Verifying \vec{A} \cdot (\vec{B} x \vec{C}) = \vec{B} \cdot (\vec{C} x \vec{A})

[FaraDazed]

Homework Statement

$\vec{A}= \hat{i} + 2 \hat{j} \\<br /> \vec{B}= \hat{j} + \hat{k} \\<br /> \vec{C} = 2 \hat{k} + \hat{i}$

Verify that $\vec{A} \cdot (\vec{B} \times \vec{C}) = \vec{B} \cdot (\vec{C} \times \vec{A})$

Homework Equations

n/a

The Attempt at a Solution

I am getting something similar but not equal, I believe I must have made a simple stupid mistake somewhere, would appreciate a second pair of eyes, thanks :)

First what I did was rewrite the three vectors like so...
$\vec{A}= (1 \,\, , 2 \,\, , 0) \\<br /> \vec{B}= (0 \,\, , 1 \,\, , 1) \\<br /> \vec{C} = (1 \,\, , 0 \,\, , 2)$

Then just did $\vec{B} \times \vec{C}$ first...
$\vec{B} \times \vec{C} = (2-0) \hat{i} - (0-1) \hat{j} + (0-1) \hat{k} = 2 \hat{i} + 1 \hat{j} - 1 \hat{k} = (2 \,\, , 1 \,\, , -1)$

And then did the dot product of that new vector with vector A
$\vec{A} \cdot (\vec{B} \times \vec{C}) = (1 \,\, , 2 \,\, , 0) \dot (2 \,\, , 1 \,\, , -1) = (2 \,\, , 2 \,\, , 0)$
So I got (2, 2, 0) for the left hand side of the equation, now moving onto the right hand side I did the same process.

First I did just C x A...
$\vec{C} \times \vec{A} = (0-4) \hat{i} - (0-2) \hat{j} + (2-0) \hat{k} = -4 \hat{i} + 2 \hat{j} + 2 \hat{k} = (-4 \,\, , 2 \,\, , 2)$
And then the dot product of that vector with B
$\vec{B} \cdot (\vec{C} \times \vec{A}) = (0 \,\, , 1 \,\, , 1) \cdot (-4 \,\, , 2 \,\, , 2) = (0 \,\, , 2 \,\, , 2)$

And as you can see, (0, 2, 2) does not equal (2, 2, 0). Although now thinking about it, the magnitude of the vectors are equal but that is not what the question asked to verify; I was expecting two vectors exactly the same, i.e. (1, 4, 5) and (1, 4, 5), is that what I should get as the answer? Or have I already done it correctly?

 

[DrClaude]

And then did the dot product of that new vector with vector A
$\vec{A} \cdot (\vec{B} \times \vec{C}) = (1 \,\, , 2 \,\, , 0) \cdot (2 \,\, , 1 \,\, , -1) = (2 \,\, , 2 \,\, , 0)$
So I got (2, 2, 0) for the left hand side of the equation

You need to review what the dot product is.

 

[Ray Vickson]

Homework Statement

[itex]\vec{A}= \hat{i} + 2 \hat}j} \\<br /> \vec{B}= \hat{j} + \hat{k} \\<br /> \vec{C} = 2 \hat{k} + \hat{i}<br /> <br /> Verify that $\vec{A} \dot (\vec{B} \times \vec{C}) = \vec{B} \dot (\vec{C} \times \vec{A})$[/itex]

$<blockquote data-attributes="" data-quote="Ray Vickson" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Ray Vickson said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You need a final " </div> </div> </blockquote>$

" in your very first paragraph.

That's not what the problem was - one of the braces was going the wrong way. It's now fixed in post #1.

 

[FaraDazed]

You need to review what the dot product is.

Ah right, of course! The final answer can't be a vector, so they are both just equal to 4?

I.e.
$\vec{A} \cdot (\vec{B} \times \vec{C}) = (1 \,\, , 2 \,\, , 0) \dot (2 \,\, , 1 \,\, , -1) = 2+2+0 = 4$

 

[haruspex]

Ah right, of course! The final answer can't be a vector, so they are both just equal to 4?

I.e.
$\vec{A} \cdot (\vec{B} \times \vec{C}) = (1 \,\, , 2 \,\, , 0) \dot (2 \,\, , 1 \,\, , -1) = 2+2+0 = 4$

Yes.