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Unformatted text preview: 1 University of California, Berkeley Physics H7A Fall 1998 ( Strovink ) SOLUTION TO PROBLEM SET 1 Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon 1. You may remember the law of cosines from trigonometry. It will be useful for several parts of this problem, so we will state it here. If the lengths of the sides of a triangle are a , b , and c , and the angle opposite the side c is ψ , then c 2 = a 2 + b 2 − 2 ab cos ψ (a.) When two vectors add up to a third vector, the three vectors form a triangle. If the angle between a and b is θ , then the angle opposite the side formed by c is 180 ◦ − θ . The law of cosines then tells us that | c | 2 = | a | 2 + | b | 2 − 2 | a || b | cos (180 ◦ − θ ) From trigonometry, remember that cos (180 ◦ − θ ) = − cos θ which gives | c | 2 = | a | 2 + | b | 2 + 2 | a || b | cos θ We know that | a | + | b | = | c | . Squaring this equation, we get | c | 2 = | a | 2 + | b | 2 + 2 | a || b | If we compare this with the equation above, we can see that cos θ has to be equal to one. This only happens when θ = 0 ◦ . What this means is that the two vectors are parallel to each other, and they point in the same direction. If the an- gle between them were 180 ◦ , then they would be parallel but point in opposite directions. (b.) This part is simple. Just subtract the vec- tor a from both sides to see that b = − b . The only way that this can happen is if b = , the zero vector. (c.) This part can also be done by the law of cosines. Like part (a.), we have the following two equations | c | 2 = | a | 2 + | b | 2 + 2 | a || b | cos θ This is just the law of cosines again, where θ is the angle between | a | and b | . The problem states that | c | 2 = | a | 2 + | b | 2 Comparing this with the equation above, we find that cos θ = 0. This happens at θ = ± 90 ◦ . This means that the vectors must be perpendicular to each other. (d.) Yet again, we can use the law of cosines. If the angle between a and b is θ , then the angle between a and − b is 180 ◦ − θ . The lengths of the sum and difference are | a + b | 2 = | a | 2 + | b | 2 + 2 | a || b | cos θ | a − b | 2 = | a | 2 + | b | 2 − 2 | a || b | cos θ For these to be equal, we need cos θ = 0, which happens when θ = ± 90 ◦ . Again, this means that the vectors are perpendicular. (e.) Guess what? Yup, law of cosines. We know that | a | = | b | = | a + b | . Adding a to b is going to look like two vectors stuck together to form two sides of a triangle. If the angle between the vectors is θ , the law of cosines gives | a + b | 2 = 2 | a | 2 + 2 | a | 2 cos θ where we have used the fact that a and b have the same length. We also know that | a + b | = | a | ....
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