Page 177 - Advanced engineering mathematics

6.2 The Dot Product 157

This is the familiar Pythagorean theorem, because the vectors F and G form the sides of a right
triangle with hypotenuse F + G (imagine Figure 6.5 with F and G forming a right angle).

EXAMPLE 6.5

Suppose two lines are defined parametrically by
L 1 : x = 2 − 4t, y = 6 + t, z = 3t
and
L 2 : x =−2 + p, y = 7 + 2p, z = 3 − 4p.
We want to know if these lines are orthogonal. Note that the question makes sense even if L 1 and
L 2 do not intersect.
The idea is to form a vector along each line and test these vectors for orthogonality. For
a vector along L 1 , take two points on this line, say (2,6,0) when t = 0 and (−2,7,3) when
t = 1. Then V 1 =−4i + j + 3k is parallel to L 1 . Similarly, (−2,7,3) is on L 2 when p = 0, and
(−1,9,−1) is on L 2 when p=1, so V 2 =i+2j−4k is parallel to L 2 . Compute V 1 ·V 2 =−14 =0.
Therefore, L 1 and L 2 are not orthogonal.

Orthogonality is also useful for determining the equation of a plane in 3-space. Any plane
has an equation of the form
ax + by + cz = d.
As suggested by Figure 6.12, if we specify a point on the plane and a vector orthogonal to
the plane, then the plane is completely determined. Example 6.6 suggests a strategy for finding
the equation of this plane.

EXAMPLE 6.6
We will find the equation of the plane containing the point (−6,1,1) and orthogonal to the
vector N =−2i + 4j + k. Such a vector N is said to be normal to and is called a normal
vector to .
Here is a strategy. Because (−6,1,1) is on , a point (x, y, z) is on exactly when the
vector between (−6,1,1) and (x, y, z) lies in . But then (x + 6)i + (y − 1)j + (z − 1)k must
be orthogonal to N,so
N · ((x + 6)i + (y − 1)j + (z − 1)k) = 0.

N

P

FIGURE 6.12 A point P and a normal
vector N determine a plane.

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October 14, 2010 14:21 THM/NEIL Page-157 27410_06_ch06_p145-186