Problem involving unit vectors

  • #1
rudransh verma
Gold Member
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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 ?
 

Answers and Replies

  • #3
rudransh verma
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i^⋅i^=−1
But cos 0=1 and magnitude of i is also 1. So how is it -1?
 
  • #4
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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}.##
 
  • #5
robphy
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It seems that this [itex]\hat i [/itex] means the complex unit , not the unit-vector along the [itex]x[/itex]-axis.
In that context, it seems the [itex]\hat{\phantom{QQ}} [/itex] 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.
 
  • #6
rudransh verma
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It seems that this [itex]\hat i [/itex] means the complex unit , not the unit-vector along the [itex]x[/itex]-axis.
In that context, it seems the [itex]\hat{\phantom{QQ}} [/itex] 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.
 
  • #7
robphy
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But I am talking about simple dot product and unit vectors only.
Is [itex]\hat \imath [/itex] akin to the complex unit? (It's hard to parse your original post.)
 
  • #9
15,565
13,677
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.
 
  • #10
rudransh verma
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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.
 
  • #11
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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.
 
  • #12
rudransh verma
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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.
 
  • #13
15,565
13,677
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 $$i \cdot i = | i |\cdot | i |\cdot (\cos (\pi/2+\pi/2)+ i \cdot \sin (\pi/2+\pi/2))=1\cdot 1\cdot (-1 + i \cdot 0)=-1$$
 
  • #14
rudransh verma
Gold Member
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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^?
 
  • #15
15,565
13,677
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.
 
  • #16
rudransh verma
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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?
 
  • #17
15,565
13,677
What if it’s dot product of i^.-i^ instead of i^.i^. What will be the answer?
$$
(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1
$$
 
  • #18
rudransh verma
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20
$$
(\hat{i}) \cdot (-\hat{i})=|\hat{i}|\cdot|-\hat{i}|\cdot\cos(\pi) =1\cdot 1\cdot (-1)=-1
$$
But the magnitude of -i^ is -1.
 
  • #19
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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.
 
  • #20
rudransh verma
Gold Member
261
20
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.
 
  • #21
15,565
13,677
Yes, assuming that the complex numbers are off the table.
 
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  • #22
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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.
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.
 
  • #23
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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.
 
  • #24
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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.
 
  • #25
15,565
13,677
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.
 

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