9/24/2002, DOT PRODUCT O. Knill, Math 21aDOT PRODUCT. The dot product of two vectors v = (v1, v2, v3) and w = (w1, w2, w3) is defined as v · w =v1w1+ v2w2+ v3w3. Other notations are v · w = (v, w) or < v|w > (quantum mechanics) or viwi(Einsteinnotation) or gijviwj(general relativity). The dot product is also called scalar product, or inner product.LENGTH. Using the dot product one can express the length of v as ||v|| =√v ·v.CHALLENGE. Express the dot product in terms of length only!SOLUTION: (v + w, v + w) = (v, v) + (w, w) + 2(v, w) can be solved for (v, w).ANGLE. Because ||v − w||2= (v − w, v − w) = ||v||2+ ||w||2− 2(v, w) isby the cos-theorem equal to ||v||2+ ||w||2− 2||v|| · ||w||cos(φ), where φis the angle between the vectors v and w, we have the formulav ·w = ||v|| · ||w||cos(φ)FINDING ANGLES BETWEEN VECTORS. Find the angle between the vectors (1, 4, 3) and (−1, 2, 3).ANSWER: cos(φ) = 16/(√26√14) ∼ 0.839. So that φ = arccos(0.839..) ∼ 33◦.ORTHOGONALITY. Two vectors are orthogonal if and only if v · w = 0.PROJECTION. The vector a = w(v·w/||w||2) is called the projection of v onto w. Its length ||a|| = (v·w/||w||2)is called the scalar projection. The vector b = v − a is the component of v orthogonal to the w-direction.WORK. If r(t) is a path and F (t) is the force, thenW =Zbar0(t) ·F(t) dtis the work done on the body. It is the antiderivative of the powerr0(t) ·F(t).CAR ACCELERATION. A car of mass m = 373kg is accelerated with constant Power P = 1Horsepower =746Watt = 746kg m2/s3and is at rest at time t = 0. What is the speed at time t?SOLUTION. The work done until time t isRt0P dt = P t which is the kinetic energy mv2/2. Therefore v(t) =p2tP/m = 2√t. The acceleration is a = v0= 1/√t, the jerk a0= v00= −3/(2t3/2). Jerk and acceleration areinfinite at t = 0!A PARADOX? A car of mass m = 100kg moving along r(t) = (x(t), 0, 0) accelerates from 0 to 10m/s. Thekinetic energy it gained was m||r0(t)||2/2 = 102m/2 − 02/2 = 5000J. Now look at this situation in a movingcoordinate system R(t) = (X(t), 0, 0) = (x(t) + vt, 0, 0) with v = −10. In that coordinate system, the caraccelerates from 10 to 20. The energy it gained is 100(202/2 − 102/2) = 150000J. Why this discrepancy inenergies? Physical laws should be invariant under Galileo transformations ~x 7→~X = (x + ut, y + vt, z +wt),˙~x 7→˙~X = ( ˙x + u, ˙y + v, ˙z + w).PRODUCT RULE. We have (v, w)0= (v0, w) + (v, w0) as you can check by expanding both sides.APPLICATION. If a curve r(t) is located on a sphere, then r0is orthogonal to r. Proof. 0 =ddt||r(t)||2=r0· r + r · r0== 2r0· r = 0. That means that r0is orthonogal to r.ENERGY. K(t) = m||r0(t)||2/2 is called the kinetic energy of a body moving along r(t) with having mass m. .ENERGY CONSERVATION. If the energy K(t) is differentiated with respect to time, we get K0(t) = m(r00(t) ·r0(t)). Therefore, by Newton’s law mr00(t) = F (t) and using the fundamental theorem of calculus, we haveK(b) − K(a) =ZbaK0(t) dt =Zbar0(t) · mr00(t) dt =Zbar0(t) · F (t) dt = WEnergy conservation: the energy difference is the amount of work which has been used.PLANES. A surface ax + by + cz = d in space is called a plane. Using the dot product we can say n · x = d,where n = (a, b, c) and x = (x, y, z).NORMAL VECTOR If x, y are two points on the plane, then v ·x = d and v ·y = d which means v ·(x −y) = 0.Every vector inside the plane is orthogonal to v, which is called the normal vector to the plane.DISTANCE POINT-PLANE IN SPACE. Let P be a point and n · x = da plane which contains the point Q. Then|(P − Q) · n|/||n||is the distance from P to the plane. A formula worthwile to remember.DISTANCE POINT-LINE IN THE PLANE. Let P be a point in the plane and n ·x = d a line which containsa point Q. Then Then |(P − Q) · n|/||n|| is the distance from P to the line.REMARK. Sometimes, distance formulas are given without the absolute sign: (P −Q) ·n/||n|| is then a signeddistance which also shows on which side of the plane (rsp. line) we are. That is useful in computer graphicsor games, where it can make a difference on which side of of a wall you are.PROBLEM. A plane has distance d = 4 from the origin and is orthogonal to the vector (2, 4, 5). Find theequation of the plane.SOLUTION. ~n = (2, 4, 5) is a normal vector to the plane. Therefore, the equation has the form 2x + 4y + 5z =~n · ~x = d. The point Q = (d/2, 0, 0) is on the plane for example. The distance of P = (0, 0, 0) to the plane is|(P − Q) · n|/||n|| = d/||n|| = d/√29 = 4 so that the still unknown constant is d =
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