The Dot Product

Transcription

1 The Dot Product If = ( 1,, 3 ) and = ( 1,, 3 ) are ectors, the dot product of and is defined algebraically as = Example. (a) Compute the dot product (,3, 7) ( 3,,0). (b) Compute the dot product (4î ĵ+7ˆk) ( î 3ĵ+11ˆk). (a) (b) (,3, 7) ( 3,,0) = ()( 3)+(3)()+( 7)(0) = 0. (4î ĵ+7ˆk) ( î 3ĵ+11ˆk) = (4)( )+( 1)( 3)+(7)(11) = 7. The dot product of to ectors is a number. Since numbers are often referred to as scalars, the dot product is often called the scalar product. The definition orks just as ell for ectors ith components, or more than 3 components. For example, here is the dot product of to 4-dimensional ectors: (3, 4, 1, 1) (0,,,5) = (3)(0)+( 4)()+( 1)( )+( 1)(5) = 11. Here are some properties of the dot product. Theorem. Let k R and let u,, R n. (a) = (b) u ( + ) = u + u (c) k( ) = (k ) = (k ) (d) = Proof. All of these results can be proed by riting the ectors in terms of components and computing. As an example, I ll proe (d) for 3-dimensional ectors. Suppose = ( 1,, 3 ). Then = =. The dot product also has a geometric interpretation. Theorem. Let and be ectors and let θ be the angle beteen them. Then = cosθ. Note: Since cos θ = cos( θ), it does not matter hether θ is measured counterclockise or clockise. 1

2 Proof. Apply the La of Cosines and use the fact that a a = a : = + cosθ ( ) ( ) = + cosθ + = + cosθ + = + cosθ = cosθ = cosθ You can often use ectors to obtain results by computing things algebraically, then interpreting the results geometrically. In this case, you can do this because there are to ays of looking at the dot product. For example, hen ill to ectors be perpendicular? This ill happen if the angle beteen them is θ = ± π. In either case, cosθ = 0, and hence = 0. Going the other ay, if and are nonzero ectors and = 0, then cosθ = 0. Therefore, cosθ = 0, so θ = ± π and the ectors are perpendicular. Notes. 1. The ord orthogonal is synonymous ith perpendicular.. The zero ector 0 is triially perpendicular to any other ector. But usually you ant a nonzero ector perpendicular to another ector, and I ll try to be careful to ask for a nonzero ector. In addition: (a) If cosθ > 0, the angle beteen the ectors is acute. (b) If cosθ < 0, the angle beteen the ectors is obtuse. Note that you can proe these geometric facts about to ectors een though it might be hard to determine them by draing the ectors. And hile e ll be primarily concerned ith ectors in and 3 dimensions, these facts about angles and dot products are true in n dimensions. Example. Determine hether the ectors make an acute angle, an obtuse angle, or are perpendicular. (a) = (1, 5,6), = (,, 7). (b) = (,1,,), = (4,0, 3, 7) (a) = 10 4 = 50. Since < 0, cosθ < 0, and the angle beteen the ectors is obtuse. The ectors are perpendicular. = = 0. Example. Find the exact alue of the cosine of the angle beteen (3,,1) and (,5,3). Tell hether the ectors are orthogonal; if not, tell hether the angle beteen them is acute or obtuse. The angle is obtuse. cosθ = (3,,1) (,5,3) (3,,1) (,5,3) = =

3 Example. Find to unit ectors hich are perpendicular to (, 5). Ho many unit ectors are perpendicular to (1, 3,3)? (5, ) is perpendicular to (,5), since the to ectors hae dot product 0 by inspection. (5, ) = 1 1 9, so the ectors (5, ) and (5, ) are unit ectors perpendicular to (,5). 9 9 There are infinitely many unit ectors perpendicular to (1, 3, 3). If (1, 3,3) is like a flagpole, think of ectors pointing along the ground aay from the base of the flagpole. Each may be diided by its length to get a unit ector. Algebraically, they are the unit ectors (a,b,c) such that a 3b+3c = 0. Example. Find a nonzero ector hich is simultaneously orthogonal to both (1,3,) and (,5, 4). Let (a,b,c) be such a ector. I ant This gies the equations (1,3,) (a,b,c) = 0 and (,5, 4) (a,b,c) = 0. a+3b+c = 0 and a+5b 4c = 0. Since I hae to equations but three ariables, I can t expect a unique solution. I ll eliminate one of the ariables to start ith. The first equation gies Substitute this into a+5b 4c = 0 to get a = 3b c. ( 3b c)+5b 4c = 0, or b 8c = 0. At this point, I can assign a alue of my choice to one of the ariables. Let c = 1. Then b 8 = 0, so b = 8. Plugging these alues into a = 3b c, I get a = 4 =. Thus, (, 8,1) is perpendicular to (1,3,) and (,5, 4). In fact, any solution must be a multiple of the ector I found, since the ectors perpendicular to both (1,3,) and (,5, 4) form a line. line of ectors perpendicular to (1,3,) and (,5,-4) (,-8,1) (1,3,) (,5,-4) 3

4 Example. Find ectors u,, and such that u = u but. There are lots of possibilities. For instance, But (,1) ( 6,3). (,4) (,1) = 0 = (,4) ( 6,3). The scalar component of in the direction of is comp =. It gies the (signed) length of the shado that makes on. It is positie if and point in the same direction (i.e. if the angle beteen them is acute) and negatie if and point in the opposite direction (i.e. if the angle beteen them is obtuse). comp comp comp > 0 comp < 0 To see this, consider the right triangle in the picture. The base of the triangle is comp, and (base) = cosθ, so (base) = cosθ. cos But Therefore, cosθ = (base) =, so cosθ =., so comp =. 4

5 Example. If = (1,4, 5) and = (,1, ), find the scalar component of in the direction of. comp = (1,4, 5) (,1, ) (,1, ) = The ector projection of in the direction of is ector hose (signed) length is comp and hose direction is the direction of. proj To obtain it, I multiply comp by the unit ector ( ) proj = (comp ) = hich has the same direction as. This gies ( )( ) = =. Thus, the formula is proj =. Example. Find the ector projection of = î ĵ+ˆk in the direction of = 4î+3ˆk. proj = (, 1,1) (4,0,3) (4,0,3) (4,0,3) (4,0,3) = (4,0,3) = 5î+ 5ˆk. c 017 by Bruce Ikenaga 5