# Vector problem: Questions about a unit vector

• WendysRules
Well, the point of the chapter is "Geometric multiplication of vectors" in which you ask what is ##\vec{v}^2## is, using the definition ##\vec{v} = |\vec{v}|\hat{v}## we see that $$\vec{v}^2 = (|\vec{v}|\hat{v})(|\vec{v}|\hat{v}) = |\vec{v}|^2\hat{v}^2 = |\vec{v}|^2$$ iff we define ##\hat{v}^2 = 1##. We can then usef

#### WendysRules

Homework Statement
Let ##\vec{e}## be a unit vector so that ##\vec{e}^2 = 1## and let ##\vec{v} = 3\vec{e}##.

Show that ##\hat{v} = \vec{e}## and ##|v| = \vec{v}\vec{e} = 5##
Relevant Equations
##\vec{x} = |x|\hat{x}##
Not sure how to show that because ##\vec{v} = |v|\hat{v} = 3|e|\hat{e}##, but since ##\vec{e}## is a unit vector we know ##|e| = 1## so our equation now becomes ##\hat{v} = \frac{3\hat{e}}{|v|}##. So, we're left to the task of showing that ##|v| = 3## in order to conclude that ##\hat{v} = \vec{e}##.

Another way I tried to do it was ##\vec{v}^2 = |v|^2\hat{v}^2= 9\vec{e}^2## which implies that ##|v|^2 = 9## or ##|v| = 3##

However, if this is true, the second part can't be true i.e ##|v| = 3 \neq 5##

So, either there is a typo in the problem, or I'm interpreting the first part that since ##\vec{e}## is a unit vector, that ##|e| = 1##

The questions that follow up on this have it being that ##|v| = 5## which makes since for the second part, but I can't see how both of these can be true.

Homework Statement:: Let ##\vec{e}## be a unit vector so that ##\vec{e}^2 = 1## and let ##\vec{v} = 3\vec{e}##.

Show that ##\hat{v} = \vec{e}## and ##|v| = \vec{v}\vec{e} = 5##
Relevant Equations:: ##\vec{x} = |x|\hat{x}##

Not sure how to show that because ##\vec{v} = |v|\hat{v} = 3|e|\hat{e}##, but since ##\vec{e}## is a unit vector we know ##|e| = 1## so our equation now becomes ##\hat{v} = \frac{3\hat{e}}{|v|}##. So, we're left to the task of showing that ##|v| = 3## in order to conclude that ##\hat{v} = \vec{e}##.

Another way I tried to do it was ##\vec{v}^2 = |v|^2\hat{v}^2= 9\vec{e}^2## which implies that ##|v|^2 = 9## or ##|v| = 3##

However, if this is true, the second part can't be true i.e ##|v| = 3 \neq 5##

So, either there is a typo in the problem, or I'm interpreting the first part that since ##\vec{e}## is a unit vector, that ##|e| = 1##

The questions that follow up on this have it being that ##|v| = 5## which makes since for the second part, but I can't see how both of these can be true.

I think you might be right about a typo. Since ##\vec e## is a unit vector and ##\vec v = 3\vec e##, then ##\vec v## is just a scalar multiple of ##\vec e##, and both point in the same direction. So ##\hat v## will also be a unit vector with the same direction as ##\vec v##.
The typo is probably in what they're saying ##| \vec v|## is, which should be 3, not 5.

The part about ##\vec e^2## doesn't make sense to me unless they actually wrote ##|\vec e|^2## instead of what you have. Is there some info missing? What's the field here -- real numbers? Complex numbers?

WendysRules
The part about ##\vec e^2## doesn't make sense to me unless they actually wrote ##|\vec e|^2## instead of what you have. Is there some info missing? What's the field here -- real numbers? Complex numbers?
##\vec e^2## likely means ##\vec e \cdot \vec e##. Similarly, ##\vec v \vec e## probably means ##\vec v \cdot \vec e##.

SammyS
The part about ##\vec e^2## doesn't make sense to me unless they actually wrote ##|\vec e|^2## instead of what you have. Is there some info missing? What's the field here -- real numbers? Complex numbers?
It's a problem from a book that has a section on hyperbolic numbers (https://www.amazon.com/dp/1704596629/?tag=pfamazon01-20), and you're correct, I should've wrote ##\vec{v}^2 = |\vec{v}|^2\hat{v}^2## and define (for euclidean vectors) that ##\hat{v}^2 = 1##.

But thanks for confirming it should be a typo.

It's a problem from a book that has a section on hyperbolic numbers (https://www.amazon.com/dp/1704596629/?tag=pfamazon01-20), and you're correct, I should've wrote ##\vec{v}^2 = |\vec{v}|^2\hat{v}^2## and define (for euclidean vectors) that ##\hat{v}^2 = 1##.

