The Dot Product. If v = a 1 i + b 1 j and w = a 2 i + b 2 j are vectors then their dot product is given by: v w = a 1 a 2 + b 1 b 2

Transcription

1 The Dot Product In this section, e ill no concentrate on the vector operation called the dot product. The dot product of to vectors ill produce a scalar instead of a vector as in the other operations that e examined in the previous section. The dot product is equal to the sum of the product of the horizontal components and the product of the vertical components. If v = a 1 i + b 1 j and = a i + b j are vectors then their dot product is given by: v = a 1 a + b 1 b Properties of the Dot Product If u, v, and are vectors and c is a scalar then: u v = v u u (v + ) = u v + u 0 v = 0 v v = v (cu) v = c(u v) = u (cv) Example 1: If v = 5i + j and = 3i 7j then find v. v = a 1 a + b 1 b v = (5)(3) + ()(-7) v = v = 1 Example : If u = i + 3j, v = 7i 4j and = i + j then find (3u) (v + ). Find 3u 3u = 3( i + 3j) 3u = 3i + 9j Find v + v + = (7i 4j) + (i + j) v + = (7 + ) i + ( 4 + 1) j v + = 9i 3j

2 Example (Continued): Find the dot product beteen (3u) and (v + ) (3u) (v + ) = ( 3i + 9j) (9i 3j) (3u) (v + ) = ( 3)(9) + (9)(-3) (3u) (v + ) = 7 7 (3u) (v + ) = 54 An alternate formula for the dot product is available by using the angle beteen the to vectors. If v and are to nonzero vectors and θ is the smallest nonnegative angle beteen them then their dot product is given: v = v cos θ This same equation could be solved for theta if the angle beteen the vectors needed to be determined. θ = cos 1 v v Example 3: If u = 6i j and v = 3i + 5j then find the angle θ beteen the vectors. Round the anser to the nearest tenth of a degree, if necessary. Find the magnitude of u u = a + b u = (6) + ( ) u = u = 40 u = 10 Find the magnitude of v v = a + b v = (3) + (5) v = 9+ 5 v = 34

3 Example 3 (Continued): Find the dot product of u and v u v = a 1 a + b 1 b u v = (6)(3) + (-)(5) u v = u v = 8 Find the angle beteen the vectors θ = cos θ = cos 1 θ θ 77.5 u v u v 1 8 ( 10)( 34) ( ) 1 cos When comparing to lines they ere described as being parallel, perpendicular, or neither depending on the values of their slopes. The same can be done ith vectors but orthogonal is used instead of the term perpendicular. Hoever, e ould be looking at the measure of the angle beteen the to vectors. Parallel vectors To vectors are parallel hen the angle beteen them is either 0 (the vectors point in the same direction) or 180 (the vectors point in opposite directions) as shon in the figures belo. Orthogonal vectors To vectors are orthogonal hen the angle beteen them is a right angle (90 ). The dot product of to orthogonal vectors is zero.

4 Example 4: Determine if u = i 3j and v = 9i 3j are orthogonal. Verify that the dot product is 0 u v = ( i 3j) (9i 3j) u v = ( 1)(9) + ( 3)( 3) u v = u v = 0 The dot product is zero so the vectors are orthogonal. There are real orld applications of vectors that ill require for the vectors to be broken don into its orthogonal components. By breaking a vector into its orthogonal components e can express a vector as the sum of vectors. The components are formed by hat is called vector projection. Vector projection involves draing a line from the terminal point of the vector e ant to project don to form a right angle ith a line passing through the other vector. If for example vector v is projected upon vector, then the projected vector ould be denoted as proj v and is given by: proj v v = The projected vector can then be subtracted from vector v to finish the decomposition process of v into its orthogonal components. Decomposition of v into vector components If vectors v and are to nonzero vectors, then vector v can be expressed as the sum of its orthogonal component vectors v 1 and v, here v 1 (the vector projected onto ) is parallel to and v is orthogonal to. v v1 = projv = and v = v v 1 Since the vector components are orthogonal there dot product must be equal to zero. You can use this as a ay of checking to make sure the vector components are correct. v 1 v = 0

5 Example 5: Let v = 5i + j and = i 4j. Decompose v into its vector components v 1 and v, here v 1 is parallel to and v is orthogonal to. Find the magnitude of vector = a + b = () + ( 4) = = 0 Find the dot product of v and v = ( 5i + j) (i 4j) v = ( 5)() + ()( 4) v = 10 8 v = 18 Use vector projection to find v 1 v1 = v 18 v = i j ( ) ( 4 ) v1 = i 4 j v1 = i+ j 5 5 ( )

6 Example 5 (Continued): Use vector subtraction to find v v = v v v = ( 5i+ j) i + j v = ( 5i+ j) + i j v = + i+ j v = i j 5 5 Check of the anser v1 v = i+ j i j v1 v = v1 v = 5 5 v v = 1 0 Since the dot product is zero, the vector v 1 ould be parallel to and v ould be orthogonal to. One example of ho the magnitude of a projected vector is used ould be in the field of physics hen calculating the ork done by a force moving an object over a distance. The force applied to the object ill not alays be applied along the line of motion. The amount of ork W done by a force F moving an object from point A to point B is given by: W = F AB W = F AB cosθ, here AB is the distance from A to B.

7 Example 6: A force of 90 pounds is used to pull an object along a level surface. The force is applied at an angle of 43 to the surface and moves the object a total of 50 feet. Ho much ork as done? Round the anser to the nearest foot-pound. W = F AB cosθ W = (90 pounds)(50 feet) cos 43 W Rounded off to the nearest foot-pound, the amount of ork done is approximately 16,455 foot-pounds.