Vector equation involving linearly independent vectors of the three dimensional space

[Robin04]

Homework Statement

Solve the following vector equation for $\vec{y}$. $\vec{a}$, and $\vec{b}$ are linearly independent vectors of the three dimensional space.

$\vec{a} \times (8\vec{y}+\vec{b}) = \vec{b}\times(-5\vec{y}+\vec{a})$

Homework Equations

The Attempt at a Solution

First I developed the brackets
$\vec{a}\times8\vec{y}+\vec{a}\times\vec{b} = \vec{b}\times(-5\vec{y})+\vec{b}\times\vec{a}$
I subtracted $\vec{b}\times(-5\vec{y})$ and $\vec{a}\times\vec{b}$ from both sides, so I get
$\vec{a}\times8\vec{y}-\vec{b}\times(-5\vec{y})=\vec{b}\times\vec{a}-\vec{a}\times\vec{b}$ which also equals to
$\vec{y}\times(-8\vec{a}-5\vec{b}) = 2\vec{b}\times\vec{a}$

Then I expressed $\vec{y}$ as the linear combination of $\vec{a}$, $\vec{b}$ and $\vec{b}\times\vec{a}$
$\vec{y} = \alpha\vec{a}+\beta\vec{b}+\gamma\vec{b}\times\vec{a}$ then I multiplied it (cross product) with $-8\vec{a}-5\vec{b}$
$\vec{y}\times(-8\vec{a}-5\vec{b}) = -8\alpha(\vec{a}\times\vec{a})-5\alpha(\vec{a}\times\vec{b})-8\beta(\vec{b}\times\vec{a})-5\beta(\vec{b}\times\vec{b}) + \gamma(\vec{b}\times\vec{a})\times(-8\vec{a}-5\vec{b})$
$=(5\alpha-8\beta)(\vec{b}\times\vec{a})-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})]$
From the equation this has to be equal to $2\vec{b}\times\vec{a}$ so $5\alpha-8\beta = 2$ and $\gamma = 0$

But if I make up an arbitrary $\vec{y}$ vector of this form and choose random $\vec{a}$ and $\vec{b}$ (linearly independent of course) it doesn't satisfy the equation. Where did I mess up?

 

[jedishrfu]

So if $\gamma = 0$ then doesn't that mean that $\vec{y}$ is a linear combination of $\vec{a}$ and $\vec{b}$ ?

When I computed it it seemed that $\gamma$ can be anything since the resultant $\gamma$ component term vectors are both perpendicular to $\vec{a}$ x $\vec{b}$

 

[Robin04]

So if $\gamma = 0$ then doesn't that mean that $\vec{y}$ is a linear combination of $\vec{a}$ and $\vec{b}$ ?

When I computed it it seemed that $\gamma$ can be anything since the resultant $\gamma$ component term vectors are both perpendicular to $\vec{a}$ x $\vec{b}$

How did you get that $\gamma$ can be anything?

 

[jedishrfu]

Both a and b are perpendicular to a x b , right? Assuming that neither a nor b are the zero vector.

 

[Robin04]

Yes

 

[jedishrfu]

so their cross products are zero and thus $\gamma$ can be any value, right?

 

[jedishrfu]

Did you test it by choosing an a and b vector, construct a y vector and evaluate your original equation?

Perhaps there's a mistake in the check...

 

[Robin04]

Let $\vec{a} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}, \vec{b} = \begin{bmatrix} 1 \\ 2 \\ 1 \\ \end{bmatrix}, \vec{y} = \begin{bmatrix} 2 \\ 1 \\ 0 \\ \end{bmatrix}, \vec{y}$ has to be a good choice because $5\cdot 2 - 8\cdot1 = 2$, and $\vec{a},\vec{b}$ are independendent.
$-8\vec{a}-5\vec{b}= \begin{bmatrix} -13 \\ -18 \\ -13 \\ \end{bmatrix}, \vec{y}\times(-8\vec{a}-5\vec{b}) = \begin{bmatrix} -13 \\ 26 \\ -23 \\ \end{bmatrix}$, but $2\vec{b}\times\vec{a} = \begin{bmatrix} 2 \\ 0 \\ -2 \\ \end{bmatrix}$

 

[jedishrfu]

Wouldn't you compute y from the the $\alpha, \beta, \gamma$ values and the a and b vectors? and then show that a, b and y satisfy the original formula?

