Question

Answer the question with words only, NO equations

1. Find an equation of the plane. The plane through the points (2,1,2), (3,-8,6), and (-2,-3,1)

I need steps in words and Matlab code for above question.

Answer 1

step 1

First we assign three points:

Code:

P1 =  [2,1,2];

P2 = [3,-8,6];

P3 = [-2,-3,1]

Step 2

we will find the normal vector using cross product between two vector lying on plane. these two vector is vector joining position vector of these three point.

Code:

normal = cross(P1-P2, P1-P3)

normal = (25, -10, 31) ; "this is direction cosine or vector perpendicular to required plane"

Step 3

Next assign x, y, z as variable, dot product of vector lying on plane (using this variable) with normal vector will become zero.

Code:

syms x y z

P = [x,y,z]

eqPlane = dot(normal, P-P1)

eqPlane = 25x-15y-40z+45, "this will be output"

Our required equation of plane is :  25x-15y-40z+45 = 0