# Problem involving unit vectors

• B
• rudransh verma
In summary, the dot product or scalar product of two vectors, in this case ##\hat{i}## and ##\hat{i}##, is equal to the magnitude of both vectors multiplied by the cosine of the angle between them. However, when considering the vector ##\hat{i}## as a complex number, the dot product is equal to ##-1## instead of ##1## due to the properties of complex numbers. The notation with ##i## can be confusing and it is important to clarify the context in which it is being used.
rudransh verma
Gold Member
TL;DR Summary
Why is i cap. -i cap producing two values?
For example is this correct : 19icap.4(-i cap) = 76(i.-i)= 76
Or is it , take - out. Then -76(icap.icap)= -76
Is it -76 or 76 ?

fresh_42 said:
i^⋅i^=−1
But cos 0=1 and magnitude of i is also 1. So how is it -1?

rudransh verma said:
But cos 0=1 and magnitude of i is also 1. So how is it -1?
Because ##\hat{i}## solves the real equation
$$\hat{i}^2 \sim \begin{bmatrix} 0\\1 \end{bmatrix}\cdot \begin{bmatrix} 0\\1 \end{bmatrix}=\begin{bmatrix} -1\\0 \end{bmatrix}$$
which makes the difference between ##\mathbb{C}## and ##\mathbb{R}^2##. We also have a direction, not only magnitude, and we have a multiplication ##\hat{i}^2 \in \mathbb{R}.##

It seems that this $\hat i$ means the complex unit , not the unit-vector along the $x$-axis.
In that context, it seems the $\hat{\phantom{QQ}}$ is unnecessary, and potentially confusing at first glance.

Furthermore, it seems this dot operation is complex multiplication, not the familiar Euclidean dot product over a real vector space.

robphy said:
It seems that this $\hat i$ means the complex unit , not the unit-vector along the $x$-axis.
In that context, it seems the $\hat{\phantom{QQ}}$ is unnecessary, and potentially confusing at first glance.

Furthermore, it seems this dot operation is complex multiplication, not the familiar Euclidean dot product over a real vector space.
But I am talking about simple dot product and unit vectors only.

rudransh verma said:
But I am talking about simple dot product and unit vectors only.
Is $\hat \imath$ akin to the complex unit? (It's hard to parse your original post.)

robphy said:
Is ı^ akin to the complex unit?
I don’t know anything about complex units.

rudransh verma said:
I don’t know anything about complex units.
The question is whether you consider ##\hat{i}## as the vector ##(0,1)## in the complex plane, or the complex number ## i ##. Not that it makes much difference, but complex numbers are more than just a real vector.

fresh_42 said:
The question is whether you consider ##\hat{i}## as the vector ##(0,1)## in the complex plane, or the complex number ## i ##. Not that it makes much difference, but complex numbers are more than just a real vector.
Yah! As a vector. I don’t know what ^ is called where you are but it’s called as cap.

rudransh verma said:
Yah! As a vector. I don’t know what ^ is called where you are but it’s called as cap.

The notations ##\hat{i}\, , \,\hat{j}\, , \,\hat{k}## are sometimes used in physics to name the three Euclidean directions in space. In that case, we only have
$$\hat{i}\cdot \hat{i} = \begin{bmatrix}0\\1\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=1$$
But if ##\hat{i}## is a complex number, a vector in the complex number plane, then we have
$$\hat{i}\cdot\hat{i}=\begin{bmatrix}0\\1\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix} -1\\ 0 \end{bmatrix}=-\hat{1}$$

So all depends on what you mean by ##\hat{i},## i.e. the context you took your formula from.

fresh_42 said:
The notations i^,j^,k^ are sometimes used in physics to name the three Euclidean directions in space. In that case, we only have
a.b= mod of a* mod of b* cos theta
Mod means magnitude in physics!
What is i^.-i^?
I am confused what will be the mod of -i^. Is it -1 or 1?
I guess magnitude of -i^ will be -1.

rudransh verma said:
a.b= mod of a* mod of b* cos theta
Mod means magnitude in physics!
What is i^.-i^?
I am confused what will be the mod of -i^. Is it -1 or 1?
I guess magnitude of -i^ will be -1.
Sounds like ##|a\cdot b|=|a|\cdot|b|\cdot\cos \theta## to me, i.e. "what is the modulus of the product" not what is the product itself which would be $$i \cdot i = | i |\cdot | i |\cdot (\cos (\pi/2+\pi/2)+ i \cdot \sin (\pi/2+\pi/2))=1\cdot 1\cdot (-1 + i \cdot 0)=-1$$

fresh_42 said:
Sounds like ##|a\cdot b|=|a|\cdot|b|\cdot\cos \theta## to me, i.e. "what is the modulus of the product" not what is the product itself which would be $$i \cdot i = | i |\cdot | i |\cdot (\cos (\pi/2+\pi/2)+ i \cdot \sin (\pi/2+\pi/2))=1\cdot 1\cdot (-1 + i \cdot 0)=-1$$
When we find dot product of two vectors what should we take ,mod of vectors ?
Also what does mod do to a vector? Does it give the magnitude? If yes, what will be the mod of -i^?

