1. Vectors in R2 and R3 Definition 1.1. A vector �v ∈ R3 is a 3-tuple of real numbers (v1,v2,v3). Hopefully the reader can well imagine the definition of a vector in R2 . Example 1.2. (1, 1, 0) and (√2,π, 1/e) are vectors in R3 . Definition 1.3. The zero vector in R3 , denoted �0, is the vector (0, 0, 0). If �v =(v1,v2,v3) and w�=(w1,w2,w3) are two vectors in R3, the sum of �v and w�, denoted �v + w�, is the vector (v1 + w1,v2 + w2,v3 + w3). If �v =(v1,v2,v3) ∈ R3 is a vector and λ ∈ R is a scalar, the scalar product of λ and v, denoted λ �v, is the vector (λv1,λv2,λv3).· Example 1.4. If �v = (2, −3, 1) and w�= (1, −5, 3) then �v + w�= (3, −8, 4). If λ = −3 then λ �v =(−6, 9, −3).· Lemma 1.5. If λ and µ are scalars and �u, �v and w�are vectors in R3 , then (1) �0+ �v = �0. (2) �u +(�v + w�)=(�u + �v)+ w�. (3) �u + �v = �v + �u. (4) λ (µ �v)=(λµ) �v.·· · (5) (λ + µ) �v = λ �v + µ �v.· ·· (6) λ (�u + �v)= λ �u + λ �v.· ·· Proof. We check (3). If �u =(u1,u2,u3) and �v =(v1,v2,v3), then �u + �v =(u1 + v1,u2 + v2,u3 + v3) =(v1 + u1,v2 + u2,v3 + u3) = �v + �u. � We can interpret vector addition and scalar multiplication geometricallly We can think of a vector as representing a displacement from the origin. Geometrically a vector �v has a magnitude (or length) |�v| =(v12 + v22 + v2)1/2 and every non-zero vector has a direction 3 �v �u = . |�v| Multiplying by a scalar leaves the direction unchanged and rescales the magnitude. To add two vectors �v and w�, think of transporting the tail of w�to the endpoint of �v. The sum of �v and w�is the vector whose tail is the tail of the transported vector. 1 One way to think of this is in terms of directed line segments. Note that given a point P and a vector �v we can add �v to P to get another point Q. If P =(p1,p2,p3) and �v =(v1,v2,v3) then Q = P + �v =(p1 + v1,p2 + v2,p3 + v3). If Q = PQ, such that Q(q1,q2,q3), then there is a unique vector −→= P + �v, namely −→PQ =(q1 − p1,q2 − p2,q3 − p3). Lemma 1.6. Let P , Q and R be three points in R3 . Then −→QR = −→PQ + −→PR. Proof. Let us consider the result of adding −→QR to P ,PQ + −→P +(−→QR)=(P + −→QRPQ + −→PQ)+ −→= Q + −→QR = R. On the other hand, there is at most one vector �v such that when we add it P we get R, namely the vector −→So −→QR = −→�PR. PQ + −→PR. Note that (1.6) expresses the geometrically obvious statement that if one goes from P to Q and then from Q to R, this is the same as going from P to R. Vectors arise quite naturally in nature. We can use vectors to repressen forces; every force has both a magnitude and a direction. The combined effect of two forces is represented by the vector sum. Similaarl we can use vectors to measure both velocity and acceleration. The equation F�= m�a, is the vector form of Newton’s famous equation. Note that R3 comes with three standard unit vectors ˆı = (1, 0, 0) jˆ= (0, 1, 0) and kˆ= (0, 0, 1), which are called the standard basis. Any vector can be written uniquely as a linear combination of these vectors, ˆ�v =(v1,v2,v3)= v1ˆı + v2jˆ+ v3k. We can use vectors to parametrise lines in R3 . Suppose we are given two different points P and Q of R3 . Then there is a unique line l containing P and Q. Suppose that R =(x,y,z) is a general point of 2 the line. Note that the vector −→PQ, so that PR is parallel to the vector −→−→PQ.PR is a scalar multiple of −→Algebraically, −→PQ, PR = t−→for some scalar t ∈ R. If P =(p1,p2,p3) and Q =(q1,q2,q3), then (x − p1,y − p2,z − p3)= t(q1 − p1,q2 − p2,q3 − p3)= t(v1,v2,v3), where (v1,v2,v3)=(q1 − p1,q2 − p2,q3 − p3). We can always rewrite this as, (x,y,z)=(p1,p2,p3)+ t(v1,v2,v3)=(p1 + tv1,p2 + tv2,p3 + tv3). Writing these equations out in coordinates, we get x = p1 + tv1 y = p2 + tv2 and z = p3 + tv3. Example 1.7. If P = (1, −2, 3) and Q = (1, 0, −1), then �v = (0, 2, −4) and a general point of the line containing P and Q is given parametricaall by (x,y,z) = (1, −2, 3) + t(0, 2, −4) = (1, −2+2t, 3 − 4t). Example 1.8. Where do the two lines l1 and l2 (x,y,z) = (1, −2+2t, 3 − 4t) and (x,y,z) = (2t − 1, −3+ t, 3t), intersect? We are looking for a point (x,y,z) common to both lines. So we have (1, −2+2s, 3 − 4s) = (2t − 1, −3+ t, 3t). Looking at the first component, we must have t =1. Looking at the second component, we must have −2+2s = −2, so that s =0. By inspection, the third component comes out equal to 3 in both cases. So the lines intersect at the point (1, −2, 3). Example 1.9. Where does the line (x,y,z) = (1 − t, 2 − 3t, 2t + 1) intersect the plane 2x − 3y + z = 6? We must have 2(1 − t) − 3(2 − 3t) + (2t +1) = 6. Solving for t we get 9t − 3=6, so that t =1. The line intersects the plane at the point (x,y,z) = (0, −1, 3). 3 Example 1.10. A cycloid is the path traced in the plane, by a point on the circumference of a circle as the circle rolls along the ground. Let’s find the parametric form of a cycloid. Let’s suppose that the circle has radius a, the circle rolls along the x-axis and the point is at the origin at time t =0. We suppose that the cylinder rotates through an angle of t radians in time t. So the circumference travels a distance of at. It follows that the centre of the circle at time t is at the point P =(at,a). Call the point on the circumference Q =(x,y) and let O be the centre of coordinates. We have (x,y)= −→OP + −→OQ = −→PQ. Now relative to P , the point Q just goes around a circle of radius a. Note that the circle rotates backwards and at time t =0, the angle 3π/2. So we have −→PQ =(a cos(3π/2 − t),a sin(3π/2 − t)) = (−a sin t, −a cos t) Putting all of this together, we have (x,y)=(at − a sin t,a − a cos t). 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.