# Find Linear Equation of a plane

• PsychonautQQ
In summary: Then draw another vector ##\langle 0,0,0\rangle## with its tip at the origin and its tail at the tip of the first vector. That vector is in the plane. You should be able to write down the coordinates of the tip of that vector.
PsychonautQQ
1. The problem estatement, all variables and given/known data
Find the Linear equation of a plane passing through the origin and points (6,2,2) and (5,-1,1)

## Homework Equations

Linear equation of a plane takes the form ax+by+cz+d = 0

## The Attempt at a Solution

So I believe I can find the direction numbers a,b,c by taking the first position minus the second position.

a = 6 - 5 = 1
b = 2 - -1 = 3
c = 2 - 1 = 2

I don't know what to do from here though. I could plug in one of the locations into the formula of format
a(x-xI) + b(y-yi) + c(z-zi) = 0
say I plug in the first position (6,2,2)
1(x-6) + 3(y-2) + 2(z-2) = 0
to get x+3y+2z-16=0
but my I'm typing that into the answer box and it's saying I'm wrong

PsychonautQQ said:
1. The problem estatement, all variables and given/known data
Find the Linear equation of a plane passing through the origin and points (6,2,2) and (5,-1,1)

## Homework Equations

Linear equation of a plane takes the form ax+by+cz+d = 0

## The Attempt at a Solution

So I believe I can find the direction numbers a,b,c by taking the first position minus the second position.

a = 6 - 5 = 1
b = 2 - -1 = 3
c = 2 - 1 = 2

I don't know what to do from here though. I could plug in one of the locations into the formula of format
a(x-xI) + b(y-yi) + c(z-zi) = 0
say I plug in the first position (6,2,2)
1(x-6) + 3(y-2) + 2(z-2) = 0
to get x+3y+2z-16=0
but my I'm typing that into the answer box and it's saying I'm wrong

It's obviously wrong, since the origin (x = 0, y = 0, z = 0) doesn't satisfy the equation.

If we call your two given points P and Q, the cross product of the vectors OP and OQ is perpendicular to both. That's a normal to your plane. Once you know a normal and a point on the plane, you can quickly get the plane equation.

BTW, it's good practice to check your work by making sure that the points that are supposed to be on the plane are actually solutions to the equation you came up with. If all of them are, you can be pretty sure that your equation is correct. If at least one is not a solution to the equation, that's a tipoff that something is wrong.

PsychonautQQ said:
1. The problem estatement, all variables and given/known data
Find the Linear equation of a plane passing through the origin and points (6,2,2) and (5,-1,1)

## Homework Equations

Linear equation of a plane takes the form ax+by+cz+d = 0

## The Attempt at a Solution

So I believe I can find the direction numbers a,b,c by taking the first position minus the second position.

a = 6 - 5 = 1
b = 2 - -1 = 3
c = 2 - 1 = 2

I don't know what to do from here though. I could plug in one of the locations into the formula of format
a(x-xI) + b(y-yi) + c(z-zi) = 0
say I plug in the first position (6,2,2)
1(x-6) + 3(y-2) + 2(z-2) = 0
to get x+3y+2z-16=0
but my I'm typing that into the answer box and it's saying I'm wrong

The equation of the plane you got does not have (5,-1,1) or (0,0,0) as a solution, I.e it does not pass through that point. The eqn a(x-xI) + b(y-yI) + c(z-zI) = 0 can be recast as ##\vec{n} \cdot \vec{r} = 0## where ##\vec{n}## is a normal to the plane and ##\vec{r}## is a vector in the plane.

I cross OP with OQ and got
0i+4j-16k and that gives me n.

n = (a,b,c) = (0,4,-16)
and then say I use the point (6,2,2)

Can i plug this values into the equation
a(x-xi) + b(y-yi) + c(z-zi) = 0
0 + 4(y-2) - 16(z-2) = 0
4y - 8 - 16z + 32 = 0
4y - 16z + 24 = 0
Web work is saying this is wrong but I don't know why it looks right?

And then you said n dot r = 0
so (0,4,-16) dot (xi+yj+zk) = 0 to get r which maybe are the real (a,b,c) values to the equation above..?
if I do this to solve for r there can be more than one correct answer.. the lowest one I guess would be x=0 y=4 and z=1 but I don't know if I'm on the right track here...

PsychonautQQ said:
I cross OP with OQ and got
0i+4j-16k and that gives me n.
That's not what I get.

When you calculate a cross product, there are lots of opportunities to make errors, especially sign errors. When you're done it's a good idea to check your work. Is <6, 2, 2> * <0, 4, -16> = 0? Is <5, -1, 1> * <0, 4, -16> = 0?
PsychonautQQ said:
n = (a,b,c) = (0,4,-16)
and then say I use the point (6,2,2)

Can i plug this values into the equation
a(x-xi) + b(y-yi) + c(z-zi) = 0
0 + 4(y-2) - 16(z-2) = 0
4y - 8 - 16z + 32 = 0
4y - 16z + 24 = 0
Web work is saying this is wrong but I don't know why it looks right?

And then you said n dot r = 0
so (0,4,-16) dot (xi+yj+zk) = 0 to get r which maybe are the real (a,b,c) values to the equation above..?
if I do this to solve for r there can be more than one correct answer.. the lowest one I guess would be x=0 y=4 and z=1 but I don't know if I'm on the right track here...

PsychonautQQ said:
And then you said n dot r = 0
so (0,4,-16) dot (xi+yj+zk) = 0 to get r which maybe are the real (a,b,c) values to the equation above..?
if I do this to solve for r there can be more than one correct answer.. the lowest one I guess would be x=0 y=4 and z=1 but I don't know if I'm on the right track here...

Once you fix the normal, you want to find a vector in the plane between some arbritary point (x,y,z) in the plane and a known point. Draw some vector ##\langle x,y,z\rangle## with its tail at the origin and the tip at a point in the plane. This will allow you to find a vector in the plane which extends between some arbitrary (x,y,z) and a known point in the plane.

## What is a linear equation of a plane?

A linear equation of a plane is a mathematical expression that represents a straight line on a two-dimensional plane. It can be written in the form of y = mx + b, where m is the slope of the line and b is the y-intercept.

## How do you find the linear equation of a plane?

To find the linear equation of a plane, you need two pieces of information: the slope and a point on the line. The slope can be calculated by finding the change in y over the change in x between two points on the line. Once you have the slope, you can plug it into the slope-intercept form of a line, y = mx + b, and solve for b using the given point.

## Can a linear equation of a plane have a negative slope?

Yes, a linear equation of a plane can have a negative slope. A negative slope simply means that the line is decreasing from left to right, instead of increasing. This is represented by a line that slopes downward from left to right.

## What is the importance of finding the linear equation of a plane?

Finding the linear equation of a plane is important in many fields, including physics, engineering, and statistics. It allows us to model and predict relationships between two variables, and can be used to make calculations and solve problems involving those variables.

## Can a linear equation of a plane be used to represent a curve?

No, a linear equation of a plane can only represent a straight line. To represent a curve, a different type of equation, such as a quadratic or exponential function, is needed.

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