Mixing
A plane $P$ contained in $mathbb R^3$ is given by the equation $ax +by +cz+d=0$. Show that the vector $v =...
$begingroup$
This problem is from do Carmo. In calculus, we are given this vector orthogonal to the plane and derive the equation $ax+by+cz+d=0$, but here the question seems to be asking the opposite.
I'm not sure how to go about the question. Any hints?
calculus differential-geometry
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
$endgroup$
add a comment |
$begingroup$
This problem is from do Carmo. In calculus, we are given this vector orthogonal to the plane and derive the equation $ax+by+cz+d=0$, but here the question seems to be asking the opposite.
I'm not sure how to go about the question. Any hints?
calculus differential-geometry
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
$endgroup$
add a comment |
$begingroup$
This problem is from do Carmo. In calculus, we are given this vector orthogonal to the plane and derive the equation $ax+by+cz+d=0$, but here the question seems to be asking the opposite.
I'm not sure how to go about the question. Any hints?
calculus differential-geometry
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
$endgroup$
This problem is from do Carmo. In calculus, we are given this vector orthogonal to the plane and derive the equation $ax+by+cz+d=0$, but here the question seems to be asking the opposite.
I'm not sure how to go about the question. Any hints?
calculus differential-geometry
calculus differential-geometry
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
asked Jan 20 at 3:52
WolfgangWolfgang
4,29743377
4,29743377
add a comment |
add a comment |
2 Answers
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$begingroup$
Let $P_1 = left(x_1, y_1, z_1right)$ and $P_2 = left(x_2, y_2, z_2right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = left(x_1 - x_2, y_1 - y_2, z_1 - z_2right)$ is a vector in this plane. Taking the dot product of this with $v = left(a, b, cright)$ gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) tag{1}label{eq1}$$
As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives
$$ax_1 + by_1 + cz_1 + d = 0 tag{2}label{eq2}$$
and
$$ax_2 + by_2 + cz_2 + d = 0 tag{3}label{eq3}$$
Subtracting eqref{eq3} from eqref{eq2} gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) = 0 tag{4}label{eq4}$$
Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
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$begingroup$
Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that:
$$(v,w) = am+bn+pc = 0.$$
Now, a vector on a plane is really the same as two points on a plane, call them $
Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.
Can you follow it from here?
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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$begingroup$
Let $P_1 = left(x_1, y_1, z_1right)$ and $P_2 = left(x_2, y_2, z_2right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = left(x_1 - x_2, y_1 - y_2, z_1 - z_2right)$ is a vector in this plane. Taking the dot product of this with $v = left(a, b, cright)$ gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) tag{1}label{eq1}$$
As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives
$$ax_1 + by_1 + cz_1 + d = 0 tag{2}label{eq2}$$
and
$$ax_2 + by_2 + cz_2 + d = 0 tag{3}label{eq3}$$
Subtracting eqref{eq3} from eqref{eq2} gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) = 0 tag{4}label{eq4}$$
Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
$endgroup$
add a comment |
$begingroup$
Let $P_1 = left(x_1, y_1, z_1right)$ and $P_2 = left(x_2, y_2, z_2right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = left(x_1 - x_2, y_1 - y_2, z_1 - z_2right)$ is a vector in this plane. Taking the dot product of this with $v = left(a, b, cright)$ gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) tag{1}label{eq1}$$
As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives
$$ax_1 + by_1 + cz_1 + d = 0 tag{2}label{eq2}$$
and
$$ax_2 + by_2 + cz_2 + d = 0 tag{3}label{eq3}$$
Subtracting eqref{eq3} from eqref{eq2} gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) = 0 tag{4}label{eq4}$$
Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
$endgroup$
add a comment |
$begingroup$
Let $P_1 = left(x_1, y_1, z_1right)$ and $P_2 = left(x_2, y_2, z_2right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = left(x_1 - x_2, y_1 - y_2, z_1 - z_2right)$ is a vector in this plane. Taking the dot product of this with $v = left(a, b, cright)$ gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) tag{1}label{eq1}$$
As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives
$$ax_1 + by_1 + cz_1 + d = 0 tag{2}label{eq2}$$
and
$$ax_2 + by_2 + cz_2 + d = 0 tag{3}label{eq3}$$
Subtracting eqref{eq3} from eqref{eq2} gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) = 0 tag{4}label{eq4}$$
Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
$endgroup$
Let $P_1 = left(x_1, y_1, z_1right)$ and $P_2 = left(x_2, y_2, z_2right)$ be any $2$ distinct points in the plane. Then $P_1 - P_2 = left(x_1 - x_2, y_1 - y_2, z_1 - z_2right)$ is a vector in this plane. Taking the dot product of this with $v = left(a, b, cright)$ gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) tag{1}label{eq1}$$
As $P_1$ and $P_2$ are on the plane, they must satisfy the equation $ax + by + cz + d = 0$. Thus, this gives
$$ax_1 + by_1 + cz_1 + d = 0 tag{2}label{eq2}$$
and
$$ax_2 + by_2 + cz_2 + d = 0 tag{3}label{eq3}$$
Subtracting eqref{eq3} from eqref{eq2} gives
$$aleft(x_1 - x_2right) + bleft(y_1 - y_2right) + cleft(z_1 - z_2right) = 0 tag{4}label{eq4}$$
Thus, since the dot product is $0$, then the $2$ vectors are perpendicular. As $P_1$ and $P_2$ are any $2$ points, this shows the vector $v$ is perpendicular to the entire plane.
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
edited Jan 20 at 4:40
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
answered Jan 20 at 4:03
John OmielanJohn Omielan
3,4801215
3,4801215
add a comment |
add a comment |
$begingroup$
Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that:
$$(v,w) = am+bn+pc = 0.$$
Now, a vector on a plane is really the same as two points on a plane, call them $
Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.
Can you follow it from here?
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
$endgroup$
add a comment |
$begingroup$
Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that:
$$(v,w) = am+bn+pc = 0.$$
Now, a vector on a plane is really the same as two points on a plane, call them $
Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.
Can you follow it from here?
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
$endgroup$
add a comment |
$begingroup$
Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that:
$$(v,w) = am+bn+pc = 0.$$
Now, a vector on a plane is really the same as two points on a plane, call them $
Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.
Can you follow it from here?
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
$endgroup$
Pick any vector $w = (m,n,p)$ on the plane $P.$ Then, you need to prove that:
$$(v,w) = am+bn+pc = 0.$$
Now, a vector on a plane is really the same as two points on a plane, call them $
Q_i = (x_i, y_i,z_i)$ with $i=1,2.$ Then, what is $m,n,p$ in terms of these coordinates? Don't also forget that $Q_i$ satisfies the equation of the plane.
Can you follow it from here?
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
answered Jan 20 at 3:58
dezdichadodezdichado
6,3691929
6,3691929
add a comment |
add a comment |
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