   Chapter 12.5, Problem 35E

Chapter
Section
Textbook Problem

# Find an equation of the plane.35. The plane that passes through the point (3, 5, −1) and contains the line x = 4 − t, y = 2t − 1, z = −3t

To determine

To find: An equation of the plane that passes through the point (3,5,1) and contains the line x=4t,y=2t1,z=3t .

Explanation

Formula:

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=4t,y=2t1,z=3t

Rewrite the parametric equations as follows.

x=4+(1)t,y=1+2t,z=0+(3)t (3)

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

a=1,2,3

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=4+(1)(0),y=1+2(0),z=0+(3)(0)x=4,y=1,z=0

Therefore, the point in the line is (4,1,0) .

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 (3,5,1) to (4,1,0) is (b) .

Calculation of vector b :

Substitute 3 for x1 , 5 for y1 , 1 for z1 , 4 for x2 , 1 for y2 , and 0 for z2 in equation (4),

b=(43),(15),[0(1)]=1,6,1

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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