By P. Fitzpatrick

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**Extra info for Advanced Calculus [Int'l Student edn]**

**Example text**

2: An obtuse angle between vectors. Noting that |a|2 = x21 + y12 , |b|2 = x22 + y22 , and θ |c|2 = (x2 x1 )2 + (y2 y1 ) 2 , 36 CHAPTER 1. MULTIVARIABLE SPACES AND FUNCTIONS we find that the Law of Cosines becomes x22 2x1 x2 + x21 + y22 2y1 y2 + y12 = x21 + y12 + x22 + y22 Subtracting x21 + y12 + x22 + y22 from both sides, and dividing by 2|a||b| cos ✓. 2, we arrive at x1 x2 + y1 y2 = |a||b| cos ✓. Note how simple the left-hand side of the above formula is; it’s just the product of the x-coordinates of the vectors a and b added to the product of the y-coordinates.

Suppose that the point-masses have masses M and m and are located at points pM and pm , respectively. Let r = pM pm , so that r is a vector which points from m to M , and whose magnitude is the distance between the masses. We assume that the masses are located at di↵erent points, so that r 6= 0. Then, Newton’s Law of Universal Gravitation states that each mass exerts a gravitational attraction force on the other, along the line between masses, and that the magnitude of the force is proportional to the product of the masses and inversely proportional to the square of the distance between the masses.

20. Let a = ( 1, 2, 0, 5), b = (3, 1, 3, 4), and c = (0, 1, 4, 7). Calculate the linear combinations 2a b + 3c 1 c. 2 3 a + 2b 2 and Solution: There’s nothing tricky to do here; you simply have to calculate. We find: 2a b + 3c = 2( 1, 2, 0, 5) ( 2, 4, 0, 10) (3, 1, 3, 4) + 3(0, 1, 4, 7) = (3, 1, 3, 4) + (0, 3, 12, 21) = ( 5, 6, 15, 15) and 3 a + 2b 2 ✓ 3 , 2 3, 0, 15 2 1 c = 2 ◆ + (6, 2, 3 ( 1, 2, 0, 5) + 2(3, 1, 3, 4) 2 ✓ 6, 8) + 0, 1 (15, 2 3, 1 , 2 2, 7 2 ◆ = 1 (0, 1, 4, 7) = 2 ✓ 15 , 2 3 , 2 8, 4 ◆ = 16, 8).