Vector Operations Vector Operations ME 202 Multiplication by a scalar Addition/subtraction Scalar multiplication (dot product) Vector multiplication (cross product) 1 2 Graphical Operations Component Operations Multiplying by a scalar Adding Subtracting A + A B = B = 2 A Multiplying by a scalar a A = aa x î + aa y ĵ + aa z ˆk Adding A + B = ( A x + )î + A y + B y Subtracting A B = ( A x )î + A y B y ( ) ĵ + A z + B z ( ) ˆk ( ) ĵ + A z B z ( ) ˆk 3 4
Properties: Commutative: Distributive: Dot Product - 1 Given: two vectors P and Q in some arbitrary plane. By definition, their dot product is P i Q = PQcosθ Q P θ Q i P = QP cosθ = PQcosθ = P i Q Q i P + R ( ) = Q i P + Q i R Dot Product - 2 Special cases: P P P Q Q Q P i Q = PQ P i Q = 0 P i Q = PQ 5 6 Dot Product - 3 î i î = î î cos( 0) = ( 1) ( 1) ( 1) = 1 ĵ i ĵ = ĵ ĵ cos( 0) = ( 1) ( 1) ( 1) = 1 ˆk i ˆk = ˆk ˆk cos( 0) = ( 1) ( 1) ( 1) = 1 î i ĵ = î ĵ cos( 90 ) = ( 1) ( 1) ( 0) = 0 ĵ i ˆk = ĵ ˆk cos( 90 ) = ( 1) ( 1) ( 0) = 0 ˆk i î = ˆk î cos ( 90 ) = ( 1) ( 1) ( 0) = 0 7 î ˆk ĵ Note: Dot Product - 4 P i Q = ( P x î + P ĵ + P ˆk y z )i Q x î + Q ĵ + Q ˆk y z = P x Q x î i î +P y Q x ĵ i î +P z Q x ˆk i î ( ) ( ) + P x Q y ( î i ĵ ) + P x Q z ( î i ˆk ) ( ) + P y Q y ( ĵ i ĵ ) + P y Q z ( ĵ i ˆk ) ( ) + P z Q y ( ˆk i ĵ) + P z Q z ( ˆk i ˆk ) = P x Q x + P y Q y + P z Q z P i P = P 2 x + P 2 y + P 2 z = P 2 P = P = P i P 8
Cross-Product -1 C = A B = Cû Cross product of two vectors yields a vector, C, with magnitude C = ABsinθ. θ is the angle between A and B such that 0 θ π. Direction of û (and C ) is perpendicular to plane containing A and B ; direction is determined by right-hand rule. Cross-Product - 2 Commutative? A B = B A Distributive? NO YES A B + D ( ) = A B + A D 9 10 Cross-Product - 3 î î = î î sin( 0) = ( 1) ( 1) ( 0) = 0 î ĵ = î ĵ sin( 90 ) ˆk = ( 1) ( 1) ( 1) ˆk = ˆk î ˆk = î ˆk sin( 90 ) ĵ ĵ î = ĵ î sin ( 90 ) ˆk ( ) = 1 ( ) = 1 ( )( 1) ( 1) ĵ ( )( 1) ( 1) ˆk ĵ ĵ = ĵ ĵ sin( 0) = ( 1) ( 1) ( 0) = 0 ĵ ˆk = ĵ ˆk sin( 90 ) î ( ) = 1 ( )( 1) ( 1) î ( ) = ĵ ( ) = ( ˆk ) ( ) = î Magnitude of C is. ABsinθ Direction of C is perpendicular to plane containing A and B. Direction is determined by right-hand rule. 11 Cross-Product - 4 + î ĵ ˆk î ĵ ĵ 12 îî + ˆk
Cross-Product - 5 ( ) ( î + B ĵ + B ˆk y z ) ( ) + A x B y ( î ĵ ) + A x B z ( î ˆk ) ( ) + A y B y ( ĵ ĵ ) + A y B z ( ĵ ˆk ) ( ) + A z B y ( ˆk ĵ) + A z B z ( ˆk ˆk ) P Q = A x î + A y ĵ + A z ˆk = A x î î +A y ĵ î +A z ˆk î = ( A y B z A z B y )î + A B A B z x x z ( ) ˆk ( ) ĵ + A x B y A y A B = Cross-Product- 6 î ĵ ˆk A x A y A z B y B z Recall: + + + + + = A y A z B y B z î A x A z B z ĵ + A x A y B y ˆk 13 14
Vector Operations 2 All four of the vector operations listed here are needed for solving problems in statics. You should become proficient with these as soon as possible. Otherwise, the rest of this course may forever remain a mystery to you. 3 A number that has no direction is called a scalar. Multiplying a vector by a scalar produces a vector with magnitude equal to the magnitude of the original vector multiplied by the absolute value of the scalar. A scalar greater or less than one produces a vector with magnitude that is respectively greater or less than the magnitude of the original vector. If the scalar is positive or negative, the direction of the product is respectively the same as or opposite to the direction of the original vector. Adding two vectors graphically is the same as placing the tail of the second vector at the head of the first vector. The sum is the vector from the tail of the first vector to the head of the second. Subtracting a vector is the same as adding its negative. 1 of 6
4 The three operations described on the previous slide are shown here using Cartesian components of the vectors. Components in the same coordinate direction ( î, ĵ or ˆk ) are combined, but components in perpendicular directions are not. 5 We will make use of two different ways of multiplying two vectors together. One way is to use the dot product, which is also called the scalar product because the result is a scalar, not a vector. The dot product s definition and two of its properties are shown here. You are expected to memorize the definition and to learn to use it. 6 In the first case on this slide, the cosine of the angle between the two vectors is one. In the second case on this slide, the cosine of the angle between the two vectors is zero. In the third case on this slide, the cosine of the angle between the two vectors is negative one. 7 Since all of the factors in these products are unit vectors, each dot product is merely the cosine of the angle between the two factors. It follows that each of the first three products shown is one and each of the last three products is zero. 2 of 6
