| The ebook Elementary Calculus is based on material originally written by H.J. Keisler. For more information please read the copyright pages. |
|
Home  Vectors Product of Vectors Inner Product Perpendiculars and Parallels - Theorem 3 |
|||||
 Search the VIAS Library | Index
|
  |
||||
Perpendiculars and Parallels - Theorem 3
DEFINITION Two nonzero vectors A and B are said to be perpendicular (or orthogonal), A ⊥ B, if the angle between them is π/2. A and B are said to be parallel, A ∥ B, if the angle between them is either 0 or π. TEST FOR PERPENDICULARS Let A and B be nonzero vectors. Then A ⊥ B if and only if A · B = 0. PROOF The following are equivalent: A · B = 0, TEST FOR PARALLELS Given two nonzero vectors A and B, the following are equivalent:
PROOF To show that (i) is equivalent to (ii). we note that the following are equivalent.
We now show that (i) implies (iii), and (iii) implies (i). Assume (i), A ∥ B. Case 1 θ = 0. Let U and V be the unit vectors of A and B. By the Law of Cosines,
Case 2 θ = π. We see, by a similar proof, that
In either case, A is a scalar multiple of B. Finally, assume (iii), say A = tB. Then
Therefore θ = 0 or θ = π, so A ∥ B.
We conclude this section with a theorem about perpendicular vectors, first in the plane and then in space. THEOREM 3 Let A = a1i + a2j be a nonzero vector in the plane. (i) The vector B = a2i - a1j is perpendicular to A. (ii) Any vector perpendicular to A is parallel to B. PROOF (i) Wecompute A · B = a1a2 + a2(-a1) = 0. (ii) If C ⊥ A, then both B and C make angles of π/2 with A, so the angle between B and C is either 0 or π. Therefore B ∥ C.
|
|||||
| Home Vectors Product of Vectors Inner Product Perpendiculars and Parallels - Theorem 3 |
|
||||
, cos θ = 0, θ = π/2.
Last Update: 2006-11-06