The Dot Product Thus far we have discussed the addition, subtraction, and scalar multiplication of vectors. Another operation on vectors is called the dot product. It is important because it can be used to compute the angle between two vectors. There are other properties of the dot product which are covered in a calculus course. We begin by defining the dot product. Dot Product Let 123vvvvijk and 123wwwwijk . The dot product of v and w, denoted by , is a real number given byvw11 2 2 33vw vw vwvw Example 1: Compute if vw 430vijk and 225 wijk . Solution: . 4( 2) ( 3)(2) (0)(5) 14  vw As we mentioned above, the dot product can be used to find the angle between two vectors. The relationship is given by the following: Angle Between Two Vectors If v and w are nonzero vectors, then cosvw v w Thus, the angle q between v and w is given by 1cosvwvw The vectors v and w are perpendicular if and only if . 0vw To see why the first formula holds, consider the following diagram. For the sake of simplicity, the vectors shown are only in two dimensions. 1xyvv1,v2ww1,w2v-wq Figure 1: Visualizing the Law of Cosines We have that 12vvvij, 1ww2wij and 11 2 2()(vw vw) vw ij. By the Law of Cosines, we have that 22 22cosvw v w vw . First off, observe that 22211 222221111222222 2 212 1 2 112222()( )22()( )2(2vw vwvvwwvvwwvv ww vwvw        vwvw vw2) Thus, our equation becomes 22 22222cos vw vwvwvw vw Subtracting ||v||2 and ||w||2 from both sides, we have: 22cosvw v w Finally, dividing both sides by -2, we get the desired formula. Example 2: Find the angle between the vectors 2 6vij and 9 3wij. Solution: Notice that , so by the note after our formula, we have that v and w are perpendicular, i.e. q = p/2. 2 ( 9) 6 3 0 vw 2Example 3: Given a unit cube (each side has length 1) with a vertex at (0, 0, 0), find the angle q between the main diagonal and one of the faces of the cube. (See the figure below.) q0,0,0111 Figure 2: The Unit Cube Solution: Let 1 1 0vijk and 1 1 1wijk . These vectors correspond with the bottom (red line) and diagonal (blue line), respectively. We want to find the angle between these two vectors, so we use the formula for the dot product. That is, cosvw v w . Here, , 11 11 01 2vw22 2110 2v and 222111 3w . So, we see that 2cos23 6vwvw 2. Thus,12cos 35.266. Looking at the geometric interpretation of the dot product, cosvw v w , we see that for two vectors, v and w, of a fixed length, the dot product changes as the angle q changes. In particular, the dot product will be largest when v and w point in the same direction (since in that case, q = 0 and so cos(q) = 1). The dot product will be smallest when v and w point in opposite directions (since in that case q = p and so cos(q) = -1). There is a simple relationship between the magnitude of a vector and its dot product with itself, as well as some useful properties concerning dot products that we record below. 3Properties of the Dot Product For any vectors a, b, and c and any scalar k, we have that 1. 2aa a 2. ab ba 3.   kk kab ab ab 4. ( )abcacbc When we first talked about the equation of a plane, we specified it in terms of the slope in the x-direction, the slope in the y-direction and the z-intercept. There is another interpretation that involves vectors. We call a vector that is perpendicular to a plane a normal vector to the plane. In Figure 3 (below), we see that n is a normal vector to the plane. Suppose that P0 = (x0, y0, z0) is a fixed point in the plane and let P = (x, y, z) be any other point in the plane. Then the vector 000()()()PP x x y y z z  i0jk0PP is a vector with its head and tail in the plane. Thus, we have that n and  are perpendicular. This means that , i.e. a(x – x0) + b(y – y0) + c(z – z0) = 0. 00PPn nP0=x0,y0,z0P=x,y, zP0P Figure 3: The plane with normal n and point (x0, y0, z0) Equation of the Plane with a Normal Vector The equation of the plane with normal vector abcnijk and containing the point P0 = (x0, y0, z0) is given by a(x – x0) + b(y – y0) + c(z – z0) = 0. Letting d = ax0 + by0 + cz0, we can rewrite the above equation as ax + by + cz = d 4Example 4: Find the equation of the plane passing through the point (1, 2, 1) that is perpendicular to the vector 2i – 3j + 4k. Solution: Here, we have that n = 2i – 3j + 4k, so we have that a = 2, b = –3, and c = 4. We are given the point (x0, y0, z0) = (1, 2, 1). Plugging those values into the formula above, we have that the equation of the plane is given by 2(x – 1) – 3(y – 2) + 4(z – 1) = 0. That is, 2x – 3y + 4z = 0. When we first introduced vectors, we mentioned how their components were the amount traveled in the x, y and z-directions respectively. There are some instances (especially in physics) where it is often useful to talk about the parallel and perpendicular parts of a vector with respect to another vector (not the standard axes). Figure 4 below shows the vectors v and w, starting at the same initial point P. If S is the point on w that is closest to the point R (that is, S is located at the foot of the perpendicular from R to the vector w), then the vector PS is called the vector projection of v onto w. PQRSvwpro jwv Figure 4: Projection of v onto w There is a clean relationship between dot products and the vector projection. To see this, notice that  is a scalar multiple of the vector w. In fact, the length of , PS PS PS, is equal to ||v||cos(q). See Figure 5. 5qPQRSvwvcosq Figure 5: The Scalar Projection of v onto w But recall that we showed above that cosvwvw. Substituting that in,