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Why does a unit vector point in the same direction? [duplicate]
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This question already has an answer here:
Quick doubt about multiplying vectors by scalars
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I know how to compute the unit vector
$$hat{textbf{u}} = left( frac{u_1}{||u||} , dots , frac{u_n}{||u||} right)$$
and I also know how to show that this will have length 1 by using the definition of a norm.
What I do not understand is why dividing each coordinate by a scalar (in this case the norm of u) maintains the direction of the original vector.
Is there some axiom that says that dividing/multiplying each coordinate by a scalar always maintains direction? Is there a way prove the direction does not change?
It's easy to see how things scale up and down in 2D, but I'm curious if there is a way to show that the direction is always maintained regardless of the dimension.
linear-algebra geometry vectors analytic-geometry
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
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marked as duplicate by David K, José Carlos Santos linear-algebra
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Jan 17 at 16:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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$begingroup$
This question already has an answer here:
Quick doubt about multiplying vectors by scalars
4 answers
I know how to compute the unit vector
$$hat{textbf{u}} = left( frac{u_1}{||u||} , dots , frac{u_n}{||u||} right)$$
and I also know how to show that this will have length 1 by using the definition of a norm.
What I do not understand is why dividing each coordinate by a scalar (in this case the norm of u) maintains the direction of the original vector.
Is there some axiom that says that dividing/multiplying each coordinate by a scalar always maintains direction? Is there a way prove the direction does not change?
It's easy to see how things scale up and down in 2D, but I'm curious if there is a way to show that the direction is always maintained regardless of the dimension.
linear-algebra geometry vectors analytic-geometry
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
$endgroup$
marked as duplicate by David K, José Carlos Santos linear-algebra
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Jan 17 at 16:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
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How do you define direction?
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– Sean Clark
Jan 17 at 16:38
add a comment |
$begingroup$
This question already has an answer here:
Quick doubt about multiplying vectors by scalars
4 answers
I know how to compute the unit vector
$$hat{textbf{u}} = left( frac{u_1}{||u||} , dots , frac{u_n}{||u||} right)$$
and I also know how to show that this will have length 1 by using the definition of a norm.
What I do not understand is why dividing each coordinate by a scalar (in this case the norm of u) maintains the direction of the original vector.
Is there some axiom that says that dividing/multiplying each coordinate by a scalar always maintains direction? Is there a way prove the direction does not change?
It's easy to see how things scale up and down in 2D, but I'm curious if there is a way to show that the direction is always maintained regardless of the dimension.
linear-algebra geometry vectors analytic-geometry
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
$endgroup$
This question already has an answer here:
Quick doubt about multiplying vectors by scalars
4 answers
I know how to compute the unit vector
$$hat{textbf{u}} = left( frac{u_1}{||u||} , dots , frac{u_n}{||u||} right)$$
and I also know how to show that this will have length 1 by using the definition of a norm.
What I do not understand is why dividing each coordinate by a scalar (in this case the norm of u) maintains the direction of the original vector.
Is there some axiom that says that dividing/multiplying each coordinate by a scalar always maintains direction? Is there a way prove the direction does not change?
It's easy to see how things scale up and down in 2D, but I'm curious if there is a way to show that the direction is always maintained regardless of the dimension.
This question already has an answer here:
Quick doubt about multiplying vectors by scalars
4 answers
linear-algebra geometry vectors analytic-geometry
linear-algebra geometry vectors analytic-geometry
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
asked Jan 17 at 16:36
ApprenticeOfMathematicsApprenticeOfMathematics
17510
17510
marked as duplicate by David K, José Carlos Santos linear-algebra
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Jan 17 at 16:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by David K, José Carlos Santos linear-algebra
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Jan 17 at 16:49
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
4
$begingroup$
How do you define direction?
$endgroup$
– Sean Clark
Jan 17 at 16:38
add a comment |
4
$begingroup$
How do you define direction?
$endgroup$
– Sean Clark
Jan 17 at 16:38
4
4
$begingroup$
How do you define direction?
$endgroup$
– Sean Clark
Jan 17 at 16:38
$begingroup$
How do you define direction?
$endgroup$
– Sean Clark
Jan 17 at 16:38
add a comment |
1 Answer
1
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Two nonzero vectors $a, b$ are parallel iff $a=lambda b$ for some $lambdainBbb R$, and if this $lambda$ is positive, they point to the same direction.
answered Jan 17 at 16:50
BerciBerci
61.1k23674
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$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Two nonzero vectors $a, b$ are parallel iff $a=lambda b$ for some $lambdainBbb R$, and if this $lambda$ is positive, they point to the same direction.
answered Jan 17 at 16:50
BerciBerci
61.1k23674
$endgroup$
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
add a comment |
$begingroup$
Two nonzero vectors $a, b$ are parallel iff $a=lambda b$ for some $lambdainBbb R$, and if this $lambda$ is positive, they point to the same direction.
answered Jan 17 at 16:50
BerciBerci
61.1k23674
$endgroup$
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
add a comment |
$begingroup$
Two nonzero vectors $a, b$ are parallel iff $a=lambda b$ for some $lambdainBbb R$, and if this $lambda$ is positive, they point to the same direction.
answered Jan 17 at 16:50
BerciBerci
61.1k23674
$endgroup$
Two nonzero vectors $a, b$ are parallel iff $a=lambda b$ for some $lambdainBbb R$, and if this $lambda$ is positive, they point to the same direction.
answered Jan 17 at 16:50
BerciBerci
61.1k23674
answered Jan 17 at 16:50
BerciBerci
61.1k23674
answered Jan 17 at 16:50
BerciBerci
61.1k23674
answered Jan 17 at 16:50
BerciBerci
61.1k23674
61.1k23674
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
add a comment |
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
$begingroup$
After looking at this and the duplicate I see now that it comes down to the definition of directions or similar triangles which makes sense. thanks for the response
$endgroup$
– ApprenticeOfMathematics
Jan 17 at 17:03
add a comment |
4
$begingroup$
How do you define direction?
$endgroup$
– Sean Clark
Jan 17 at 16:38