Page 176 - Advanced engineering mathematics

156 CHAPTER 6 Vectors and Vector Spaces

EXAMPLE 6.2
The angle θ between F =−i + 3j + k and G = 2j − 4k is given by
(−i + 3j + k) · (2j − 4k)
cos(θ) =
−i + 3j + k 2j − 4k
(−1)(0) + (3)(2) + (1)(−4) 2
= √ √ = √ .
2
2
1 + 3 + 1 2 2 + 4 2 220
2
Then θ ≈ 1.436 radians.
EXAMPLE 6.3
Lines L 1 and L 2 have parametric equations
L 1 : x = 1 + 6t, y = 2 − 4t, z =−1 + 3t
and
L 2 : x = 4 − 3p, y = 2p, z =−5 + 4p.
The parameters t and p can take on any real values. We want an angle θ between these lines.
The strategy is to take a vector V 1 along L 1 and a vector V 2 along L 2 and find the angle
between these vectors. For V 1 , find two points on L 1 ,say (1,2,−1) when t = 0 and (7,−2,2)
when t = 1, and form
V 1 = (7 − 1)i + (−2 − 2)j + (2 − (−1))k = 6i − 4j + 3k.
On L 2 ,take (4,0,−5) with p = 0 and (1,2,−1) with p = 1, forming
V 2 = 3i − 2j − 4k.
Now compute
6(3) − 4(−2) + 3(−4) 14
cos(θ) = √ √ = √ .
36 + 16 + 9 9 + 4 + 16 1769
√
An angle between L 1 and L 2 is arccos(14/ 1769), which is approximately 1.23 radians.

Two nonzero vectors F and G are orthogonal (perpendicular) when the angle θ between
them is π/2 radians. This happens exactly when
F · G
cos(θ) = 0 =
F G
which occurs when F · G = 0. It is convenient to also agree that O is orthogonal to every
vector. With this convention, two vectors are orthogonal if and only if their dot product is
zero.

EXAMPLE 6.4

Let F =−4i + j + 2k, G = 2i + 4k and H = 6i − j − 2k. Then F · G = 0, so F and G are orthog-
onal. But F · H and G · H are not zero, so F and H are not orthogonal and G and H are not
orthogonal.

Property (6) of the dot product has a particularly simple form when the vectors are
orthogonal. In this case, F · G = 0, and upon setting α = β = 1, we have
2
2
2
F + G = F + G .

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October 14, 2010 14:21 THM/NEIL Page-156 27410_06_ch06_p145-186