If |v X w| = 3 then |(v+w) X (v-w)| = 6

[mill]

Homework Statement

1. If |v x w | = 3, then |(v + w) X (v - w)| = 6.

Homework Equations

definitions for cross product and dot product

The Attempt at a Solution

The statement is true, but I am unsure how to test this. Would I break anything I find within the || into v1 v2 or w1 w2 etc? How would that work for the second part?

 

[Curious3141]

Why do you say the statement is true?

What is the cross product of a vector with itself?

 

[mill]

The answer was provided, but I'm not sure how the answer was reached.

 

[Curious3141]

The answer was provided, but I'm not sure how the answer was reached.

As it is written, the assertion that $\displaystyle|(v+w) \times (v+w)| = 6$ is wrong.

 

[mill]

I typed in the wrong sign. I have changed it.

 

[Curious3141]

I typed in the wrong sign. I have changed it.

Well, that changes everything (why?).

You don't have to break the individual vectors down into their components. Just multiply out the brackets. Be very careful, as cross products are not commutative (in fact, they are anticommutative). However, the cross product obeys distributivity, which is very helpful here.

What are the terms you get after the expansion?

 

[HallsofIvy]

For any vectors, v and w, what is "(v+ w) x (v- w)"?

 

[mill]

I'm looking at the table for properties but I don't quite see where things are connecting.

If |v x w | = 3 this means that the area of that parallelogram is 3.

then |(v + w) X (v - w)| = 6 is another area of a parallelogram.

(v+ w) x (v- w) Assuming v+w=y and v-w=-y This means that it is the cross product of these two vectors.

Would anyone have a more in-depth cross product properties reference? By multiplying out the brackets according to (v+w) x u = v x u + w x u (from my textbook) I think I would get
|(v^2 x vw) + (vw + w^2)|. Which still doesn't mean anything to me.

 

[mill]

Furthermore, for this

if i x v = 0 and i dot v = -1, then |v| = 1

I don't see how the first part connects to the second. i dot v = i1v1 + i2v2 etc. So assuming i dot v = 1(-1), i is 1 and v=-1 then |v|=1

 

[SammyS]

I'm looking at the table for properties but I don't quite see where things are connecting.

If |v x w | = 3 this means that the area of that parallelogram is 3.

then |(v + w) X (v - w)| = 6 is another area of a parallelogram.

(v+ w) x (v- w) Assuming v+w=y and v-w=-y This means that it is the cross product of these two vectors.

Would anyone have a more in-depth cross product properties reference? By multiplying out the brackets according to (v+w) x u = v x u + w x u (from my textbook) I think I would get
|(v^2 x vw) + (vw + w^2)|. Which still doesn't mean anything to me.

For vectors, we don't write \ \bf{v}\times\bf{v}\ as $\ \text{v}^2\ .$


If a, b, c, and d are real variables, what is (a+b)(c-d) for arithmetic multiplication?

 

[D H]

Would anyone have a more in-depth cross product properties reference? By multiplying out the brackets according to (v+w) x u = v x u + w x u (from my textbook) I think I would get
|(v^2 x vw) + (vw + w^2)|. Which still doesn't mean anything to me.

No!

You are missing the picture. You are focusing on that absolute value sign and you are ignoring two key things: For any two vectors $\vec a$ and $\vec b$,

• $\vec a \times \vec a = \vec b \times \vec b = 0$
• $\vec a \times \vec b = - \vec b \times \vec a$ .

 

[mill]

For vectors, we don't write \ \bf{v}\times\bf{v}\ as $\ \text{v}^2\ .$


If a, b, c, and d are real variables, what is (a+b)(c-d) for arithmetic multiplication?

(ac - ad + bc - bd)

 

[SammyS]

(ac - ad + bc - bd)

Right.

So, if they were vectors and the multiplication was the cross-product, what is $\ (\bf{a}+\bf{b})\times(\bf{c}-\bf{d})\ ?$

 

[Curious3141]

(ac - ad + bc - bd)

Now replace a with $\bf{v}$, b with $\bf{w}$, c with $\bf{v}$, d with $\bf{w}$, and put the cross product symbol between each variable. Preserve the order of the variables.

Use the laws of cross products :

1) a vector cross itself is the null vector (magnitude zero).

2) anti-commutativity. If $\bf{a}$ and $\bf{b}$ are vectors, $\bf{a} \times \bf{b} = -(\bf{b} \times \bf{a})$

 

[STEMucator]

It might help you see it easier if you make a substitution.

Let $u = v + w$. Then:

$|(v + w) \times (v - w)| = |u \times (v - w)| = |(u \times v) - (u \times w)|$

Subbing back:

$|((v + w) \times v) - ((v + w) \times w)|$

 

[mill]

Thanks. Got it. I didn't realize I could substitute. Are we generally allowed to substitute in all cross products?

Thank you for all of the help. Does anyone happen to know where I can look at some solved equations for cross product problems like these? Or solved problems in general? I would ask over at the math learning materials area, but I'm not allowed to make a thread there for some reason. I'm trying to do practice problems, but it is not too efficient to spend a day for one. Other than looking at old homework with solutions from here and there online, I don't really know where to look.

 

[STEMucator]

I do see about the substitution. I didn't realize I could substitute. Are we generally allowed to substitute in all cross products?

But I'm am still confused about how one equals 3 and that equals 6.

So assuming ^ I would get $|((v + w) \times v) - ((v + w) \times w)|$ If |v+w| = 3 then something between this and the second equation should equal 2.

The anti-commutativity property says a x b = -b x a but how would that work out if those two are not exactly the same? Even if I substitute u, |uxv - uxw|. For the property where crossing itself is zero, aren't v and w different vectors?

Thank you for all of the help. Does anyone happen to know where I can look at some solved equations for cross product problems like these? Or solved problems in general? I'm trying to do practice problems, but it is not too efficient to spend a day for one.

Expand the cross terms in the expression, I'll do the left hand term as an example:

$|((v + w) \times v) - ((v + w) \times w)|$

$= |(v \times v + w \times v) - ((v + w) \times w)|$

You know something about $v \times v$, right? In fact the property applies to any vector crossed with itself. The angle between a vector crossed with itself is 0.

 

[mill]

I got it thanks. I didn't see that I had to substitute twice.

 

[mill]

Thanks. Got it. I didn't realize I could substitute. Are we generally allowed to substitute in all cross products?

Thank you for all of the help. Does anyone happen to know where I can look at some solved equations for cross product problems like these? Or solved problems in general? I would ask over at the math learning materials area, but I'm not allowed to make a thread there for some reason. I'm trying to do practice problems, but it is not too efficient to spend a day for one. Other than looking at old homework with solutions from here and there online, I don't really know where to look.

Reposting this.