   Chapter 12, Problem 20RE

Chapter
Section
Textbook Problem

# Find an equation of the plane.20. The plane through (1, 2, −2) that contains the line x = 2t, y = 3 − t, z = 1 + 3t

To determine

To find: An equation of the plane that passes through the point (1,2,2) and contains the line x=2t,y=3t,z=1+3t.

Explanation

Formula used:

Write the expression to find 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)

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

Write the expressions for the parametric equations for a line through the point (x0,y0,z0) and parallel to the direction vector a,b,c.

x=x0+at,y=y0+bt,z=z0+ct (2)

Write the parametric equations of the line as follows.

x=2t,y=3t,z=1+3t

Rearrange the parametric equations as follows.

x=0+(2)t,y=3+(1)t,z=1+3t (3)

Compare equation (3) with (2) and write the direction vector (a) of a line as follows.

a=2,1,3

The direction vector (b) in the plane is determined by finding any one point in the line.

Consider the value of scalar term t is 0.

Substitute 0 for t in equation (3) and find the coordinates of point in the line,

x=0+(2)(0),y=3+(1)(0),z=1+3(0)x=0+0,y=3+0,z=1+0x=0,y=3,z=1

Therefore, the point in the line is (0,3,1).

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

PQ=(x2x1),(y2y1),(z2z1) (4)

Consider the vector from the point (1,2,2) to (0,3,1) is (b).

Calculation of vector b:

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

b=(01),(32),[1(2)]=1,1,3

As both the 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 normal vector

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