Lesson 8: The Cross Prodct of Vectors IBHL - SANTOWSKI In this lesson yo will learn how to find the cross prodct of two ectors how to find an orthogonal ector to a plane defined by two ectors how to find the area of a parallelogram gien two ectors how to find the olme of a parallelepiped gien three ectors CROSS PRODUCT IN PHYSICS The idea of a cross prodct occrs often in physics.
CROSS PRODUCT IN PHYSICS In particlar, we consider a force F acting on a rigid body at a point gien by a position ector r. CROSS PRODUCT IN PHYSICS For instance, if we tighten a bolt by applying a force to a wrench, we prodce a trning effect. TORQUE The torqe τ (relatie to the origin) is defined to be the cross prodct of the position and force ectors τ = r x F It measres the tendency of the body to rotate abot the origin.
TORQUE The direction of the torqe ector indicates the axis of rotation. TORQUE The magnitde of the torqe ector is τ = r x F = r F sin θ where θ is the angle between the position and force ectors. TORQUE Obsere that the only component of F that can case a rotation is the one perpendiclar to r that is, F sin θ. The magnitde of the torqe is eqal to the area of the parallelogram determined by r and F.
TORQUE A bolt is tightened by applying a 40-N force to a 0.5-m wrench, as shown. Find the magnitde of the torqe abot the center of the bolt. TORQUE The magnitde of the torqe ector is: τ = r x F = r F sin 75 = (0.5)(40) sin75 = 0 sin75 9.66 N m TORQUE If the bolt is right-threaded, then the torqe ector itself is τ = τ n 9.66 n where n is a nit ector directed down into the slide. 4
Objectie FINDING THE CROSS PRODUCT OF TWO VECTORS The Cross Prodct The cross prodct of two ectors, denoted as a b, nlike the dot prodct, represents a ector. The cross prodct is defined to be for a a, a, a a b and b b, b, b a b a b, a b a b, a b a b Yo are probably wondering if there is an easy way to remember this. The easy way is to se determinants of size x. a a, a, a a b and b b, b, b a b a b, a b a b, a b Let s set p a x determinant as follows:. First se the nit ectors i, j, and k as the first row of the determinant. a b. Use row for the components of a and row for the components of b. We will expand by the first row sing minors. a a a i a b a b j a b a b k a b a b b b b a b a b i a b a b j a b a b k Don t forget to change the sign to for the j expansion. 5
Find the cross prodct for the ectors below. a,4,5 and b,, Find the cross prodct for the ectors below. a,4,5 and b,, 4 5 4 0 i ( 5) j 4 4 k 6i 7j8k CROSS PRODUCT EXAMPLE If a = <,, 4> and b = <, 7, 5>, then 6
CROSS PRODUCT EXAMPLE If a = <,, 4> and b = <, 7, 5>, then ab 4 7 5 4 4 i j k 7 5 5 7 ( 5 8) i ( 5 8) j (7 6) k 4i j k CROSS PRODUCT Show that a x a = 0 for any ector a in V. If a = <a, a, a >, then CROSS PRODUCT Show that a x a = 0 for any ector a in V. If a = <a, a, a >, then aa a a a a a a ( a a a a ) i ( a a a a ) j 0i 0 j 0k 0 ( a a a a ) k 7
Since the cross prodct is determined by sing determinants, we can nderstand the algebraic properties of the Cross Prodct which are: Which wold come from the fact that if yo interchange two rows of a determinant yo negate the determinant. w w c c c 0 0 0 0 w w Objectie FIND AN ORTHOGONAL VECTOR TO A PLANE DEFINED BY TWO VECTORS Now that yo can do a cross prodct the next step is to see why this is sefl. Let s look at the ectors from the last problem a,4,5, b,, and a b 6,7, 8 What is the dot prodct of a,4,5 with a b 6,7, 8 And b,, with a b 6,7, 8? 8
Now that yo can do a cross prodct the next step is to see why this is sefl. Let s look at the ectors from the last problem a,4,5, b,, and a b 6,7, 8 What is the dot prodct of a,4,5 with a b 6,7, 8 And b,, with a b 6,7, 8? If yo answered 0 in both cases, yo wold be correct. Recall that wheneer two non-zero ectors are perpendiclar, their dot prodct is 0. Ths the cross prodct creates a ector perpendiclar to the ectors a and b. Orthogonal is another name for perpendiclar. Show that w,,, w,,0,and w 5,0, is tre for Soltion: Show that w,,, w,,0,and w 5,0, is tre for Soltion:,,,,,0,and w 5,0, The left hand side 0 i j 5k 5 0,,,,5 5 4 The right hand side 0,, 5,0, 5 0 4 9
Geometric Properties of the Cross Prodct Let and be nonzero ectors and let (the Greek letter theta) be the angle between and. is orthogonal to both and Remember: orthogonal means perpendiclar to CROSS PRODUCT Let a and b be represented by directed line segments with the same initial point, as shown. CROSS PRODUCT The cross prodct a x b points in a direction perpendiclar to the plane throgh a and b. 0
CROSS PRODUCT It trns ot that the direction of a x b is gien by the right-hand rle, as follows. RIGHT-HAND RULE If the fingers of yor right hand crl in the direction of a rotation (throgh an angle less than 80 ) from a to b, then yor thmb points in the direction of a x b. Geometric Properties of the Cross Prodct We know the direction of the ector a x b. The remaining thing we need to complete its geometric description is its length a x b. Let and be nonzero ectors and let be the angle between and, then sin
. 0 if and only if and are mltiples of each other 4. area of the paralleogram haing and as adjacent sides. y Proof: The area of a parallelogram is base times height. A = bh sin = y/ sin = y = height y = sin = Example problem for property Find nit ectors perpendiclar to a = i - j + k and b = -4i + j - k. Soltion. Example problem for property Find nit ectors perpendiclar to a = i - j + k and b = -4i + j - k. Soltion The two gien ectors define a plane. The Cross Prodct of the ectors is perpendiclar to the plane and is proportional to one of the desired nit ectors. To make its length eqal to one, we simply diide by its magnitde:.
