Problem involving unit vectors

[rudransh verma]

TL;DR

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]

$\hat{i} \,\cdot\,\hat{i} =-1$ so you have two minus signs, hence $+76$ is the answer.

Here is explained how you can type formulas on PF: https://www.physicsforums.com/help/latexhelp/

 

[rudransh verma]

i^⋅i^=−1

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

 

[fresh_42]

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 $\hat{i}^2 \in \mathbb{R}.$

 

[robphy]

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.

 

[rudransh verma]

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.

 

[robphy]

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.)

 

[rudransh verma]

Is ı^ akin to the complex unit?

I don’t know anything about complex units.

 

[fresh_42]

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.

 

[rudransh verma]

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.

 

[fresh_42]

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.

 

[rudransh verma]

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.

 

[fresh_42]

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 $$i \cdot i = | i |\cdot | i |\cdot (\cos (\pi/2+\pi/2)+ i \cdot \sin (\pi/2+\pi/2))=1\cdot 1\cdot (-1 + i \cdot 0)=-1$$

 

[rudransh verma]

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^?

 

[fresh_42]

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.

 

[rudransh verma]

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?

 

[fresh_42]

What if it’s dot product of i^.-i^ instead of i^.i^. What will be the answer?

$$
(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1
$$

 

[rudransh verma]

$$
(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1
$$

But the magnitude of -i^ is -1.

 

[fresh_42]

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.

 

[rudransh verma]

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.

 

[fresh_42]

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

 

[Mark44]

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

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.

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]

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.

 

[Mark44]

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:

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.

 

[fresh_42]

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.

 

[Mark44]

There was still the possibility of quaternions

In which case you would have needed four coordinates, not two.

 

[fresh_42]

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.