The Dot Product. Properties of the Dot Product If u and v are vectors and a is a real number, then the following are true:

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1 The Dot Prodct Tesday, 2// Section 8.5, Page 67 Definition of the Dot Prodct The dot prodct is often sed in calcls and physics. Gien two ectors = <a, b > and = <a 2, b 2 >, then their dot prodct (represented as ) is defined as = a a 2 + b b 2 NOTICE: The dot prodct is not another ector it is a real nmber (a scalar). Find the dot prodct of each pair of ectors. = <2, -3> & = <-2, -5> = 6i + 4j & = 3j Properties of the Dot Prodct If and are ectors and a is a real nmber, then the following are tre: = 2. a() = a( ) = (a) 3. ( + ) w = w + w 4. 2 =

2 The Dot Prodct Theorem If and are ectors with the same initial point and the initial points are located on the origin, then the angle θ between and is defined as the angle created in between the two ectors. The angle θ will always be between 0 and π θ The Dot Prodct Theorem says: If θ is the angle between two nonzero ectors and, then = cos θ Angle Between Two Vectors The Dot Prodct Theorem can be sed to find the angle between two ectors by soling for cos θ. Yo can then se the arccosine to find the ale of θ. If θ is the angle between two nonzero ectors and, then cos Find the angle between ectors = <6, 7> and = <-2, > Orthogonal Vectors Two ectors and are called orthogonal, or perpendiclar, if the angle between the ectors is π/2 (90 ) Yo can determine if two ectors are orthogonal by finding their dot prodcts, becase two nonzero ectors and are orthogonal (perpendiclar) if and only if = 0. Determine if the ectors are orthogonal. = <3, 5> and = <2, -8> a = <2, > and b = <-, 2> 2

3 The Component of Along The component of along (or component of in the direction of ) is the length of the part of that points in the direction of The component of along In the second drawing the component is negatie. This happens if π/2 < θ π. The component of along is defined to be cos θ, where θ is the angle between and. Applications of The Component of Along When analyzing forces in physics and engineering, it is often helpfl to express a ector as the sm of two ectors lying in perpendiclar directions. In the pictre the weight of the car is a ector w that points directly downward. If the car is on an incline the ector w still points downward bt we can write w in terms of and w where points along the grond and is perpendiclar to so that w = +. w is the force that tends to roll the car down and is the force experienced by the srface of the drieway. Applications of The Component of Along The magnitdes of the two ectors and are the component of w along and the component of w along w A car weighing 2000 lbs. is parked on a hill that is inclined 2. A) Find the magnitde of the force reqired to preent the car from rolling down the hill (). B) Find the magnitde of the force experienced by the drieway de to the weight of the car (). 3

4 Calclating Components An alternate way of calclating the component of along is to se the dot prodct and the following formla The component of along is Let = <2, 6> and = <-4, 9>. Find the component of along. The Projection of onto The projection of onto (denoted by proj ) is the ector whose direction is the same as and whose length is the component of along The ble arrow is the projection of onto The projection of onto is the ector proj gien by proj 2 Resoling a Vector Occasionally it is necessary to resole a ector into the sm of two ectors, one that is parallel to a ector and one that is orthogonal (perpendiclar) to. What this means is yo will write = + 2 where is parallel to and 2 is orthogonal to. To find and 2 se the formlas = proj and 2 = proj Let = <-3, 8> and = <2, 5>. Find proj and then resole into and

5 Work Another se of the dot prodct is in calclating work. In general terms work is defined as the amont of effort reqired to perform a task. In physics, if a constant force of magnitde F moes an object throgh a distance d along a straight line, then the work done is W = Fd. If F is measred in ponds and d in feet, then the nit of work is a foot-pond (ft-lb). The formla W = Fd applies only when the force is directed along the direction of motion. When the force is a ector F that is at an angle to the direction of motion (from P to Q) then only the component of the force in the direction of motion affects the object. P F θ F cos θ Q Work To find the work W done by a force F moing in a direction D is W = F D A force ector F = <3, 8> moes an object from the point (, 2) to the point (5, 2). Find the work done. A man plls a wagon horizontally by exerting a force of 30 lbs. on the handle. If the handle makes an angle of 60 with the horizontal, find the work done in moing the wagon 250 feet Assignment Pg. 624: 3 odds, De Thrsday, 2/3/ 5