,868
Maybe this is obvious but how do we know that $\{\mathbf{b}, \mathbf{c}, \mathbf{c} \times \mathbf{b} \}$ spans our vector space? What if these are just lin ind but don't span? If it spans, yes v is a lin combination.
The cross product is and can only be defined for 3 dimensions.
Since your problem contains $\mathbf{b} \times \mathbf{v}$, your vector space is 3-dimensional.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
For this approach to work, you do need to know if $\mathbf{b}$ and $\mathbf{c}$ are collinear or not. If they are collinear, then the proposed basis does not work. If they are not collinear, the proposed basis does work.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\mathbf{v} = c_1\mathbf{b} + c_2\mathbf{c}+ c_3(\mathbf{b}\times\mathbf{c})
$$
So what now? How do I get lambda incorporated?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,868
We have:
$\mathbf{b} \cdot \mathbf{v} = \lambda$​

Substitute your expression for $\mathbf{v}$.
$\mathbf{b} \cdot (c_1\mathbf{b} + c_2\mathbf{c}+ c_3(\mathbf{b}\times\mathbf{c})) = \lambda$

$c_1 \mathbf{b}^2 + c_2 \mathbf{b} \cdot \mathbf{c}+ c_3 \mathbf{b} \cdot (\mathbf{b}\times\mathbf{c}) = \lambda$​


Since $\mathbf{c}$ is the result of a cross product with $\mathbf{b}$ involved, they are perpendicular, so their dot product is zero.
Furthermore, $\mathbf{b}\times\mathbf{c}$ is perpendicular to $\mathbf{b}$, so their dot product is also zero.

So:
$c_1 b^2 = \lambda$​
$c_1 = {\lambda \over b^2}$​


It helps if you draw a picture of the vectors involved.
From the picture you should be able to see that $\mathbf{v}$ will be a linear combination of $\mathbf{b}$ and $\mathbf{c} \times \mathbf{b}$.

Btw, as Ackbach remarked, it is possible that $\mathbf{c}$ and $\mathbf{c} \times \mathbf{b}$ are both null vectors, but the calculations will still hold.
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
We have:
$\mathbf{b} \cdot \mathbf{v} = \lambda$​

Substitute your expression for $\mathbf{v}$.
$\mathbf{b} \cdot (c_1\mathbf{b} + c_2\mathbf{c}+ c_3(\mathbf{b}\times\mathbf{c})) = \lambda$

$c_1 \mathbf{b}^2 + c_2 \mathbf{b} \cdot \mathbf{c}+ c_3 \mathbf{b} \cdot (\mathbf{b}\times\mathbf{c})) = \lambda$​


Since $\mathbf{c}$ is the result of a cross product with $\mathbf{b}$ involved, they are perpendicular, so their dot product is zero.
Furthermore, $\mathbf{b}\times\mathbf{c}$ is perpendicular to $\mathbf{b}$, so their dot product is also zero.

So:
$c_1 b^2 = \lambda$​
$c_1 = {\lambda \over b^2}$​


It helps if you draw a picture of the vectors involved.
From the picture you should be able to see that $\mathbf{v}$ will be a linear combination of $\mathbf{b}$ and $\mathbf{c} \times \mathbf{b}$.

Btw, as Ackbach remarked, it is possible that $\mathbf{c} \times \mathbf{b}$ is the null vector, but the calculations will still hold.
I don't think b and c can be collinear since c is normal to b.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,868
I don't think b and c can be collinear since c is normal to b.
What can happen is that b and v are collinear.
In that case c is the null vector.
I'm not sure if the definition of collinearity includes the null vector or not...
 
Last edited: