Mixing
Find the equation of the plane that passes through the points
$begingroup$
Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$
The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$
But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.
linear-algebra orthogonality
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
$endgroup$
|
show 1 more comment
$begingroup$
Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$
The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$
But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.
linear-algebra orthogonality
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
$endgroup$
$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
2
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29
|
show 1 more comment
$begingroup$
Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$
The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$
But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.
linear-algebra orthogonality
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
$endgroup$
Find the equation of the plane that passes through the points:
$$P_1=(1,1,2)\P_2=(2,3,3) \P_3=(3,-3,3)$$
The answer writes:
Let $x=vec{P_1P_2}=begin{bmatrix}1\2\1end{bmatrix}$ and $y=vec{P_1P_3}=begin{bmatrix}2\-4\1end{bmatrix}$
The normal vector $N$ must be orthogonal to both $X$ and $Y$. If we set $N=Xtimes Y=begin{bmatrix}6\1\-8end{bmatrix}$
Then $N$ will be a normal vector to the plane that passes through the given points. Using point $P_1$, we see that the equation of the plane is
$$6(x-1)+(y-1)-8(z-1)=0$$
But I don't know what the answer means, especially how it calculates X×Y and get $begin{bmatrix}6\1\-8end{bmatrix}$, I really have no idea.
linear-algebra orthogonality
linear-algebra orthogonality
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
edited Jan 5 at 4:24
Mostafa Ayaz
15.4k3939
15.4k3939
asked Jan 5 at 4:04
Shadow ZShadow Z
396
asked Jan 5 at 4:04
Shadow ZShadow Z
396
asked Jan 5 at 4:04
Shadow ZShadow Z
396
396
$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
2
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29
|
show 1 more comment
$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
2
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29
$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
2
2
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).
$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.
$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$
The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
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1 Answer
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votes
$begingroup$
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).
$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.
$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$
The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).
$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.
$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$
The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
add a comment |
$begingroup$
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).
$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.
$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$
The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
$endgroup$
According to geometry, a plane can be uniquely determined given the normal vector of it and a point lying on it (the reason is intuitively simple). To solve this problem, a number of arguments are employed:
$A_1$: The cross product of two linearly independent vectors, is always orthogonal on them (linearly independent in the case of two vectors means that they are not any multiple of each other).
$A_2$: If two points belong to a plane, then so is the corresponding vector whose endpoints are the given points of the plane.
$A_3$: The cross product of two vectors $v_1=ahat i+bhat j+chat k$ and $v_2=dhat i+ehat j+fhat k$ can be calculated as $$v_1times v_2=(bf-ce)hat i+(cd-af)hat j+(ae-bd)hat k$$
The normal vector then would be$$v=v_1times v_2=begin{bmatrix}1\2\1end{bmatrix}times begin{bmatrix}2\-4\1end{bmatrix}=begin{bmatrix}2times 1-1times(-4)\2times 1-1times 1\1times(-4)-2times 2end{bmatrix}=begin{bmatrix}6\1\-8end{bmatrix}$$therefore $$(x,y,z)cdot v=k$$for some constant $k$. By substitution we finally obtain$$6x+y-8z=-1$$
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
answered Jan 5 at 5:04
Mostafa AyazMostafa Ayaz
15.4k3939
15.4k3939
add a comment |
add a comment |
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$begingroup$
Have you computed the vectors $x$ and $y$ and their cross product $xtimes y$?
$endgroup$
– Dave
Jan 5 at 4:06
$begingroup$
@Dave yes, but I get (2 -8 1)^T instead of (6 1 -8)^T
$endgroup$
– Shadow Z
Jan 5 at 4:24
$begingroup$
I get $[6~1~-8]^T$. What steps are you taking to compute this (include these steps in the body of your question for us to see)?
$endgroup$
– Dave
Jan 5 at 4:25
$begingroup$
I just simply times the the first, second, third number of the vectors....I have no idea how to compute..
$endgroup$
– Shadow Z
Jan 5 at 4:27
2
$begingroup$
Found the leak! That is not what $xtimes y$ means. Rather, $xtimes y$ is the cross product of $x$ and $y$, which is computed as $$xtimes y=detbegin{bmatrix}e_1&e_2&e_3\x_1&x_2&x_3\y_1&y_2&y_3end{bmatrix}$$ where $e_i$ are the standard basis vectors in $mathbb R^3$ and $x_i$ and $y_i$ are the components of $x$ and $y$, respectively. Have you not covered cross product yet in your course, because it seems that this question is assuming you have.
$endgroup$
– Dave
Jan 5 at 4:29