rudransh verma said:
When we find dot product of two vectors what should we take ,mod of vectors ?
Also what does mod do to a vector? Does it give the magnitude? If yes, what will be the mod of -i^?
Yes, the dot product (scalar product, inner product) goes like ##\vec{a}\cdot\vec{b}=|\vec{a}|\cdot|\vec{b}|\cdot\cos(\sphericalangle (\vec{a},\vec{b})).## In that case we have
$$\hat{i}^2=(1,0)\cdot (1,0)= |(1,0)|\cdot|(1,0)|\cdot\cos(0)=1$$
The result is a number. In that case the problem has nothing to do with complex numbers. ##\hat{i}## is only the short form for ##(1,0,0)## or whatever the number of dimensions is.

fresh_42 said:
Yes, the dot product (scalar product, inner product) goes like a→⋅b→=|a→|⋅|b→|⋅cos⁡(∢(a→,b→)). In that case we have
What if it’s dot product of i^.-i^ instead of i^.i^. What will be the answer?

rudransh verma said:
What if it’s dot product of i^.-i^ instead of i^.i^. What will be the answer?
$$(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1$$

fresh_42 said:
$$(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1$$
But the magnitude of -i^ is -1.

rudransh verma said:
But the magnitude of -i^ is -1.
The magnitude is the length. And the lengths of ##\hat{i}## and ##-\hat{i}## is ##1## in both cases. Only the direction is opposite of the other.

The notation with ## i ## is very confusing in this context.

fresh_42 said:
The magnitude is the length. And the lengths of ##\hat{i}## and ##-\hat{i}## is ##1## in both cases. Only the direction is opposite of the other.

The notation with ## i ## is very confusing in this context.
So back to OP. It will be -76 in both cases, whether we operate minus with dot product or we take out minus in the beginning.

Yes, assuming that the complex numbers are off the table.

rudransh verma
rudransh verma said:
But cos 0=1 and magnitude of i is also 1. So how is it -1?

fresh_42 said:
The notations ##\hat{i}\, , \,\hat{j}\, , \,\hat{k}## are sometimes used in physics to name the three Euclidean directions in space. In that case, we only have
$$\hat{i}\cdot \hat{i} = \begin{bmatrix}0\\1\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=1$$
No, in that case it would be
$$\hat{i}\cdot \hat{i} = \begin{bmatrix}1 \\0\\0\end{bmatrix}\cdot \begin{bmatrix}1\\0\\0\end{bmatrix}=1$$
You're still thinking of i as being the imaginary unit, but being represented as a vector.
fresh_42 said:
But if ##\hat{i}## is a complex number, a vector in the complex number plane, then we have
$$\hat{i}\cdot\hat{i}=\begin{bmatrix}0\\1\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix} -1\\ 0 \end{bmatrix}=-\hat{1}$$

So all depends on what you mean by ##\hat{i},## i.e. the context you took your formula from.

Mark44 said:
No, in that case it would be
$$\hat{i}\cdot \hat{i} = \begin{bmatrix}1 \\0\\0\end{bmatrix}\cdot \begin{bmatrix}1\\0\\0\end{bmatrix}=1$$
You're still thinking of i as being the imaginary unit, but being represented as a vector.
No, I did not. I only changed the order to make the product look equal to the second one, emphasizing the different RHS by equal LHS. Since nobody ever used the words basis or dimension, I felt free to choose one where ##\hat{i}=(0,1)## and dimension is two. I mentioned the possibly three-dimensional case in another post.

fresh_42 said:
No, I did not. I only changed the order to make the product look equal to the second one, emphasizing the different RHS by equal LHS.
Leading up to what I quoted, you wrote this:
fresh_42 said:
The notations i^,j^,k^ are sometimes used in physics to name the three Euclidean directions in space. In that case, we only have $$\hat{i}\cdot \hat{i} = \begin{bmatrix}0\\1\end{bmatrix}\cdot \begin{bmatrix}0\\1\end{bmatrix}=1$$
So the context here was ##\mathbb R^3## with the standard basis vectors i, j, and k, all of which have three components. And as I'm sure you know, i has 1 in its first component, and 0 in the other two.

There was still the possibility of quaternions, in which case ##\hat{i}^2=\hat{j}^2=-1.##
I decided to demonstrate the different multiplications, not a specific use. Mainly because I was still guessing what exactly has been meant.

fresh_42 said:
There was still the possibility of quaternions
In which case you would have needed four coordinates, not two.

Mark44 said:
In which case you would have needed four coordinates, not two.
Such things happen when you need more than a dozen posts only to figure out what might have been meant.

## 1. What is a unit vector?

A unit vector is a vector that has a magnitude of 1 and points in a specific direction. It is commonly used in mathematics and physics to represent a specific direction or orientation.

## 2. How do you find the unit vector of a given vector?

To find the unit vector of a given vector, you need to divide the vector by its magnitude. This will result in a vector with the same direction, but a magnitude of 1.

## 3. What is the significance of unit vectors in problem solving?

Unit vectors are important in problem solving because they allow us to represent and manipulate directions and orientations in a simplified manner. They also help us to break down complex problems into smaller, more manageable parts.

## 4. How are unit vectors used in vector operations?

Unit vectors are commonly used in vector operations, such as addition, subtraction, and multiplication. They help to simplify the calculations and make them easier to understand.

## 5. Can unit vectors be negative?

No, unit vectors cannot be negative. They always have a magnitude of 1 and therefore cannot have a negative value. However, they can point in any direction, including negative directions.

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