   Chapter 12.5, Problem 36E

Chapter
Section
Textbook Problem

# Find an equation of the plane.36. The plane that passes through the point (6, −1, 3) and contains the line with symmetric equations x/3 = y + 4 = z/2

To determine

To find: An equation of the plane that passes through the point (6,1,3) and contains the line with symmetric equations x3=y+4=z2 .

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

xx0a=yy0b=zz0c (2)

Write the symmetric equations of the line as follows.

x3=y+4=z2

Rewrite the symmetric equations as follows.

x03=y(4)1=z02 (3)

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

(x0,y0,z0)=(0,4,0)a=3,1,2

Therefore the plane passes through the points (6,1,3) and (0,4,0) .

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 (0,4,0) to (6,1,3) is (b) .

Calculation of vector b :

Substitute 0 for x1 , 4 for y1 , 0 for z1 , 6 for x2 , 1 for y2 , and 3 for z2 in equation (4),

b=(60),[1(4)],(30)=6,3,3

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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