finding the equation of a plane?

Question

given 4 points ( 0 0 0) ( 1 0 0) ( 0 2 0 ) and ( 0 0 3), please explain how to find the equation of the plane which they form

the equation is 2z + 6x + 3y = 6

is there a quick way of finding this equation instead of finding an orthogonal vector and plugging it into the equation of a plane?

Answer 1

( 0 0 0) is not on the plane. Check it!
Given three non-linear point, you can find a plane. The equation can be obtained by A∙(B x C) = 0.

Answer 2

Given 4 points O( 0 0 0); P(1 0 0); Q( 0 2 0 ) and R( 0 0 3), please explain how to find the equation of the plane containing the points.
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From your answer it is immediately apparent that the first point O(0,0,0) is NOT in the plane. The remaining three points define the plane.

First create two directional vectors in the plane.

u = PQ = = <0-1, 2-0, 0-0> = <-1, 2, 0>
v = PR = = <0-1, 0-0, 3-0> = <-1, 0, 3>

The normal vector n, of the plane is perpendicular to both directional vectors. Take the cross product.

n = u X v = <-1, -2, 0> X <-1, 0, 3> = <6, 3, 2>

With a point on the plane and the normal vector we can write the equation of the plane. Let's choose point P(1, 0, 0).

6(x - 1) + 3(y - 0) + 2(z - 0) = 0
6x - 6 + 3y - 0 + 2z - 0 = 0
6x + 3y + 2z = 6
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Usually the above is the quickest way. In this particular instance, since the points P(1,0,0); Q(0,2,0) and R(0,0,3) are on the three coordinate axes, you can take a shortcut.

First find the least common multiple of 1, 2, and 3. It is 6. Now divide 6 by the coordinate of each axis.

(6/1)x + (6/2)y + (6/3)z = 6
6x + 3y + 2z = 6

This approach only works when you have one point on each coordinate axis.

Answer 3

first you have to show those 4 points determine a part
Actually only 3 points determine a plane. You find the equation of the plane and check if the forth verifies it.

To find the equation of the plane, there are at least 3 ways:
either make a system satisfied by all 3 points
or you use the normal vector to the plane
or the plane is the linear combination of 2 vectors on plane


"the equation is 2z + 6x + 3y = 6 "
is that a new question because it is not the equation of the plane that passes through those 4 points

"is there a quick way of finding this equation instead of finding an orthogonal vector and plugging it into the equation of a plane?"
no, to me that is the fastest way. But, depending on problem, one other way could be more useful.

Answer 4

you may the two use a in basic terms algebraic recommendations-set to sparkling up it as 3 equations in 3 unknowns or use a geometrical recommendations-set. I choose the latter. we've the three given factors A(3,-a million,2); B(8,2,4): and C(-a million, -2, -3). 2 directional vectors in the needed aircraft are: AB = = <8 - 3, 2 - -a million, 4 - 2> = <5, 3, 2> AC = = <8 - -a million, 2 - -2, 4 - -3> = <9, 4, 7> the traditional vector n, of the aircraft is orthogonal to the two directional vectors. Take the circulate product. n = AB X AC = <5, 3, 2> X <9, 4, 7> = <13, -17, -7> With a evaluate the aircraft and the traditional vector to the aircraft we are able to write the equation of the aircraft. enable's opt for factor B(8, 2, 4). keep in mind, the traditional vector is orthogonal to any vector that lies in the aircraft. And the dot made of orthogonal vectors is 0. define R(x,y,z) to be an arbitrary evaluate the aircraft. Then vector BR lies in the aircraft. n • BR = 0 n • = 0 <13, -17, -7> • = 0 13(x - 8) - 17(y - 2) - 7(z - 4) = 0 13x - 104 - 17y + 34 - 7z + 28 = 0 13x - 17y - 7z - 40 two = 0