Page 178 - Advanced engineering mathematics

158 CHAPTER 6 Vectors and Vector Spaces

Then
−2(x + 6) + 4(y − 1) + (z − 1) = 0,
or

−2x + 4y + z = 17.
This is the equation of .

Following this reasoning in general, the equation of a plane containing a point P 0 :(x 0 , y 0 , z 0 )
and having a normal vector N = ai + bj + ck is

N ·[(x − x 0 )i + (y − y 0 )j + (z − z 0 )k]= 0
or
a(x − x 0 ) + b(y − y 0 ) + c(z − z 0 ) = 0. (6.2)
It is also sometimes convenient to notice that the vector ai + bj + ck is always a normal
vector to a plane ax + by + cz = d, for any d. Changing the value of d moves the plane in 3-
space but does not change its orientation with respect to the axes, so the normal vector remains
the same and is determined by the coefficients a, b, and c only.
Another use for the dot product is in forming vector projections.
Let u and v be given, nonzero vectors, represented as arrows from a common point (for
convenience). The projection of v onto u is a vector proj v in the direction of u having magnitude
u
equal to the length of the perpendicular projection of the arrow representing v onto the line along
the arrow representing u. This projection is done by constructing a perpendicular line from the
tip of v onto the line through u. The base of the right triangle having v as hypotenuse is the length
d of proj v (Figure 6.13).
u
If θ is the angle between u and v, then
d
cos(θ) = .
v
Then
u · v u · v
d = v cos(θ) = v = .
u v u

z

v
u
θ
d
y

x
FIGURE 6.13 Orthogonal projection of v
onto u.

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October 14, 2010 14:21 THM/NEIL Page-158 27410_06_ch06_p145-186