10.4 The Cross Product

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1 Math 172 Chapter 10B notes Page 1 of The Cross Product The cross product, or vector product, is defined in 3 dimensions only. Let aa = aa 1, aa 2, aa 3 bb = bb 1, bb 2, bb 3 then aa bb = aa 2 bb 3 aa 3 bb 2, aa 3 bb 1 aa 1 bb 3, aa 1 bb 2 aa 2 bb 1 Each component is obtained from the previous one by cyclic permutation of indices: 1 2, 2 3, 3 1 Recall 2 2 determinants aa 1 aa 2 bb 1 bb 2 = aa 1 bb 2 aa 2 bb 1 and 3 3 determinants aa 1 aa 2 aa 3 bb 1 bb 2 bb 3 = aa 1 bb 2 bb 3 aa cc 1 cc 2 cc cc 2 cc 2 bb 1 bb 3 + aa 2 cc 1 cc 3 bb 1 bb 2 3 cc 1 cc 2 3 Cross product as a symbolic determinant ii jj kk aa bb = aa 1 aa 2 aa 3 bb 1 bb 2 bb 3 = ii aa 2 aa 3 bb 2 bb jj aa 1 aa 3 3 bb 1 bb + kk aa 1 aa 2 3 bb 2 bb 1 Example. Let aa = 2, 1,3 and bb = 1, 2,4. Find aa bb. ii jj kk aa bb = = ii jj kk = ii ( 4 + 6) jj (8 + 3) + kk ( 4 1)

2 Math 172 Chapter 10B notes Page 2 of 9 = 2, 11, 5 Example. Find ii jj. ii jj kk ii jj = = ii jj kk = kk sketch 3D axes, ii, jj, kk [Fig. 1.] Three algebraic properties of the cross product Let aa, bb and cc be vectors in 3D 1. The cross product is anti-commutative bb aa = aa bb 2. The cross product distributes over vector addition aa (bb + cc) = aa bb + aa cc (aa + bb) cc = aa cc + bb cc 3. In addition let kk be a scalar (kkaa) bb = kk(aa bb) = aa (kkbb) Geometrical properties of the cross product Theorem. aa bb is orthogonal to both aa and bb. The direction is given by the right hand rule. sketch aa, bb, aa bb [Fig. 2.] Theorem. If θθ is the angle between aa and bb then aa bb = aa bb sin (θθ)

3 Math 172 Chapter 10B notes Page 3 of 9 Geometric Interpretation of aa bb sketch aa, bb, θθ, extend to give parallelogram, proj h from the tip of bb to aa [Fig. 3] Area of parallelogram = aa h = aa bb sin(θθ) Example. (a) Find a vector orthogonal to the plane through points PP, QQ and RR. (b) Find the area of triangle PPPPPP. PP(1,0, 1), QQ(2,4,5), RR(3,1,7) sketch origin, PP, QQ, RR, vectors, PPPP PPPP [Fig. 4] (a) PPPP PPPP will be orthogonal to the plane PPPP = OOOO OOOO = 2,4,5 1,0, 1 = 1,4,6 PPPP = OOOO OOOO = 3,1,7 1,0, 1 = 2,1,8 ii jj kk PPPP PPPP = = ii (32 6) jj (8 12) + kk (1 8) = 26, 4, 7 is an answer for part (a) (b) PPPP PPPP = area of parallelogram defined by PP, QQ, RR sketch PP, QQ, RR,, PPPP PPPP, parallelogram [Fig. 5] 1 PPPP PPPP = area of triangle defined by PP, QQ, RR 2 =

4 Math 172 Chapter 10B notes Page 4 of 9 = = Equations of Lines and Planes Lines in Space Lines in 2D sketch 0 xx-, 0 yy-, PP(xx 0, yy 0 ), line segment through PP with slope mm [Fig. 6] determined by point and slope Lines in 3D 0 xx-, 0 yy-, 0 zz-, PP(xx 0, yy 0, zz 0 ), line LL through PP [Fig. 7] determined by point and direction add direction vector vv Let QQ be any other point on the line add QQ(xx, yy, zz), rr 0 =, OOOO vvtt from PP to QQ, rr = OOOO [Fig. 8] rr = rr 0 + vvtt vector equation of line LL rr = xx, yy, zz rr 0 = xx 0, yy 0, zz 0 Let the direction vector be vv = aa, bb, cc, where aa, bb, cc are direction numbers. Substitute into the vector equation to obtain: xx = xx 0 + aaaa,

