   Chapter 12.5, Problem 38E

Chapter
Section
Textbook Problem

# Find an equation of the plane.38. The plane that passes through the points (0, −2, 5) and (−1, 3, 1) and is perpendicular to the plane 2z = 5x + 4y

To determine

To find: An equation of the plane that passes through the points (0,2,5) and (1,3,1), and perpendicular to the plane 2z=5x+4y.

Explanation

Formula used:

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

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

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

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

Calculation:

The normal vector (n) to the plane is the cross product of two direction vectors in the plane.

Let the vector from the point (0,2,5) to (1,3,1) be a.

Calculation of vector a:

Substitute 0 for x1, 2 for y1, 5 for z1, 1 for x2, 3 for y2, and 1 for z2 in equation (2),

a=(10),[3(2)],(15)=1,5,4

The normal vector of the perpendicular plane is the direction vector which is parallel to the required plane.

Write the expression of perpendicular plane.

2z=5x+4y

Rearrange the expression as follows

5x+4y2z=0

Write the normal vector from the plane 5x+4y2z=0 and consider it as b (direction vector that is parallel to the required plane).

b=5,4,2

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

Find the normal vector (n).

n=a×b

Substitute 1,54 for a and 5,4,2 for b,

n=1,54×5,4,2

Calculate the cross product as follows

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