Basic question from vector analysis - Louis Brand Ch1, problem 2

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I am an open university second-year math undergrad student. I am taking a first course on multivariable calculus - I am solving problems from Vector analysis by Louis Brand to gain a physics-based intuition to what's going on.

I would like someone to verify my simple proof - or possible suggest alternate line of thought.

Show that $vecAB+vecCD=2vecMN$, where $M$ and $N$ are the mid-points of $AC$ and $BD$.

Solution.

$beginaligned
vecMN &= vecAN-vecAM\
&= (vecAB + vecBN) - frac12vecAC\
&= vecAB + frac12vecBD - frac12 vecAC\
&= vecAB + frac12(vecBC + vecCD) - frac12(vecAB + vecBC) \
&= frac12(vecAB + vecCD)\
2vecMN &= vecAB + vecCD
endaligned$

asked Jul 14 at 14:03

Quasar

697412

2

Looks good to me:)
â€“Â Tim Dikland
Jul 14 at 14:08

add a commentÂ |Â

up vote
2
down vote

favorite

I would like someone to verify my simple proof - or possible suggest alternate line of thought.

Show that $vecAB+vecCD=2vecMN$, where $M$ and $N$ are the mid-points of $AC$ and $BD$.

Solution.

asked Jul 14 at 14:03

Quasar

697412

2

Looks good to me:)
â€“Â Tim Dikland
Jul 14 at 14:08

add a commentÂ |Â

up vote
2
down vote

favorite

I would like someone to verify my simple proof - or possible suggest alternate line of thought.

Show that $vecAB+vecCD=2vecMN$, where $M$ and $N$ are the mid-points of $AC$ and $BD$.

Solution.

asked Jul 14 at 14:03

Quasar

697412

I would like someone to verify my simple proof - or possible suggest alternate line of thought.

Show that $vecAB+vecCD=2vecMN$, where $M$ and $N$ are the mid-points of $AC$ and $BD$.

Solution.

asked Jul 14 at 14:03

Quasar

697412

asked Jul 14 at 14:03

Quasar

697412

asked Jul 14 at 14:03

Quasar

697412

asked Jul 14 at 14:03

Quasar

697412

2

Looks good to me:)
â€“Â Tim Dikland
Jul 14 at 14:08

add a commentÂ |Â

2

Looks good to me:)
â€“Â Tim Dikland
Jul 14 at 14:08

Looks good to me:)
â€“Â Tim Dikland
Jul 14 at 14:08

add a commentÂ |Â

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

The posted proof is correct.

For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $,x=vecOX,$, and using that $,m = (a+c)/2,$ and $,n = (b+d)/2,$:

$$
beginalign
vecAB+vecCD &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2vecMN
endalign
$$

answered Jul 15 at 2:52

dxiv

54.2k64797

1

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

1

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

add a commentÂ |Â

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1 Answer
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active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
2
down vote

accepted

The posted proof is correct.

For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $,x=vecOX,$, and using that $,m = (a+c)/2,$ and $,n = (b+d)/2,$:

$$
beginalign
vecAB+vecCD &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2vecMN
endalign
$$

answered Jul 15 at 2:52

dxiv

54.2k64797

1

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

1

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

add a commentÂ |Â

up vote
2
down vote

accepted

The posted proof is correct.

For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $,x=vecOX,$, and using that $,m = (a+c)/2,$ and $,n = (b+d)/2,$:

$$
beginalign
vecAB+vecCD &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2vecMN
endalign
$$

answered Jul 15 at 2:52

dxiv

54.2k64797

1

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

1

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

add a commentÂ |Â

up vote
2
down vote

accepted

The posted proof is correct.

For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $,x=vecOX,$, and using that $,m = (a+c)/2,$ and $,n = (b+d)/2,$:

$$
beginalign
vecAB+vecCD &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2vecMN
endalign
$$

answered Jul 15 at 2:52

dxiv

54.2k64797

The posted proof is correct.

For an alternative, express everything in terms of position vectors relative to an arbitrary origin $O$. With the notation $,x=vecOX,$, and using that $,m = (a+c)/2,$ and $,n = (b+d)/2,$:

$$
beginalign
vecAB+vecCD &= (b-a)+(d-c) = (b+d)-(a+c) = 2n -2m =2(n-m)=2vecMN
endalign
$$

answered Jul 15 at 2:52

dxiv

54.2k64797

answered Jul 15 at 2:52

dxiv

54.2k64797

answered Jul 15 at 2:52

dxiv

54.2k64797

answered Jul 15 at 2:52

dxiv

54.2k64797

1

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

1

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

add a commentÂ |Â

1

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

1

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

It occurred to me later, that this is the mid-point formula, but relative to vector $a$. Enjoying doing this.
â€“Â Quasar
Jul 15 at 2:55

@Quasar Yes, you may look at it that way as well. You can find more properties of the bimedians at https://www.cut-the-knot.org/Curriculum/Geometry/AnyQuadri.shtml for example.
â€“Â dxiv
Jul 15 at 3:03

add a commentÂ |Â

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