find an equation of the plane passing through points (4,2,1) and (-3,5,7) and is parrallel to z-axis?

Answer 1

This is very doable.

A unique plane can be defined by
Three non-colliear points, or
One point and two non-parallel vectors, or
Two points and a vector other than the one between the two points as we have here.

You might ask, where is the vector? It is implied in the statement that the plane is parallel to the z-axis. Such a vector is of the form

ak

where a is any constant. We will select a = 1.

We have the vector v, which comes from the two given points, and k.

v = <-3-4,5-2,7-1> = <-7,3,6>
k = <0,0,1>

The cross product of v and k will yield a vector perpendicular to both of them, the normal vector of the plane.

v X k = 3i + 7j + 0k = 3i + 7j = <3,7,0>

Now with the normal vector and any point in the plane we can write the equation of the plane. Let's choose the point (4,2,1).

3(x - 4) + 7(y - 2) + 0(z - 1) = 0
3x - 12 + 7y - 14 = 0
3x + 7y - 26 = 0

Please note, the variable z does not appear in the equation. This is because for a given x and y, z can be anything. This is what you would expect for a plane parallel to the z-axis. Furthermore, this solution is unique.

Answer 2

One directional vector of the airplane is AB. the different is v = <0, 0, a million> that's parallel to the z-axis. AB = = = <3, -2, a million> the traditional vector n, of the airplane is orthogonal to the two directional vectors. Take the pass product. n = v X AB = <0, 0, a million> X <3, -2, a million> = <2, 3, 0> With a factor interior the airplane and the traditional vector of the airplane we are able to write the equation of the airplane. permit's use A(2, 2, a million). bear in recommendations, the traditional vector is orthogonal to any vector that lies interior the airplane. And the dot made of standard vectors is 0. permit R(x,y,z) be an arbitrary factor interior the airplane. Then vector PR lies interior the airplane. n • PR = 0 n • = 0 <2, 3, 0> • = 0 2(x + 2) + 3(y - 2) + 0(z - a million) = 0 2x + 4 + 3y - 6 = 0 2x + 3y - 2 = 0

Answer 3

For the given points its impossible to find the plane.

For a plane parallel to z axis, all the points in atleast one plane, either x or y axis should be constant. But in the given 2 points, there are no constant points.

I would be eager to know, if somebody finds a solution for this and have added this question to my watch list.