8 If we write two three-dimensional vectors in Cartesian components and compute their dot product, we find nine partial products. From the previous slide, we know that six of these partial products are zero. Note that any of the scalar components that appear in the result could be negative, so the overall result could be positive, negative or zero. The last two equations show that the magnitude of any vector is the square root of the dot product of the vector with itself. 9 In addition to the dot product, we will also need to use the cross product, also called the vector product, which is another way of multiplying two vectors together. The cross product of two vectors is a vector that is perpendicular to both of the product s factors, and so is also perpendicular to the plane containing the factors. The magnitude of the cross product is the product of the two factors magnitudes and the sine of the angle between the factors. The angle is always chosen so that its sine is a non-negative number. (In other words, we always use an angle in the range 0 θ π.) The direction of the cross product is determined by the right hand rule. First, straighten the index finger of the right hand and open the thumb of the right hand so that it is roughly perpendicular to the index finger. Next, point the index finger of the right hand in the direction of the first factor and rotate the hand with the palm leading until the fingers points in the direction of the second factor. The direction of the extended thumb in this process is the direction of the cross product. 3 of 6
10 Reversing the order of the factors in a cross product reverses the direction of the result. So, the order of the factors matters. 11 Since all of the factors in these products are unit vectors, each cross product is merely the sine of the angle between the two factors multiplied by a unit vector. When we compute the cross product of a unit vector with itself, the angle between the two factors is zero. It follows that the sine of the angle is zero and the cross product is zero. When we compute the cross product of two perpendicular unit vectors, the angle between them is 90, the sine is one and the magnitude of the cross product is one. So, the cross product of two perpendicular unit vectors is also a unit vector. 12 In the previous slide, we saw that when we compute the cross product of any one of the three unit vectors î, ĵ or ˆk with another one of those, we get plus or minus the third one of those. When is the result positive and when is it negative? Whenever the two factors in the cross product are in alphabetical order, the result is positive. When the factors are in reverse alphabetical order, the result is negative. How can we remember this? Two commonly used methods are shown on this slide. It may help you to commit one of these methods to memory. 4 of 6
13 If we write two three-dimensional vectors in Cartesian components and compute their cross product, we find nine partial products. From results on the previous two slides, we find that in general, three of the nine partial products are zero, and the other six are not. Most people do not memorize this result. Instead, they use the method shown on the following slide. 14 Evaluating the determinant of a three-by-three array is done by the process referred to as expanding the determinant. We do this in three steps using any row or any column of the array. (It can be proven that the result is the same no matter which row or column we choose.) Once we have chosen the row or column, the first step is to multiply each element of that row or column by the two-by-two sub-determinant built from the rows and columns that do not contain that element. The second step is to multiply each such product of an element and a sub-determinant by a plus or minus sign. The diagram shown at the upper right indicates which sign is to be used depending on which element of the array is being multiplied by a sub-determinant. The third step is to add the three products. The cross product shown on the previous slide can be evaluated by expanding the determinant shown here. To construct the determinant, we put the unit vectors î, ĵ and ˆk into the first row. Then we put 5 of 6
the components of the first factor in the cross product into the second row. Finally, we put the components of the second factor in the third row. The expansion shown uses the first row of the array. Note that only the second term is preceded by a minus sign, which is consistent with the plus-minus-plus pattern shown in the first row at the upper right. When we expand the three two-by-two sub-determinants shown here, we get the final equation on the previous slide. Using this method, we need to remember only how to construct the array and how to expand the determinant. Most people find this easier that remembering all six terms in the final equation on the previous slide. 6 of 6