11.4 Dot Product Contemporary Calculus 111.4 DOT PRODUCTIn the previous sections we looked at the meaning of vectors in two and three dimensions, but the onlyoperations we used were addition and subtraction of vectors and multiplication by a scalar. Some of theapplications of 2–dimensional vectors used the angles that the vectors made with the coordinate axes andwith each other, but, so far, in three dimensions we have not used angles. This section addresses both ofthose situations. It introduces a way to multiply two vectors, in two and three dimensions, called the dotproduct, and this dot product provides us with a relatively easy way to determine angles between vectors.Section 11.5 introduces a different method of multiplying two vectors, the cross product, in threedimensions that has other useful applications.Since we will soon have three different types of multiplications for a vector (scalar, dot, and cross), it isimportant that you distinguish among them and call each multiplication operation by its full name.Definition: Dot ProductTwo dimensions: The dot product of A = 〈 a1, a2 〉 and B = 〈 b1, b2 〉is A•B = a1b1 + a2b2 .Three dimensions: The dot product of A = 〈 a1, a2, a3 〉 and B = 〈 b1, b2, b3 〉is A•B = a1b1 + a2b2 + a3b3 .Both vectors in the dot product must have the same number of components, and the result of the dotproduct U•V is a scalar.Example 1: For A = 〈 4, 1, 8 〉 and B = 〈 2, –4, 4 〉 , calculate A•B , A•A , B•B , and(A–B)•(A+2B) .Solution: A•B = 〈 4, 1, 8 〉•〈 2, –4, 4 〉 = (4)(2)+ (1)(–4) + (8)(4) = 8 – 4 + 32 = 36.A•A = 〈 4, 1, 8 〉•〈 4, 1, 8 〉 = (4)(4) + (1)(1) + (8)(8) = 81.B•B = 〈 2, –4, 4 〉•〈 2, –4, 4 〉 = (2)(2) + (–4)(–4) + (4)(4) = 36.You should notice that A•A = |A|2 and B•B = |B|2 .Finally, A–B = 〈 2, 5, 4 〉 and A + 2B = 〈 8, –7, 16 〉so (A–B)•(A+2B) = 〈 2, 5, 4 〉•〈 8, –7, 16 〉 = (2)(8) + (5)(–7) + (4)(16) = 45.Practice 1: For U = 〈 2, 6, –3  and V = 〈 –1, 2, 2  , calculate U•V , U•U , V•V , U•(U+V) ,and U•U + U•V . Does U•U = |U|2 ? Does U•V = V•U ?11.4 Dot Product Contemporary Calculus 2As you might have noticed in Example 1 and Practice 1, the dot product seems to have some of theproperties of ordinary multiplication of numbers.Properties of the Dot Product: (1) A•A = |A|2(2) A•B = B•A(3) k( A•B ) = ( kA )•B = A•( kB )(4) A•(B + C) = A•B + A•CAll of these properties can be proved using the definition of the dot product.Proof of (1): If A = 〈 a1, a2, a3 〉 then A•A = (a1)2 + (a2)2 + (a3)2 and|A|2 = ( (a1)2 + (a2)2 + (a3)2 )2 = (a1)2 + (a2)2 + (a3)2 so A•A = |A|2 .Proof of (3): k( A•B ) = k( a1b1 + a2b2 + a3b3 ) = ka1b1 + ka2b2 + ka3b3 .( kA )•B = 〈 ka1, ka2, ka3 〉•〈 b1, b2, b3 〉 = ka1b1 + ka2b2 + ka3b3 . AndA•( kB ) = 〈 a1, a2, a3 〉•〈 kb1, kb2, kb3 〉 = a1(kb1) + a2(kb2) + a3(kb3)= ka1b1 + ka2b2 + ka3b3 so k( A•B ) = ( kA )•B = A•( kB )Practice 2: Prove Property (2) for 3–dimensional vectors.The next result about dot products is very important, and much of the usefulness of dot products follows fromit. It enables us to easily determine the angle between two vectors in two or three (or more) dimensions.Angle Property of Dot ProductsA•B = |A| |B| cos( θ ) where θ is the angle between A and B . Equivalently,if A and B are nonzero vectors, then the angle θ between A and B satisfies cos( θ ) = A•B|A| |B| .Proof of the Angle Property: The proof uses the Law of Cosines and several ofthe properties of the dot product.The vectors A, B and A–B can be arranged to form a triangle (Fig. 1)with the angle θ between A and B. Applying the Law of Cosines tothis triangle, we have| A – B |2= |A|2 + |B|2 – 2|A||B| cos( θ ) .xyzABA – BθFig. 111.4 Dot Product Contemporary Calculus 3We can also use the properties of the dot product to expand | A – B |2 in a different way:| A – B |2= ( A – B )•( A – B ) = A•A – A•B – B•A + B•B = |A|2 – 2A•B + |B|2.From these two representations for | A – B |2we have that|A|2 – 2A•B + |B|2 = |A|2 + |B|2 – 2|A||B| cos( θ ) so– 2A•B = – 2|A||B| cos( θ ) and A•B = |A| |B| cos( θ ) .Example 2: Let A = 〈 2, 5, 14 〉 and B = 〈 2, 1, –2 〉 . Find the angles(a) between A and B and (b) between A and the positivey–axis (Fig. 2).Solution: (a) |A| = 4+25+196 = 15 , |B| = 4+1+4 = 3 ,A•B = 4+5–28 = –19 , andcos( θ ) = A•B|A| |B| = –19(15)(3) ≈ –0.4222 so θ ≈ 2.01 or about 115.0°.(b) The basis vector j = 〈 0, 1, 0  points along the positive y–axis so the angle betweenA and the positive y–axis is the same as the angle between A and j.|A| = 15, |j | = 1, and A•j = (2)(0) + (5)(1) + (14)(0) = 5 socos( θ ) = A•j|A| |j| = 5(15)(1) ≈ 0.333 and ϕ ≈ 1.23 or about 70.5°.Practice 3: Let A = 〈 2, –6, 3 〉 , B = 〈 4, 8, –1 〉 and C = 〈 3, 0, –4 〉 anddetermine the angles between the vectors (a) A and B, (b) A and C, (c)B and the negative x–axis, and (d) C and the positive y–axis.Example 3: Find the angle of intersection of the graphs of f(x) = x3 + 1 andg(x) = 3 – x2 at the point (1, 2). (The angle of intersection is the angle betweenthe tangent vectors to the graphs at the point.) Fig. 3.Solution: f '(x) = 3x2 , so the slope of f at (1,2) is f '(1) = 3(1)2 = 3 : riserun = 31 and a vector, A = 1i + 3j , is shown in Fig. 4. Similarly, g '(x) = –2x, so theslope of g at (1,2) is g '(1) = –2(1) = –2: riserun = –21 and vector B = 1i – 2j hasthe same slope. Then |A| = 10 , |B| = 5 , and A•B = –5, socos(θ) = –510 5 ≈ –0.707 . Then θ = arccos( –0.707 ) ≈ 2.356 ( or 135 o ).If a function is given parametrically by x(t) and y(t), then the tangent vector isT = x '(t)i + y '(t)j .ϕBxyzFig. 2θA2xy1x + 13Fig. 323 – xBA2x1x + 13Fig. 423 – x11.4 …