Cartesian equation of the plane through the given points

In summary, the task is to find the cartesian equation of a plane through three given points. This can be done by using two pairs of the given points to get two vectors in the plane, and then finding the cross product of these vectors to get the normal vector. The equation of the plane can then be written as A(x-x_0)+ B(y-y_0)+ C(z-z_0)=0, where A, B, C are the components of the normal vector and (x_0, y_0, z_0) is one of the given points.
  • #1
ezsmith
16
0

Homework Statement


For each part, find the cartesian equation of the plane through the given points.
(1,0,3), (2,-4,3),(4,-1,2)

The Attempt at a Solution



No attempt. Dunno how to do :(
 
Physics news on Phys.org
  • #2
Haven't been paying attention in class?:tongue: Use two pairs of those points to get two vectors in the plane: the vector from [itex](x_0, y_0, z_0)[/itex] to [itex](x_1, y_1, z_1)[/itex] is [itex](x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}[/itex].

The cross product of two vectors is perpendicular to the two vectors so perpendicular to the plane they lie in.

The equation of a plane with normal vector [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] containing point [itex](x_0, y_0, z_0)[/itex] is [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].
 
  • #3
HallsofIvy said:
Haven't been paying attention in class?:tongue: Use two pairs of those points to get two vectors in the plane: the vector from [itex](x_0, y_0, z_0)[/itex] to [itex](x_1, y_1, z_1)[/itex] is [itex](x_1- x_0)\vec{i}+ (y_1- y_0)\vec{j}+ (z_1- z_0)\vec{k}[/itex].

The cross product of two vectors is perpendicular to the two vectors so perpendicular to the plane they lie in.

The equation of a plane with normal vector [itex]A\vec{i}+ B\vec{j}+ C\vec{k}[/itex] containing point [itex](x_0, y_0, z_0)[/itex] is [itex]A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0[/itex].

I did listen in class but the lecturer only gave notes to copy during class and there is no example given in the notes so that is why. Anyway, I managed to solved it. Thanks a lot sir.. Really appreciate it :)
 

What is the Cartesian equation of the plane through the given points?

The Cartesian equation of the plane through the given points is a mathematical representation of the plane using the coordinates of the points. It can be written in the form ax + by + cz = d, where a, b, and c are the coefficients of the variables x, y, and z, and d is a constant.

How do I find the Cartesian equation of a plane through three given points?

To find the Cartesian equation of a plane through three given points, you can use the formula (x - x1)(y2 - y1) - (x2 - x1)(y - y1) = (x - x1)(z2 - z1) - (x2 - x1)(z - z1), where (x1, y1, z1), (x2, y2, z2), and (x, y, z) are the coordinates of the three points. This formula can be rearranged to get the Cartesian equation of the plane.

Can the Cartesian equation of a plane be written in different forms?

Yes, the Cartesian equation of a plane can be written in different forms. One common form is the point-normal form, which is written as (x - x0) · n = 0, where (x0, y0, z0) is a point on the plane and n is the normal vector of the plane. Another form is the intercept form, which is written as x/a + y/b + z/c = 1, where a, b, and c are the intercepts of the plane on the x, y, and z axes, respectively.

Can the Cartesian equation of a plane be used to find the distance between a point and the plane?

Yes, the Cartesian equation of a plane can be used to find the distance between a point and the plane. The formula for this is |ax0 + by0 + cz0 + d| / √(a2 + b2 + c2), where (x0, y0, z0) is the coordinates of the point and a, b, and c are the coefficients of the variables in the Cartesian equation of the plane.

Can the Cartesian equation of a plane be used in three-dimensional geometry?

Yes, the Cartesian equation of a plane can be used in three-dimensional geometry. It is a fundamental equation that is used to describe and analyze planes in three-dimensional space. It is also used in many other fields, such as physics, engineering, and computer graphics.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
7K
  • Calculus and Beyond Homework Help
Replies
24
Views
1K
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
995
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Replies
4
Views
618
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
Back
Top