Mixing
Subtraction of perpendicular vectors
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Okay, so I'm reading a proof which shows that the dot product of two perpendicular vectors is 0.
I am kinda confused by the beginning of the proof, the part where it says that for two perpendicular vectors v and w, the hypotenuse is v-w. Why is this so? Am I drawing the sketches for addition and subtraction incorrectly?
The Figure 1.7 shows that the hypotenuse is the addition of the two vectors, but the proof says that it's the subtraction.
Of course, if the hypotenuse is represented as v+b the proof wouldn't stand
What am I doing wrong?
linear-algebra
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
$endgroup$
add a comment |
$begingroup$
Okay, so I'm reading a proof which shows that the dot product of two perpendicular vectors is 0.
I am kinda confused by the beginning of the proof, the part where it says that for two perpendicular vectors v and w, the hypotenuse is v-w. Why is this so? Am I drawing the sketches for addition and subtraction incorrectly?
The Figure 1.7 shows that the hypotenuse is the addition of the two vectors, but the proof says that it's the subtraction.
Of course, if the hypotenuse is represented as v+b the proof wouldn't stand
What am I doing wrong?
linear-algebra
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
$endgroup$
$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29
add a comment |
$begingroup$
Okay, so I'm reading a proof which shows that the dot product of two perpendicular vectors is 0.
I am kinda confused by the beginning of the proof, the part where it says that for two perpendicular vectors v and w, the hypotenuse is v-w. Why is this so? Am I drawing the sketches for addition and subtraction incorrectly?
The Figure 1.7 shows that the hypotenuse is the addition of the two vectors, but the proof says that it's the subtraction.
Of course, if the hypotenuse is represented as v+b the proof wouldn't stand
What am I doing wrong?
linear-algebra
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
$endgroup$
Okay, so I'm reading a proof which shows that the dot product of two perpendicular vectors is 0.
I am kinda confused by the beginning of the proof, the part where it says that for two perpendicular vectors v and w, the hypotenuse is v-w. Why is this so? Am I drawing the sketches for addition and subtraction incorrectly?
The Figure 1.7 shows that the hypotenuse is the addition of the two vectors, but the proof says that it's the subtraction.
Of course, if the hypotenuse is represented as v+b the proof wouldn't stand
What am I doing wrong?
linear-algebra
linear-algebra
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
asked Jan 27 at 18:08
zafirzaryazafirzarya
424
424
$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29
add a comment |
$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29
$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29
add a comment |
1 Answer
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Your sketch shows perfectly that the hypotenuse is what the author claims it is. The problem is that you are on a vector space, so every vector is pinned to the origin (otherwise you would be working on an affine space). This may be the confusion point. When the author says that the third side is $v-w$, he or she means that this side corresponds to the vector $v-w$ (which is necessarily pinned at the origin) translated to some position in the affine plane.
The hypotenuse of the right triangle defined by $v$ and $w$ has in any case the same direction as $v-w$ (and not the same direction as $v+w$). Note that $w-v$ would also work, which is something you should expect (since the roles of $v$ and $w$ are symmetric).
answered Jan 27 at 18:22
PedroPedro
2,9441720
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add a comment |
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$begingroup$
Your sketch shows perfectly that the hypotenuse is what the author claims it is. The problem is that you are on a vector space, so every vector is pinned to the origin (otherwise you would be working on an affine space). This may be the confusion point. When the author says that the third side is $v-w$, he or she means that this side corresponds to the vector $v-w$ (which is necessarily pinned at the origin) translated to some position in the affine plane.
The hypotenuse of the right triangle defined by $v$ and $w$ has in any case the same direction as $v-w$ (and not the same direction as $v+w$). Note that $w-v$ would also work, which is something you should expect (since the roles of $v$ and $w$ are symmetric).
answered Jan 27 at 18:22
PedroPedro
2,9441720
$endgroup$
add a comment |
$begingroup$
Your sketch shows perfectly that the hypotenuse is what the author claims it is. The problem is that you are on a vector space, so every vector is pinned to the origin (otherwise you would be working on an affine space). This may be the confusion point. When the author says that the third side is $v-w$, he or she means that this side corresponds to the vector $v-w$ (which is necessarily pinned at the origin) translated to some position in the affine plane.
The hypotenuse of the right triangle defined by $v$ and $w$ has in any case the same direction as $v-w$ (and not the same direction as $v+w$). Note that $w-v$ would also work, which is something you should expect (since the roles of $v$ and $w$ are symmetric).
answered Jan 27 at 18:22
PedroPedro
2,9441720
$endgroup$
add a comment |
$begingroup$
Your sketch shows perfectly that the hypotenuse is what the author claims it is. The problem is that you are on a vector space, so every vector is pinned to the origin (otherwise you would be working on an affine space). This may be the confusion point. When the author says that the third side is $v-w$, he or she means that this side corresponds to the vector $v-w$ (which is necessarily pinned at the origin) translated to some position in the affine plane.
The hypotenuse of the right triangle defined by $v$ and $w$ has in any case the same direction as $v-w$ (and not the same direction as $v+w$). Note that $w-v$ would also work, which is something you should expect (since the roles of $v$ and $w$ are symmetric).
answered Jan 27 at 18:22
PedroPedro
2,9441720
$endgroup$
Your sketch shows perfectly that the hypotenuse is what the author claims it is. The problem is that you are on a vector space, so every vector is pinned to the origin (otherwise you would be working on an affine space). This may be the confusion point. When the author says that the third side is $v-w$, he or she means that this side corresponds to the vector $v-w$ (which is necessarily pinned at the origin) translated to some position in the affine plane.
The hypotenuse of the right triangle defined by $v$ and $w$ has in any case the same direction as $v-w$ (and not the same direction as $v+w$). Note that $w-v$ would also work, which is something you should expect (since the roles of $v$ and $w$ are symmetric).
answered Jan 27 at 18:22
PedroPedro
2,9441720
answered Jan 27 at 18:22
PedroPedro
2,9441720
answered Jan 27 at 18:22
PedroPedro
2,9441720
answered Jan 27 at 18:22
PedroPedro
2,9441720
2,9441720
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$begingroup$
Welcome to Maths SX! Your sketch is quite correct, but probably you misinterpret what you see. Your $v-w$ is indeed a hypotenuse – of the right triangle with right angle at the end of vector $-w$. It is also the hypotenuse of the triangle with right angle at the origin in the first quadrant.
$endgroup$
– Bernard
Jan 27 at 18:21
$begingroup$
Hi and welcome to the Math.SE. As you can see from your draft, the length of the vectors $boldsymbol{v+w}$ and $boldsymbol{v-w}$ is the same when $boldsymbol{vperp w}$. the only thing that changes is the angle of the vector, so you can conveniently assume that the hypothenuse is $boldsymbol{v-w}$ .
$endgroup$
– Daniele Tampieri
Jan 27 at 18:29