PROBLEMS, SECTION 3
18. Find z by Cramer’s rule:
cosθ 1 0
1 2 cosθ 1
0 1 2 cosθ
˛˛˛˛
˛˛= cos 3θ.
14. Show that the n-rowed determinant
˛˛˛˛
˛˛˛˛
˛˛˛˛
˛˛˛˛
˛
cosθ 1 0 0 0
1 2 cosθ 1 0 · · · · · · 0
0 1 2 cosθ 1 0
0 0 1 2 cosθ 0
... . .. ...
... 2 cosθ 1
0 0 0 0 · · · 1 2 cosθ
˛˛˛˛
˛˛˛˛
˛˛˛˛
˛˛˛˛
˛
= cosnθ.
Hint: Expand using elements of the last row or column. Use mathematical induction and the trigonometric addition formulas.
15. Use Cramer’s rule to solve Problems 2.3 and 2.11.
16. In the following set of equations (from a quantum mechanics problem), A and B are the unknowns,k and K are given, and i =√
−1. Use Cramer’s rule to find A and show that|A|2= 1.
A − B = −1 ikA − KB = ik
17. Use Cramer’s rule to solve for x and t the Lorentz equations of special relativity:
x=γ(x − vt)
t=γ(t − vx/c2) where γ2(1− v2/c2) = 1 Caution: Arrange the equations in standard form.
18. Find z by Cramer’s rule:
8<
:
(a − b)x − (a − b)y + 3b2z = 3ab (a + 2b)x − (a + 2b)y − (3ab + 3b2)z = 3b2 bx + ay − (2b2+a2)z = 0
Figure 4.1
4. VECTORS
Notation We shall indicate a vector by a boldface letter (for example, A) and a component of a vector by a subscript (for example Axis the x component ofA), as in Figure 4.1. Since it is not easy to handwrite boldface letters, you should write a vec-tor with an arrow over it (for example, A). It is very important to indicate clearly whether a letter represents a vector, since, as we shall see below, the same letter in italics (not boldface) is often used with a different meaning.
Magnitude of a Vector The length of the arrow representing a vector A is called the length or the magnitude of A (written |A| or A) or (see Section 10) the norm ofA (written ||A||). Note the use of A to mean the magnitude of A; for this reason it is important to make it clear whether you mean a vector or its magnitude (which is a scalar). By the Pythagorean theorem, we find
Section 4 Vectors 97
A =|A| =
A2x+ A2y in two dimensions, or A =|A| =
A2x+ A2y+ A2z in three dimensions.
(4.1)
Example 1. In Figure 4.2 the forceF has an x component of 4 lb and a y component of 3 lb. Then we write
Figure 4.2 Fx= 4 lb,
Fy = 3 lb,
|F| = 5 lb, θ = arc tan34.
Addition of Vectors There are two ways to get the sum of two vectors. One is by the parallelogram law: To findA + B, place the tail of B at the head of A and draw the vector from the tail ofA to the head of B as shown in Figures 4.3 and 4.4.
Figure 4.3 Figure 4.4
The second way of finding A + B is to add components: A + B has components Ax+ Bxand Ay+ By. You should satisfy yourself from Figure 4.3 that these two methods of finding A + B are equivalent. From Figure 4.4 and either definition of vector addition, it follows that
A + B = B + A (commutative law for addition);
(A + B) + C = A + (B + C) (associative law for addition).
In other words, vectors may be added together by the usual laws of algebra.
It seems reasonable to use the symbol 3A for the vector A + A + A. By the methods of vector addition above, we can say that the vectorA + A + A is a vector three times as long asA and in the same direction as A and that each component of 3A is three times the corresponding component of A. As a natural extension of these facts we define the vector cA (where c is any real positive number) to be a vector c times as long asA and in the same direction as A; each component of cA is then c times the corresponding component ofA (Figure 4.5).
The negative of a vector is defined as a vector of the same magnitude but in the opposite direction. Then (Figure 4.6) each component of−B is the negative of the corresponding component of B. We can now define subtraction of vectors by
Figure 4.5 Figure 4.6
saying thatA − B means the sum of the vectors A and −B. Each component of A − B is then obtained by subtracting the corresponding components of A and B, that is, (A − B)x = Ax− Bx, etc. Like addition, subtraction of vectors can be done geometrically (by the parallelogram law) or algebraically by subtracting the components (Figure 4.6).
The zero vector (which might arise asA = B−B = 0, or as A = cB with c = 0) is a vector of zero magnitude; its components are all zero and it does not have a direction. A vector of length or magnitude 1 is called a unit vector. Then for any A = 0, the vector A/|A| is a unit vector. In Example 1, F/5 is a unit vector.
We have just seen that there are two ways to combine vectors: geometric (head to tail addition), and algebraic (using components). Let us look first at an example of the geometric method; then we shall consider the algebraic method. Example 2 below illustrates the geometric method. By similar proofs, many of the facts of elementary geometry can be easily proved using vectors, with no reference to com-ponents or a coordinate system. (See Problems 3 to 8.)
Example 2. Prove that the medians of a triangle intersect at a point two-thirds of the way from any vertex to the midpoint of the opposite side.
To prove this, we call two of the sides of the triangleA and B. The third side of the triangle is then A + B by the parallelogram law, with the directions of A, B, and A + B as indicated in Figure 4.7. If we add the vector 12B to the vector A (head to tail as in Figure 4.7b), we have a vector from point O to the midpoint of the opposite side of the triangle, that is, we have the median to sideB. Next, take two-thirds of this vector; we now have the vector 23(A +12B) = 23A +13B extending from O to P in Figure 4.7b. We want to show that P is the intersection point of the three medians and also the “23 point” for each. We prove this by showing that P is the “23 point” on the median to sideA; then since A and B represent any two sides of the triangle, the proof holds for all three medians. The vector from R to Q (Figure 4.7c) is 12A + B; this is the median to A. The “23 point” on this median is the point P (Figure 4.7d); the vector from R to P is equal to 13(12A + B). Then the vector from O to P is 12A +13(12A + B) = 23A +13B. Thus P and P are the same point and all three medians have their “23 points” there. Note that we have made no reference to a coordinate system or to components in this proof.
Section 4 Vectors 99
Figure 4.7