2. Dot productDefinition 2.1. Let �v =(v1,v2,v3) and w�=(w1,w2,w3) be two vectors in R3 . The dot product of �v and w�, denoted �v w�, is the scalar · v1w1 + v2w2 + v3w3. Example 2.2. The dot product of �v = (1, −2, −1) and w�= (2, 1, −3) is 1 2+(−2) 1+(−1) (−3) = 2 − 2+3=3.··· Lemma 2.3. Let �u, �v and w�be three vectors in R3 and let λ be a scalar. (1) (�u + �v) w�= �u w�+ �v w�.· ·· (2) �v w�= w��v.·· (3) (λ�v) w�= λ(�v w�).·· (4) �v �v =0 if and only if �v = �0.· Proof. (1–3) are straightforward. To see (4), first note that one direction is clear. If �v = �0, then �v �v = 0. For the other direction, suppose that �v �v = 0. Then· · v12 + v22 + v32 = 0. Now the square of a real number is non-negative and if a sum of non-negative numbers is zero, then each term must be zero. It follows that v1 = v2 = v3 = 0 and so �v = �0. � Definition 2.4. If �v ∈ R3, then the norm or length of �v =(v1,v2,v3) is the scalar �v� = √�v �v =(v12 + v22 + v32)1/2 .· It is interesting to note that if you know the norm, then you can calculate the dot product: (�v + w�)(�v + w�)= �v �v +2�v w�+ w�w�· ··· (�v − w�)(�v − w�)= ��vw + ��vv − 2��ww. · ··· Subtracting and dividing by 4 we get 1 �v w�= ((�v + w�)(�v + w�) − (�v − w�)(�v − w�))· 4 ·· 1 = 4(��v + w��2 −��v − w��2). Given two non-zero vectors �v and w�in space, note that we can define the angle θ between �v and w�. �v and w�lie in at least one plane (which is in fact unique, unless �v and w�are parallel). Now just measure the angle θ between the �v and w�in this plane. By convention we always take 0 ≤ θ ≤ π. 1 Theorem 2.5. If �v and w�are any two vectors in R3, then �v · w�= ��v�� w�� cos θ. Proof. If �v is the zero vector, then both sides are equal to zero, so that they are equal to each other and the formula holds (note though, that in this case the angle θ is not determined). By symmetry, we may assume that �v and w�are both non-zero. Let �u = w�−�v and apply the law of cosines to the triangle with sides parallel to �u, �v and w�: ��u�2 = ��v�2 + �w��2 − 2��v��w�� cos θ. We have already seen that the LHS of this equation expands to �v �v − 2�v w�+ w�w�= ��v�2 − 2�v w�+ �w��.··· · Cancelling the common terms ��v�2 and �w��2 from both sides, and dividing by 2, we get the desired formula. � We can use (2.5) to find the angle between two vectors: Example 2.6. Let �v = −ˆı + kˆand w�=ˆı +ˆj. Then −1= �v · w�= ��v��w�� cos θ = 2 cos θ. Therefore cos θ = −1/2 and so θ =2π/3. Definition 2.7. We say that two vectors �v and w�in R3 are orthogonna if �v w�=0.· Remark 2.8. If neither �v nor w�are the zero vector, and �v w�=0 then· the angle between �v and w�is a right angle. Our convention is that the zero vector is orthogonal to every vector. Example 2.9. ˆı, jˆand kˆare pairwise orthogonal. Given two vectors �v and w�, we can project �v onto w�. The resulting vector is called the projection of �v onto w�and is denoted projw��v. For example, if F�is a force and w�is a direction, then the projection of F�onto w�is the force in the direction of w�. As projw��v is parallel to w�, we have projw��v = λ�w, for some scalar λ. Let’s determine λ. Let’s deal with the case that λ ≥ 0 (so that the angle θ between �v and w�is between 0 and π/2). If we take the norm of both sides, we get � projw��v� = �λ�w� = λ�w��, 2 (note that λ ≥ 0), so that λ = � projw��v�. �w�� But cos θ = � projw��v�, ��v� so that � projw��v� = ��v� cos θ. Putting all of this together we get λ = ��v� cos θ �w��= ��v��w�� cos θ�w��2 �v w�= · . �w��2 There are a number of ways to deal with the case when λ< 0 (so that θ > π/2). One can carry out a similar analysis to the one given above. Here is another way. Note that the angle φ between w�and �u = −�v is equal to π − θ < π/2. By what we already proved �u w�projw��· w.u = ��w��2 But proj��u = − proj��v and �u w�= −�v w�, so we get the same formulaww ·· in the end. To summarise: Theorem 2.10. If �v and w�are two vectors in R3, where w�is not zero, then �� �v w�projw��· w.v = ��w��2 Example 2.11. Find the distance d between the line l containing the points P1 = (1, −1, 2) and P2 = (4, 1, 0) and the point Q = (3, 2, 4). Suppose that R is the closest point on the line l to the point Q. Note that −→P1P2 of the line. So we want theQR is orthogonal to the direction −−→length of the vector −−P1Q − proj−−−−P1P→2 P1Q, that is, we want d = �−−P1Q − proj−−−−P1Q�.P1P→2 Now −−→ and P1P2 = (3, 2, −2). P1Q = (2, 3, 2) −−→ 3 We have �−−→2 =32 +22 +22 = 17 and −−→P1Q =6+6 − 4=8.P1P2�P1P2 · −−→It follows that proj−−−−→8 (3, 2, −2).P1P→2 P1Q = 17Subtracting, we get −−→−−→8 (3, 2, −2)= 1(10, 35, 50)= 5(2, 7, 10).P1Q−proj−−P1Q = (2, 3, 2)−P1P→2 171717Taking the length, we get 5 (22 +72 + 102)1/2 ≈ 3.64. 17Theorem 2.12. The angle subtended on the circumference of a circle by a diameter of the circle is always a right angle. Proof. Suppose that P and Q are the two endpoints of a diameter of the circle and that R is a point on the circumference. We want to show that the angle between −→QR is a right angle.PR and −→Let O be the centre of the circle. Then−→PO + −→and QR = −→OR.PR = −→OR −→QO + −→Note that −→PO.QO = −−→Therefore −→−→PO + −→(−→OR)PR QR =(−→OR) QO + −→·· =(−→PO + OR−→) OR − −→(−→PO)· = �−→2 − −−→�PO�2OR�= r 2 − r 2 =0, where r is the radius of the circle. PR and −→It follows that −→QR are indeed orthogonal. � 4 MIT OpenCourseWarehttp://ocw.mit.edu 18.022 Calculus of Several Variables Fall 2010 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.