Scalar Equations Of Planes [on hold] The Next CEO of Stack OverflowLines and perpendicular planesVector planes - equation of the planeEquation perpendicular to 2 non-parallel planesFind a set of scalar parametric equations.Find the equation of a planeDetermine a scalar equation is that perpendicular to the line of intersection of the planes $2x + y – z + 5 = 0$ and $x + y + 2z + 7 = 0.$Vector Algebra and planesPlanes, lines, and perpendiculars!About the equations of planesFinding an equation of a plane through the origin that is parallel to a given plane and parallel to a line.
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Scalar Equations Of Planes [on hold]
The Next CEO of Stack OverflowLines and perpendicular planesVector planes - equation of the planeEquation perpendicular to 2 non-parallel planesFind a set of scalar parametric equations.Find the equation of a planeDetermine a scalar equation is that perpendicular to the line of intersection of the planes $2x + y – z + 5 = 0$ and $x + y + 2z + 7 = 0.$Vector Algebra and planesPlanes, lines, and perpendiculars!About the equations of planesFinding an equation of a plane through the origin that is parallel to a given plane and parallel to a line.
$begingroup$
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.
calculus vectors
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Saad, John Omielan, Delta-u, Tianlalu, YiFan Mar 28 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, Delta-u, Tianlalu, YiFan
add a comment |
$begingroup$
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.
calculus vectors
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
put on hold as off-topic by Saad, John Omielan, Delta-u, Tianlalu, YiFan Mar 28 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, Delta-u, Tianlalu, YiFan
2
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago
add a comment |
$begingroup$
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.
calculus vectors
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Find the scalar equation of the plane through the points M(1,2,3) and N(3,2,-1) that is perpendicular to the plane with equation 3x + 2y + 6z +1 = 0.
calculus vectors
calculus vectors
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 1:06
lea archiboldlea archibold
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 28 at 1:06
lea archiboldlea archibold
1
asked Mar 28 at 1:06
lea archiboldlea archibold
1
1
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
lea archibold is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Saad, John Omielan, Delta-u, Tianlalu, YiFan Mar 28 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, Delta-u, Tianlalu, YiFan
put on hold as off-topic by Saad, John Omielan, Delta-u, Tianlalu, YiFan Mar 28 at 12:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, John Omielan, Delta-u, Tianlalu, YiFan
2
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago
add a comment |
2
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago
2
2
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago
add a comment |
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2
$begingroup$
Welcome to MSE, lea. What have you done so far on the problem?
$endgroup$
– amsmath
Mar 28 at 1:16
$begingroup$
I found one direction vector between the two points and I got another second direction vector by using the normal form of the plane. I know that the normal of the plane will be perpendicular to the given plane
$endgroup$
– lea archibold
Mar 28 at 23:42
$begingroup$
Correct. So, you have all the information that is needed, right?
$endgroup$
– amsmath
2 days ago
$begingroup$
Yeah I figured it out, then you do the cross product of the two direction vectors to get the new normal vector and then set up the scalar equation, finally sub in the point to isolate for D
$endgroup$
– lea archibold
2 days ago