 

[DrClaude]

$\vec{y}$ has to be a good choice because $5\cdot 2 - 8\cdot1 = 2$

If $\alpha = 2$ and $\beta=1$, then you have the wrong $\vec{y}$.

 

[Robin04]

Wouldn't you compute y from the the $\alpha, \beta, \gamma$ values and the a and b vectors? and then show that a, b and y satisfy the original formula?

If $\alpha = 2$ and $\beta=1$, then you have the wrong $\vec{y}$.

Aah what a mistake I made there... I should have written the final form of y, I've mistaken $\alpha,\beta,\gamma$ for its components, but they are coefficients of $\vec{a},\vec{b}$. It works now, thank you!

But $\gamma$ has to be 0, otherwise it doesn't work for me.

 

[jedishrfu]

Aah what a mistake I made there... I should have written the final form of y, I've mistaken $\alpha,\beta,\gamma$ for its components, but they are coefficients of $\vec{a},\vec{b}$. It works now, thank you!

But $\gamma$ has to be 0, otherwise it doesn't work for me.

My feeling is that $\gamma$ can be any value but perhaps @DrClaude can comment on it.

 

[Robin04]

Could you show it with equations, please?

 

[jedishrfu]

let a = (1,0,0) and b=(0,1,0) and so a x b = ( 0,0,1)

then y=(8,-5,10) where $\beta = (-5/8) \alpha$.

I chose $\alpha = 8$ and so $\beta = -5$ and arbitrarily chose $\gamma = 10$

a x (8y) + a x b = (0,80,-40) + (0,0,1) = (0,80,-39)

b x (-5y) + b x a = (-50,0,40) + (0,0,-1) = (-50,0,39)

Well, I didn't succeed! I see where I got $\gamma$ wrong now and I think you're right, it must be zero.

----------------------

Let y = $\alpha$ a + $\beta$ b + $\gamma$ (a x b)

then $\alpha$ a x (8a + 5b) = 0 + 5 $\alpha$ ( a x b)

and $\beta$ b x ( 8a + 5b) = -8 $\beta$ (a x b) + 0

and $\gamma$ (a x b) x (8a + 5b) = 8$\gamma$ (a x b) x a + 5$\gamma$ (a x b) x b

...

 

[DrClaude]

$\vec{y}\times(-8\vec{a}-5\vec{b}) = -8\alpha(\vec{a}\times\vec{a})-5\alpha(\vec{a}\times\vec{b})-8\beta(\vec{b}\times\vec{a})-5\beta(\vec{b}\times\vec{b}) + \gamma(\vec{b}\times\vec{a})\times(-8\vec{a}-5\vec{b})$
$=(5\alpha-8\beta)(\vec{b}\times\vec{a})-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})]$
From the equation this has to be equal to $2\vec{b}\times\vec{a}$ so $5\alpha-8\beta = 2$ and $\gamma = 0$

You found
$$
-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
from which either $\gamma = 0$ or
$$
8[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]+5[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
With a bit of algebra, you can transform this into the equation $c_1 \vec{a} = c_2 \vec{b}$, which contradicts $\vec{a}$ and $\vec{b}$ being linearly independent (except if $c_1 = c_2 = 0$, but you can eliminate that possibility for real vectors).

 

[jedishrfu]

Thanks @DrClaude

 

[Robin04]

You found
$$
-8\gamma[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]-5\gamma[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
from which either $\gamma = 0$ or
$$
8[\vec{a}(\vec{b}\vec{a})-\vec{b}(\vec{a}\vec{a})]+5[\vec{a}(\vec{b}\vec{b})-\vec{b}(\vec{a}\vec{b})] = 0
$$
With a bit of algebra, you can transform this into the equation $c_1 \vec{a} = c_2 \vec{b}$, which contradicts $\vec{a}$ and $\vec{b}$ being linearly independent (except if $c_1 = c_2 = 0$, but you can eliminate that possibility for real vectors).

Aah, so this would have been the precise way of saying this, I should have known!

Thank you all for you help!