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EXAMPLE 6
Graphing planes whose Cartesian equations involve three
variables, D 0
Sketch the plane defined by the equation p 1 : 2x 3y z 4.
Solution
To sketch the plane, we calculate the coordinates of the points where the plane
intersects each of the three coordinate axes (that is, we determine the three intercepts
for the plane). This is accomplished by setting 2 of the 3 variables equal to zero
and solving for the remaining variable. The x-, y-, and z-intercepts are 2, 4 and 4,
3 ,
respectively. These three points form a triangle that forms part of the required
plane.
z
(0, 4, 0)
O(0, 0, 0) 3
y
(2, 0, 0) p 1 2x +3y–z=4
(0, 0, –4)
x
EXAMPLE 7
Reasoning about direction vectors of planes
Determine two direction vectors for the planes p 1 : 3x 4y 12 and
p 2 : x y 5z 0.
Solution
The plane p 1 : 3x 4y 12 crosses the x-axis at the point 14, 0, 02 and
the y-axis at the point 10, 3, 02. One direction vector is thus
!
m 1 14 0, 0 3, 0 02 14, 3, 02. Since the plane can be written
as 3x 4y 0z 12, this implies that it does not intersect the z-axis, and
!
therefore has m 2 10, 0, 12 as a second direction vector.
The plane p 2 : x y 5z 0 passes through O10, 0, 02 and cuts the
xz-plane along the line x 5z 0. Convenient choices for x and z are 5 and 1,
respectively. This means that A15, 0, 12 is on p 2 . Similarly, the given plane cuts
the xy-plane along the line x y 0. Convenient values for x and y are 1 and 1.
This means that B11, 1, 02 is on p 2 .
!
Possible direction vectors for p2 are m and
! 1 15 0, 0 0, 1 02 15, 0, 12
m 2 11, 1, 02.
NEL
CHAPTER 8 475