Dr. Z’s Math251 Handout #12.4 [The Cross Product]By Doron ZeilbergerProblem Type 12.4a: Find the cross product a × b and verify that it is orthogonal to both aand b.a = ha1, a2, a3i , b = hb1, b2, b3i .Example Problem 12.4a: Find the cross product a × b and verify that it is orthogonal to botha and b.a = h4, 5, −3i , b = h2, 1, −3i .Steps Example1. Form the determinant with i, j, k inthe first row, the vector a in the secondrow and the vector b in the third row.i j ka1a2a3b1b2b3.1.i j k4 5 −32 1 −3.2. Evaluate the determinant step-by-step:i j ka1a2a3b1b2b3=ia2a3b2b3− ja1a3b1b3+ ka1a2b1b2= i(a2b3−b2a3)−j(a1b3−b1a3)+k(a1b2−b1a2) .2.i j k4 5 −32 1 −3=i5 −31 −3− j4 −32 −3+ k4 52 1= i(5·(−3)−(1)·(−3))−j(4·(−3)−(2)·(−3))+k(4·(1)−(2)·(5)) =−12i + 6j − 6kFirst Part of the Answer:−12i + 6j − 6k or h−12, 6, −6i.13. Take the dot product of this vectorwith a and b and verify that they are both0. If you get something else, somethingwent wrong.3. h−12, 6, −6i.h4, 5, −3i = (−12)(4) +(6)(5) + (−6)(−3) = −48 + 30 + 18 = 0andh−12, 6, −6i.h2, 1, −3i = (−12)(2)+(6)(1)+(−6)(−3) = −24 + 6 + 18 = 0.Second Part of Ans.: Both are zero asexpected.Problem Type 12.4b: (a) Find a vector orthogonal to the plane through the points P =(p1, p2, p3), Q = (q1, q2, q3),R = (r1, r2, r3). (b) Find the area of the triangle P QR.Example Problem 12.4b: (a) Find a vector orthogonal to the plane through the points P =(2, 1, 5), Q = (−1, 3, 4),R = (3, 0, 6). (b) Find the area of the triangle P QR.Steps Example1. Compute the vector PQ by subtract-ing Q from P and the vector PR by sub-tracting R from P .1.PQ = (−1, 3, 4)−(2, 1, 5) = h−3, 2, −1i ,PR = (3, 0, 6) − (2, 1, 5) = h1, −1, 1i .2. Take the cross product of PQ and PR.PQ × PR =i j k−3 2 −11 −1 1.= i2 −1−1 1−j−3 −11 1+k−3 21 −1= i + 2j + k = h1, 2, 1i .Ans. to First Part: A vector orthog-onal to the plane of the triangle P QR ish1, 2, 1i.23. Take the magnitude of the vector youfound in step 2. This is the area of theparralelogram. To get the area of thetriangle, divide it by 2.3.|h1, 2, 1i| =p12+ 22+ 12=√6 .So the area of the parallelogram formedby P Q and P R is√6 and the area of tri-angle PQR is half of that,√6/2.Ans. to Second Part: The area of tri-angle P QR is√6/2.Problem Type 12.4c: Find the volume of the parallelopiped with adjacent edges P Q,P R, andP S, where P, Q, R, S are given points.Example Problem 12.4c: Find the volume of the parallelopiped with adjacent edges P Q,P R,and P S, where P (0, 1, 2), Q(2, 4, 5), R(−1, 0, 1), S(6, −1, 4).Steps Example1. Compute the vectors PQ, PR,PS bythe appropriate subtractions.1.PQ = Q−P = (2, 4, 5)−(0, 1, 2) = h2, 3, 3i ,PR = R−P = (−1, 0, 1)−(0, 1, 2) = h−1, −1, −1i ,PS = S−P = (6, −1, 4)−(0, 1, 2) = h6, −2, 2i .32. Compute the cross product PQ ×PR. 2.PQ × PR =i j k2 3 3−1 −1 −1.= i3 3−1 −1−j2 3−1 −1+k2 3−1 −1= 0·i−(1)·j+(1)·k = −j+k = h0, −1, 1i3. Compute the dot product of what youfound in step 2 with the vector PS.3.(PQ × PR) · PS = h0, −1, 1i.h6, −2, 2i =(0) · (6) + (−1) · (−2) + (1) · (2) = 4 .Ans.: The volume of the parallelopipedis