Mixing
finding out if two vectors are perpendicular or parallel
$begingroup$
I'm not sure if I quite get this. For example,
(1, -1) and (-3, 3)
take the cross product, you will end up with
-3 + (-3)
This doesn't equal 0, so it's not perpendicular. So that leaves me with it being parallel. When are two vectors parallel?
calculus vectors
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
$endgroup$
add a comment |
$begingroup$
I'm not sure if I quite get this. For example,
(1, -1) and (-3, 3)
take the cross product, you will end up with
-3 + (-3)
This doesn't equal 0, so it's not perpendicular. So that leaves me with it being parallel. When are two vectors parallel?
calculus vectors
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
$endgroup$
$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31
add a comment |
$begingroup$
I'm not sure if I quite get this. For example,
(1, -1) and (-3, 3)
take the cross product, you will end up with
-3 + (-3)
This doesn't equal 0, so it's not perpendicular. So that leaves me with it being parallel. When are two vectors parallel?
calculus vectors
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
$endgroup$
I'm not sure if I quite get this. For example,
(1, -1) and (-3, 3)
take the cross product, you will end up with
-3 + (-3)
This doesn't equal 0, so it's not perpendicular. So that leaves me with it being parallel. When are two vectors parallel?
calculus vectors
calculus vectors
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
asked Jun 13 '15 at 17:27
Kevin R.Kevin R.
238148
238148
$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31
add a comment |
$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31
$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You can find the angle between the two vectors
$$theta=cos^{-1}(frac{v_1.v_2}{|v_1||v_2|})$$
if $theta=0$or $180$ the two vectors are parallel
if $theta=90$ the two vetors are perpendicular
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
$endgroup$
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
add a comment |
$begingroup$
Two vectors $v1(x1,y1)$ and $v2(x2,y2)$ are parallel iff $x1y2 = x2y1$.
edited Jan 31 at 18:57
paulmelnikow
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
$endgroup$
add a comment |
$begingroup$
They are parallel if and only if they are different by a factor i.e. (1,3) and (-2,-6).
The dot product will be 0 for perpendicular vectors i.e. they cross at exactly 90 degrees.
When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular.
answered Jun 13 '15 at 17:30
user167289user167289
11911
$endgroup$
add a comment |
$begingroup$
Two vectors are parallel when they are scalar multiples of each other. In other words, if you can multiply one vector by a constant and end up with the other vector.
The rough reason for this is that multiplying by a scalar doesn't rotate the vector at all (it can stretch or flip the vector, but it doesn't change the direction).
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can find the angle between the two vectors
$$theta=cos^{-1}(frac{v_1.v_2}{|v_1||v_2|})$$
if $theta=0$or $180$ the two vectors are parallel
if $theta=90$ the two vetors are perpendicular
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
$endgroup$
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
add a comment |
$begingroup$
You can find the angle between the two vectors
$$theta=cos^{-1}(frac{v_1.v_2}{|v_1||v_2|})$$
if $theta=0$or $180$ the two vectors are parallel
if $theta=90$ the two vetors are perpendicular
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
$endgroup$
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
add a comment |
$begingroup$
You can find the angle between the two vectors
$$theta=cos^{-1}(frac{v_1.v_2}{|v_1||v_2|})$$
if $theta=0$or $180$ the two vectors are parallel
if $theta=90$ the two vetors are perpendicular
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
$endgroup$
You can find the angle between the two vectors
$$theta=cos^{-1}(frac{v_1.v_2}{|v_1||v_2|})$$
if $theta=0$or $180$ the two vectors are parallel
if $theta=90$ the two vetors are perpendicular
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
edited Jun 13 '15 at 19:53
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
answered Jun 13 '15 at 17:46
E.H.EE.H.E
16.1k11969
16.1k11969
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
add a comment |
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
In OP's problem, you get $$ cos^{-1} left( frac{-6}{ sqrt{2} cdot 3 sqrt{2}} right) = cos^{-1} (-1) = 180º , $$ so you may want to mention the case of anti-parallelism. (Some folks do use "parallel" to refer to any vectors that are collinear.)
$endgroup$
– colormegone
Jun 13 '15 at 19:39
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@RecklessReckoner Thanks
$endgroup$
– E.H.E
Jun 13 '15 at 19:53
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
$begingroup$
@colormegone: Please don't confuse (°, U+00B0: DEGREE SIGN) and (º, U+00BA: MASCULINE ORDINAL INDICATOR)!