But thanks for confirming it should be a typo.
That still doesn't look right to me, the parts about ##\vec{v}^2## and ##\hat{v}^2##. Multiplication of two vectors isn't usually defined unless the scalar product (AKA dot product) or cross product is meant.

While there may be no calculus,
the topic of "hyperbolic numbers"
from a reference entitled "Matrix Gateway to Geometric Algebra, Spacetime and Spinors"
is not your typical "precalculus topic".

That still doesn't look right to me, the parts about ##\vec{v}^2## and ##\hat{v}^2##. Multiplication of two vectors isn't usually defined unless the scalar product (AKA dot product) or cross product is meant.
Yes, unless it's all in 2 dimensions. Then the vectors are just like complex numbers. I suspect there's a backstory here we haven't heard yet.

That still doesn't look right to me, the parts about ##\vec{v}^2## and ##\hat{v}^2##. Multiplication of two vectors isn't usually defined unless the scalar product (AKA dot product) or cross product is meant.
Well, the point of the chapter is "Geometric multiplication of vectors" in which you ask what is ##\vec{v}^2## is, using the definition ##\vec{v} = |\vec{v}|\hat{v}## we see that $$\vec{v}^2 = (|\vec{v}|\hat{v})(|\vec{v}|\hat{v}) = |\vec{v}|^2\hat{v}^2 = |\vec{v}|^2$$ iff we define ##\hat{v}^2 = 1##. We can then use this to define an "multiplication inverse for vectors" and then it goes into bivectors. Which I don't think is needed for this problem, since you reach the same conclusion on this problem (a sub problem we are asked to show that ##\vec{v}^{-1} = \frac{1}{5}\vec{e}##)

The next two chapters are inner then outer products, but I'm self studying the book, so wanted to make sure I wasn't making some obvious error on this question before moving on.

@WendysRules Can you post (insert an image of) the section of the page where the problem is posed?

Well, the point of the chapter is "Geometric multiplication of vectors" in which you ask what is ##\vec{v}^2## is, using the definition ##\vec{v} = |\vec{v}|\hat{v}##
This makes sense, as it is multiplying a unit vector, ##\hat v## by the scalar |v|, the magnitude of ##\vec v##.
WendysRules said:
we see that $$\vec{v}^2 = (|\vec{v}|\hat{v})(|\vec{v}|\hat{v}) = |\vec{v}|^2\hat{v}^2 = |\vec{v}|^2$$ iff we define ##\hat{v}^2 = 1##.
The above is the part I have trouble with, namely the product of a vector with itself or another vector. What would you get for ##\vec u \vec v## if u and v are different vectors?

Of course, if one defines operations, then you can work from these definitions, but in ordinary vector spaces, the usual operations are addition of vectors and multiplication of vectors by scalars. When you extend these to spaces with an inner product, then a definition for multiplying vectors comes into play.
WendysRules said:
We can then use this to define an "multiplication inverse for vectors" and then it goes into bivectors. Which I don't think is needed for this problem, since you reach the same conclusion on this problem (a sub problem we are asked to show that ##\vec{v}^{-1} = \frac{1}{5}\vec{e}##)

The next two chapters are inner then outer products, but I'm self studying the book, so wanted to make sure I wasn't making some obvious error on this question before moving on.

@WendysRules Can you post (insert an image of) the section of the page where the problem is posed?

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robphy
What would you get for ##\vec u \vec v## if u and v are different vectors?
Well, so far all the book has said on this is that ##\vec u \vec v = -\vec v \vec u## due to an argument with Pythagorean's theorem.

I skipped ahead a little bit, but according to the book ##\vec{u} \vec{v} = \vec{u} \cdot \vec{v} + \vec{u} \wedge \vec{v}## which makes sense from an algebraic standpoint, but id have to look into the geometric meaning. (where ##\vec{u} \cdot \vec{v} = \frac{1}{2} (\vec{u}\vec{v}+\vec{v} \vec{u})## and ##\vec{u} \wedge \vec{v} = \frac{1}{2} (\vec{u}\vec{v}-\vec{v}\vec{u})##)

If ##V## is a vector space equipped with the geometric product, defined such that for ##\mathbf{u}, \mathbf{v} \in V## then ##\mathbf{u} \mathbf{v} \equiv \mathbf{u} \cdot \mathbf{v} + \mathbf{u} \wedge \mathbf{v}##, then ##V## becomes a (geometric) algebra.

@WendysRules, how familiar are you with geometric algebras? (the question is still wrong, btw.)

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@WendysRules, how familiar are you with geometric algebras? (the question is still wrong, btw.)
Not familiar at all, but section 2.3 is on it (##G_1, G_2, G_3##)!