   Chapter 12, Problem 19RE

Chapter
Section
Textbook Problem

# Find an equation of the plane.19. The plane through (3, −1, 1), (4, 0, 2), and (6, 3, 1)

To determine

To find: An equation of the plane that passes through the points (3,1,1), (4,0,2) and (6,3,1).

Explanation

Formula used:

Write the expression to find the equation of the plane through the point P0(x0,y0,z0) with normal vector n=a,b,c as follows:

a(xx0)+b(yy0)+c(zz0)=0 (1)

Write the expression to find vector from the point P(x1,y1,z1) to Q(x2,y2,z2).

PQ=(x2x1),(y2y1),(z2z1) (2)

Consider the vector from the point (3,1,1) to (4,0,2) as (a) and the vector from the point (3,1,1) to (6,3,1) as (b).

Calculation of vector a:

Substitute 3 for x1, –1 for y1, 1 for z1, 4 for x2, 0 for y2, and 2 for z2 in equation (2),

a=(43),(0+1),(21)=1,1,1

Calculation of vector b:

Substitute 3 for x1, –1 for y1, 1 for z1, 6 for x2, 3 for y2, and 1 for z2 in equation (2),

b=(63),(3+1),(11)=3,4,0

As both vectors a and b lie on the plane, the cross product of a and b is the orthogonal of the plane and it is considered as a normal vector.

Find the normal vector (n)

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