$endgroup$
– Andreas Rejbrand
Feb 24 at 14:35
add a comment |
$begingroup$
Two vectors $v1(x1,y1)$ and $v2(x2,y2)$ are parallel iff $x1y2 = x2y1$.
edited Jan 31 at 18:57
paulmelnikow
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
$endgroup$
add a comment |
$begingroup$
Two vectors $v1(x1,y1)$ and $v2(x2,y2)$ are parallel iff $x1y2 = x2y1$.
edited Jan 31 at 18:57
paulmelnikow
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
$endgroup$
add a comment |
$begingroup$
Two vectors $v1(x1,y1)$ and $v2(x2,y2)$ are parallel iff $x1y2 = x2y1$.
edited Jan 31 at 18:57
paulmelnikow
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
$endgroup$
Two vectors $v1(x1,y1)$ and $v2(x2,y2)$ are parallel iff $x1y2 = x2y1$.
edited Jan 31 at 18:57
paulmelnikow
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
edited Jan 31 at 18:57
paulmelnikow
1032
edited Jan 31 at 18:57
paulmelnikow
1032
edited Jan 31 at 18:57
paulmelnikow
1032
1032
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
answered Jun 13 '15 at 17:38
Amir NaseriAmir Naseri
959516
959516
add a comment |
add a comment |
$begingroup$
They are parallel if and only if they are different by a factor i.e. (1,3) and (-2,-6).
The dot product will be 0 for perpendicular vectors i.e. they cross at exactly 90 degrees.
When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular.
answered Jun 13 '15 at 17:30
user167289user167289
11911
$endgroup$
add a comment |
$begingroup$
They are parallel if and only if they are different by a factor i.e. (1,3) and (-2,-6).
The dot product will be 0 for perpendicular vectors i.e. they cross at exactly 90 degrees.
When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular.
answered Jun 13 '15 at 17:30
user167289user167289
11911
$endgroup$
add a comment |
$begingroup$
They are parallel if and only if they are different by a factor i.e. (1,3) and (-2,-6).
The dot product will be 0 for perpendicular vectors i.e. they cross at exactly 90 degrees.
When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular.
answered Jun 13 '15 at 17:30
user167289user167289
11911
$endgroup$
They are parallel if and only if they are different by a factor i.e. (1,3) and (-2,-6).
The dot product will be 0 for perpendicular vectors i.e. they cross at exactly 90 degrees.
When you calculate the dot product and your answer is non-zero it just means the two vectors are not perpendicular.
answered Jun 13 '15 at 17:30
user167289user167289
11911
answered Jun 13 '15 at 17:30
user167289user167289
11911
answered Jun 13 '15 at 17:30
user167289user167289
11911
answered Jun 13 '15 at 17:30
user167289user167289
11911
11911
add a comment |
add a comment |
$begingroup$
Two vectors are parallel when they are scalar multiples of each other. In other words, if you can multiply one vector by a constant and end up with the other vector.
The rough reason for this is that multiplying by a scalar doesn't rotate the vector at all (it can stretch or flip the vector, but it doesn't change the direction).
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
$endgroup$
add a comment |
$begingroup$
Two vectors are parallel when they are scalar multiples of each other. In other words, if you can multiply one vector by a constant and end up with the other vector.
The rough reason for this is that multiplying by a scalar doesn't rotate the vector at all (it can stretch or flip the vector, but it doesn't change the direction).
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
$endgroup$
add a comment |
$begingroup$
Two vectors are parallel when they are scalar multiples of each other. In other words, if you can multiply one vector by a constant and end up with the other vector.
The rough reason for this is that multiplying by a scalar doesn't rotate the vector at all (it can stretch or flip the vector, but it doesn't change the direction).
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
$endgroup$
Two vectors are parallel when they are scalar multiples of each other. In other words, if you can multiply one vector by a constant and end up with the other vector.
The rough reason for this is that multiplying by a scalar doesn't rotate the vector at all (it can stretch or flip the vector, but it doesn't change the direction).
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
answered Jun 13 '15 at 17:31
ptrsinclairptrsinclair
38628
38628
add a comment |
add a comment |
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$begingroup$
You mean dot product, not cross product.
$endgroup$
– user26486
Jun 13 '15 at 17:29
$begingroup$
I think did you mean the "dot product".
$endgroup$
– Mann
Jun 13 '15 at 17:29
$begingroup$
Vectors $(a_1,a_2)$ and $(b_1,b_2)$ are parallel iff $frac{a_1}{b_1}=frac{a_2}{b_2}$
$endgroup$
– user26486
Jun 13 '15 at 17:31