5 Math 172 Chapter 10B notes Page 5 of 9 yy = yy 0 + bbbb, zz = zz 0 + cccc, parametric equations of the line. Eliminate t xx xx 0 aa = yy yy 0 bb = zz zz 0 cc symmetric equations of the line. Example. Find the vector, parametric and symmetric equations of the line passing through PP( 1,0,5) and QQ(4, 3,3). Let rr 0 = OOOO = 1,0,5 vv = OOOO OOPP = 4, 3,3 1,0,5 = 5, 3, 2 The vector equation is rr = rr 0 + vvtt where rr 0 and vv are given above and rr = xx, yy, zz. Parametric equation are obtained from components of the vector equation xx = 1 + 5tt, yy = 3tt, zz = 5 2tt. Symmetric equations are obtained by eliminating t xx+1 = yy = zz Example. Show that the line through (0,1,1) and (1, 1,6) LL 1 is perpendicular to the line through ( 4,2,1) and ( 1,6,2) LL 2

6 Math 172 Chapter 10B notes Page 6 of 9 Solution. Find direction vectors for each line. LL 1 : vv 1 = 1, 1,6 0,1,1 = 1, 2,5 LL 2 : vv 2 = 1,6,2 4,2,1 = 3,4,1 The lines are perpendicular if their direction vectors are perpendicular. Note that: vv 1 vv 2 = 1, 2,5 3,4,1 = = 0 Thus LL 1 and LL 2 are perpendicular. Example. Determine whether the lines LL 1 and LL 2 are parallel, intersecting or skew. LL 1 : xx = 1 + tt, yy = 2 tt, zz = 3tt LL 2 : xx = 2 ss, yy = 1 + 2ss, zz = 4 + ss Solution. Direction vectors are: vv 1 = 1, 1,3, vv 2 = 1,2,1 vv 1 and vv 2 are not parallel (their components are not in proportion), therefore LL 1 and LL 2 are not parallel. vv 1 vv 2 = = 0, therefore lines LL 1 and LL 2 are perpendicular. If the lines intersected, there would be an ss and a tt such that 1 + tt = 2 ss, [1] 2 tt = 1 + 2ss, [2] 3tt = 4 + ss, [3] Solve [1] and [2] for ss and tt. Adding [1] and [2] obtain 3 = 3 + ss ss = 0 Substitute ss = 0 into [1] to obtain tt = 1. Does ss = 0, tt = 1 satisfy [3]? No. Since the lines are not parallel but do not intersect, we conclude that they are skew.

7 Math 172 Chapter 10B notes Page 7 of 9 Planes in Space A plane is also determined by a point and a direction. OO, PP 0, nn drawn with tail at PP 0 [Fig. 9] There is a unique plane passing through PP 0 perpendicular to the vector nn. add plane To find an equation for the plane let PP 0 (xx 0, yy 0, zz 0 ) be the original point on the plane PP(xx, yy, zz) be any other point on the plane add PP nn = aa, bb, cc is the normal vector Let rr 0 = OOPP 0 rr = OOOO add add rr rr 0 = PP 0 PP add [Fig. 10] PP 0 is perpendicular to nn nn (rr rr 0 ) = 0 Vector equation of the plane In components aa, bb, cc xx xx 0, yy yy 0, zz zz 0 aa(xx xx 0 ) + bb(yy yy 0 ) + cc(zz zz 0 ) = 0 Scalar equation of the plane Example. Find an equation of the plane that passes through the point PP( 5,1,2) with normal vector

8 Math 172 Chapter 10B notes Page 8 of 9 nn = 3, 5,2 Solution. Let rr 0 = OOOO = 5,1,2 Let rr = xx, yy, zz be any other point on the plane. Vector equation of plane nn (rr rr 0 ) = 0 3, 5,2 ( xx, yy, zz 5,1,2 ) 3, 5,2 xx + 5, yy 1, zz 2 Scalar equation of the plane 3(xx + 5) 5(yy 1) + 2(zz 2) = 0 3xx yy zz 4 = 0 3xx 5yy + 2zz = 16 Linear equation of the plane. Example. Find an equation of the plane passing through the 3 points PP(1,0, 1) QQ(2,4,5) RR(3,1,7) sketch 0, PP, QQ, RR [see Fig. 4] OOOO + PPPP = OOOO add OOOO, PPPP, OOOO PPPP = OOOO OOOO = 2,4,5 1,0, 1 = 1,4,6 PPPP = OOOO OOOO = 3,1,7 1,0, 1 = 2,1,8 add PPPP Construct a vector perpendicular to PP, QQ and RR

9 Math 172 Chapter 10B notes Page 9 of 9 ıı ȷȷ kk PPPP PPPP = = ıı ȷȷ kk = ıı (32 6) ȷȷ (8 12) + kk (1 8) = 26,4, 7 Point on plane rr 0 = 1,0, 1 Normal vector nn = 26,4, 7 Vector equation of plane nn (rr rr 0 ) = 0 where rr = xx, yy, zz is any other point on the plane. 26,4, 7 xx 1, yy, zz + 1 = 0 26(xx 1) + 4yy 7(zz + 1) = 0 Scalar equation of the plane 26xx yy 7zz 7 = 0 26xx + 4yy 7zz = 33 Linear equation of the plane.

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