- Thread starter
- #1

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?

- Thread starter dwsmith
- Start date

- Thread starter
- #1

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?

- Admin
- #2

- Mar 5, 2012

- 9,416

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.

- Thread starter
- #3

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.

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.

- Admin
- #4

- Mar 5, 2012

- 9,416

The cross product is and can only be defined for 3 dimensions.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.

Since your problem contains $\mathbf{b} \times \mathbf{v}$, your vector space is 3-dimensional.

- Admin
- #5

- Jan 26, 2012

- 4,198

- Thread starter
- #6

- Admin
- #7

- Mar 5, 2012

- 9,416

We have:

Substitute your expression for $\mathbf{v}$.

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:

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.

$\mathbf{b} \cdot \mathbf{v} = \lambda$

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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- Thread starter
- #8

I don't think b and c can be collinear since c is normal to b.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.

- Admin
- #9

- Mar 5, 2012

- 9,416

What can happen is that b and v are collinear.I don't think b and c can be collinear since c is normal to b.

In that case c is the null vector.

I'm not sure if the definition of collinearity includes the null vector or not...

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