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NDA Exam Preparation & Study Guide41 chapters | 343 lessons | 32 flashcard sets

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Lesson Transcript

Instructor:
*Russell Frith*

In this lesson we will learn the technique for computing the equation of a plane when the coordinates of three noncollinear points in three dimensional space are given.

In this lesson we'll learn the technique for computing the equation of a plane when the coordinates of three **noncollinear points** - or points that aren't on the same line - in three dimensional space are given. Let the points be designated as *P(x1,y1,z1)*,*Q(x2,y2,z2)*, and *R(x3,y3,z3)*.

We'll then show that the equation of the plane through those points is:

*a(x - x0) + b(y - y0) + c(z - z0) = 0*, where *( x0, y0, z0)* are the coordinates of any one of the points *P*, *Q*, or *R*, and *<a,b,c>* is a vector perpendicular to the plane.

The following method outlines the general procedure for computing the equation of a plane passing through three given points in space:

(1) Let the three points be designated as *P(x1,y1,z1)*, *Q(x2,y2,z2)*, and *R(x3,y3,z3)*.

(2) Let *(x0,y0,z0)* be a point on the plane. You can use the coordinates from either *P*,*Q*, or *R*.

(3) Let the equation of the plane be *a(x - x0) + b(y - y0) + c(z - z0) = 0* . *<a,b,c>* is a vector perpendicular to the plane. We need to find values for *a*,*b*, and *c*.

(4) Define the following vectors that you can see below:

(5) Find the vector perpendicular to those two vectors by taking their **cross product**, which is the product of two values, represented by perpendicular lines.

Evaluating the cross product gives coefficients for a vector perpendicular to the plane.

The coefficients of the vector computed from the cross product are determinants which must be simplified into numbers.

The numbers from the vector *<a b c>* are the coefficients of the vector computed from the cross product.

(6) Using any of the three given points, write the equation of the plane.

*a(x - x0) + b(y - y0) + c(z - z0) = 0*

Find the equation of the plane that passes through the points *P(1,0,2)*, *Q(-1,1,2)*, and *R(5,0,3)*.

Find the vector perpendicular to those two vectors by taking the cross product.

Using any of the three given points (*P*, *Q*, or *R*), write the equation of the plane. In this case we use *P*.

*1(x - 1) + 2(y - 0) - 4(z - 2) = 0*

*x - 1 + 2y - 4z + 8 = 0*

*x + 2y - 4z = -7*

The graph of the plane is shown below:

In this lesson, you learned how to construct the equation of a plane when given the coordinates of three **noncollinear points** - or points that aren't on the same line - in space. In order to construct this equation, you must first calculate two different vectors that connect two of the given points. Once you find those two vectors, you must calculate the **cross product** of them, which is the product of two values represented by perpendicular lines. The coefficients *a*, *b*, and *c* of the cross product of:

give the coordinates of a vector that's perpendicular to the plane, which equation we are trying to calculate. Using those coefficients, we formulate the equation of the plane as:

*a(x - x0) + b(y - y(0) + c(z - z0) = 0*, where *<a b c>* is the vector perpendicular to the plane.

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11 in chapter 7 of the course:

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NDA Exam Preparation & Study Guide41 chapters | 343 lessons | 32 flashcard sets

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