Find nit ectors perpendiclar to a= i - j + k and b = -4i + j - k. Soltion ( 5i 0j) 4 5i 0j 5 00 5 5 5 Diide the ector by its magnitde 5 0 5 5 5i 0j i j i j i j 5 5 5 5 5 5 5 5 5 5 For a second nit ector simply mltiply the answer by - 5 5 i 5 5 j CROSS PRODUCT Find a ector perpendiclar to the plane that passes throgh the points P(, 4, 6), Q(-, 5, -), R(, -, ) Objectie FIND THE AREA OF A PARALLELOGRAM GIVEN TWO VECTORS
Example for property 4 4. area of the paralleogram haing and as adjacent sides. Area of a Parallelogram ia the Cross Prodct z Show that the following 4 points define a parallelogram and then find the area. A A(4,4,6), B(4,4,6), C(,,), D(,,) B x D C y Soltion: First we find the ectors of two adjacent sides: AB 4 4, 4 4,6 6 0, 0,0 AD 4, 4,6,,4 Now find the ectors associated with the opposite sides: DC,, 0, 0,0 AB DB 4,4, 6,, 4 AD This shows that opposite sides are associated with the same ector, hence parallel. Ths the figre is of a parallelogram. Contined The area is eqal to the magnitde of the cross prodct of ectors representing two adjacent sides: Area = AB XAD 0 0 0 40 i 0j 0k 4 40 i 0k ( 40) 0 600 900 500 50 The area of the parallelogram is 50 sqare nits. 4
Area of a Triangle ia the Cross Prodct Since the area of a triangle is based on the area of a parallelogram, it follows that the area wold be ½ of the cross prodct of ectors of two adjacent sides. Find the area of the triangle whose ertices are P(4,4,6), Q(5,6,-) R(,,) Area parallelogram) = PQ x PR. The area of the triangle is half of this. z P R x Q Area of a Triangle ia the Cross Prodct Contined P(4,4,6), Q(5,6,-) R(,,) Area parallelogram) = PQ x PR Bt what if we choose RP and RQ? Wold the reslt be the same? z Let s do both and see! P R PQ 5 4,6 4, 6,, 8 PR 4, 4, 6,, 4 x Q RP 4,4,6,,4 RQ 5,6, 4,5, 4 PQ 5 4,6 4, 6,, 8 PR 4, 4, 6,, 4 RP 4,4,6,,4 RQ 5,6, 4,5, 4 8 48 4 i 4 4 j 6 k 7 i 8j k 4 7 i 8j k ( 7) 8 584 784 089 7075 4 60 i 6 j 45 k 7 i 8j k 4 5 4 Since this is the same ector, the magnitde wold be the same also. 5
z P R x Q The area of the triangle is ½ 7057 4 sqare nits. CROSS PRODUCT Find the area of the triangle with ertices P(, 4, 6), Q(-, 5, -), R(, -, ) Objectie 4 HOW TO FIND THE VOLUME OF A PARALLELEPIPED GIVEN THREE VECTORS 6
7 The triple scalar prodct is defined as: w,, w w w w k w j w w i w and k j i k j i For The geometric significance of the scalar triple prodct can be seen by considering the parallelepiped determined by the ectors a, b, and c. SCALAR TRIPLE PRODUCTS The area of the base parallelogram is: A = b x c SCALAR TRIPLE PRODUCTS
SCALAR TRIPLE PRODUCTS If θ is the angle between a and b x c, then the height h of the parallelepiped is: h = a cos θ We mst se cos θ instead of cos θ in case θ > π/. SCALAR TRIPLE PRODUCTS Hence, the olme of the parallelepiped is: V = Ah = b x c a cos θ = a (b x c) Ths, we hae proed the following formla. Formla SCALAR TRIPLE PRODUCTS The olme of the parallelepiped determined by the ectors a, b, and c is the magnitde of their scalar triple prodct: V = a (b x c) 8
Volme of a Parallelepiped ia the Scalar Triple Prodct: Find the olme of the parallelepiped with adjacent edges AB, AC, and AD, where the points are A(4, -, -), B(, 0, 5), C(-,, ), and D(,, ). Volme of a Parallelepiped ia the Scalar Triple Prodct: Find the olme of the parallelepiped with adjacent edges AB, AC, and AD, where the points are A(4, -, -), B(, 0, 5), C(-,, ), and D(,, ). The olme is gien by the scalar triple prodct: AB (AC X AD). First we need the three ectors: AB = [ - 4]i + [0 - (-)]j + [5 - (-)]k = -i + j + 7k. AC = [- - 4]j + [ - (-)]j + [ - (-)]k = -7i + 5j + k AD = [ - 4]i + [ - (-)]j + [ - (-)]k = -i + 6j + 4k First, find the cross prodct: = i[0-8] - j[-8 - (-9)] + k[-4 - (-5)] = i + 9j - 7k Now form the dot prodct to get the olme: Volme = AB (i + 9j - 7k) = (-i + j + 7k) (i + 9j - 7k) = 4 + 57-89 = 6 cbic nits 9