Find the line perpendicular to these two vectors
$begingroup$
A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.
What is the equation of the line?
The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.
I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.
I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?
Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?
linear-algebra
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:35
MadisonMadison
31
$endgroup$
add a comment |
$begingroup$
A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.
What is the equation of the line?
The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.
I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.
I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?
Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?
linear-algebra
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:35
MadisonMadison
31
$endgroup$
1
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27
add a comment |
$begingroup$
A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.
What is the equation of the line?
The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.
I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.
I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?
Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?
linear-algebra
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:35
MadisonMadison
31
$endgroup$
A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.
What is the equation of the line?
The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.
I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.
I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?
Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?
linear-algebra
linear-algebra
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:35
MadisonMadison
31
edited Jan 18 at 13:52
idriskameni
641319
asked Jan 18 at 12:35
MadisonMadison
31
edited Jan 18 at 13:52
idriskameni
641319
edited Jan 18 at 13:52
idriskameni
641319
edited Jan 18 at 13:52
idriskameni
641319
641319
asked Jan 18 at 12:35
MadisonMadison
31
asked Jan 18 at 12:35
MadisonMadison
31
asked Jan 18 at 12:35
MadisonMadison
31
31
1
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27
add a comment |
1
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27
1
1
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
When you find a cross product,
$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$
you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).
That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.
Hence, what you have done is OK.
edited Jan 18 at 13:20
John Hughes
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
$endgroup$
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
add a comment |
$begingroup$
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you find a cross product,
$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$
you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).
That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.
Hence, what you have done is OK.
edited Jan 18 at 13:20
John Hughes
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
$endgroup$
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
add a comment |
$begingroup$
When you find a cross product,
$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$
you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).
That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.
Hence, what you have done is OK.
edited Jan 18 at 13:20
John Hughes
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
$endgroup$
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
add a comment |
$begingroup$
When you find a cross product,
$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$
you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).
That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.
Hence, what you have done is OK.
edited Jan 18 at 13:20
John Hughes
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
$endgroup$
When you find a cross product,
$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$
you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).
That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.
Hence, what you have done is OK.
edited Jan 18 at 13:20
John Hughes
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
edited Jan 18 at 13:20
John Hughes
64k24191
edited Jan 18 at 13:20
John Hughes
64k24191
edited Jan 18 at 13:20
John Hughes
64k24191
64k24191
answered Jan 18 at 13:08
idriskameniidriskameni
641319
answered Jan 18 at 13:08
idriskameniidriskameni
641319
answered Jan 18 at 13:08
idriskameniidriskameni
641319
641319
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
add a comment |
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
1
1
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
$begingroup$
Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
$endgroup$
– Madison
Jan 18 at 13:20
add a comment |
$begingroup$
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
$endgroup$
add a comment |
$begingroup$
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
$endgroup$
add a comment |
$begingroup$
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
$endgroup$
Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.
Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.
The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:
$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$
and $$3x + z = 3(9+t)+(31-3t) = 58$$
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
answered Jan 18 at 13:44
steven gregorysteven gregory
18.1k32258
18.1k32258
add a comment |
add a comment |
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1
$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00
$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05
$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15
$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20
$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27