Planes in R3 (§12.5)Cross Product (§12.4)Planes in R3(§12.5)Cross Product (§12.4)Math 241, Spring 2014Jayadev S. AthreyaSpring 2014Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Describing a PlaneA point P0on the plane, a normal vector n (perpendicular to theplane).For any other point P on the plane, the vector−−→P0P = r −r0isperpendicular to n.Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Equations for a planen = (a, b, c), P0= (x0, y0, z0). Then P = (x, y, z) is on theplane if and only if:0 = (a, b, c) · (x − x0, y −y0, z − z0)That is,ax + by + cz = ax0+ by0+ cz0Writing d = (ax0+ by0+ cz0), we have the equation of theplaneax + by + cz = dJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Equations for a planen = (a, b, c), P0= (x0, y0, z0). Then P = (x, y, z) is on theplane if and only if:0 = (a, b, c) · (x − x0, y −y0, z − z0)That is,ax + by + cz = ax0+ by0+ cz0Writing d = (ax0+ by0+ cz0), we have the equation of theplaneax + by + cz = dJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Equations for a planen = (a, b, c), P0= (x0, y0, z0). Then P = (x, y, z) is on theplane if and only if:0 = (a, b, c) · (x − x0, y −y0, z − z0)That is,ax + by + cz = ax0+ by0+ cz0Writing d = (ax0+ by0+ cz0), we have the equation of theplaneax + by + cz = dJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Example: Tangent PlaneJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Intersections and AnglesGeneral Question: Given two planes P1and P2, how can wesee:1If they intersect?2If so, where do they intersect?3What is the angle of their intersection?Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)ExampleJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)ExamplePlanes P1is the plane given by z = 1, P2is set of pointssatisfying x + z = y.Normals n1= (0, 0, 1), n2= (1, −1, 1)Intersection is a line L, consisting of points satisfying bothequations, i.e., (x, y, z) so thatx + z = y and z = 1For example (−1, 0, 1), (0, 1, 1).Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)ExamplePlanes P1is the plane given by z = 1, P2is set of pointssatisfying x + z = y.Normals n1= (0, 0, 1), n2= (1, −1, 1)Intersection is a line L, consisting of points satisfying bothequations, i.e., (x, y, z) so thatx + z = y and z = 1For example (−1, 0, 1), (0, 1, 1).Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)ExamplePlanes P1is the plane given by z = 1, P2is set of pointssatisfying x + z = y.Normals n1= (0, 0, 1), n2= (1, −1, 1)Intersection is a line L, consisting of points satisfying bothequations, i.e., (x, y, z) so thatx + z = y and z = 1For example (−1, 0, 1), (0, 1, 1).Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)AngleAngle between planes is the angle between their normalvectors. In our example, we havecos θ =n1· n2kn1kkn2k=(0, 0, 1) · (1, −1, 1)k(0, 0, 1)kk(1, −1, 1)k=1√3Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Vectors in R3Cross product is a way to combine vectors u, v in R3to get anew vector u × v, which is perpendicular to both of them.It’s very useful for describing planes.Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)DeterminantGiven a 2 × 2 matrixa bc d, we define the determinant bydeta bc d=a bc d= ad − bcJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant and AreaJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant, Angle, and Areav = (a, b)w = (c, d)θ·v·whArea A = base × height = kvkkwksin θJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant, Angle, and AreaA2= kvk2kwk2sin2θ = kvk2kwk2(1−cos2θ) = kvk2kwk2−(v·w)2v = (a, b), w = (c, d), so we getA2= (a2+b2)(c2+d2)−(ac+bd)2= a2d2+b2c2−2abcd = (ad−bc)2(ad −bc)2=a bc d2Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant, Angle, and AreaA2= kvk2kwk2sin2θ = kvk2kwk2(1−cos2θ) = kvk2kwk2−(v·w)2v = (a, b), w = (c, d), so we getA2= (a2+b2)(c2+d2)−(ac+bd)2= a2d2+b2c2−2abcd = (ad−bc)2(ad −bc)2=a bc d2Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant, Angle, and AreaA2= kvk2kwk2sin2θ = kvk2kwk2(1−cos2θ) = kvk2kwk2−(v·w)2v = (a, b), w = (c, d), so we getA2= (a2+b2)(c2+d2)−(ac+bd)2= a2d2+b2c2−2abcd = (ad−bc)2(ad −bc)2=a bc d2Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinants and RowsAlso, for any t ∈ R,a + tc b + tdc d=a bc d=a bc + ta d + tbExercise: think of what this means in terms of parallelograms,reduce area computation to the case of a rectangle.Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinants and RowsAlso, for any t ∈ R,a + tc b + tdc d=a bc d=a bc + ta d + tbExercise: think of what this means in terms of parallelograms,reduce area computation to the case of a rectangle.Jayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Signed Area(3, 1)(1, 2)1 23 1= 1 − 6 = −5. Why negative?3 11 2= 6 − 1 = 5Reason: sin(−θ) = −sin θJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Signed Area(3, 1)(1, 2)1 23 1= 1 − 6 = −5. Why negative?3 11 2= 6 − 1 = 5Reason: sin(−θ) = −sin θJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Signed Areaa bc d=AREA if (a, b) to (c, d) is counterclockwise−AREA if (a, b) to (c, d) is clockwiseJayadev S. Athreya Math 241, Spring 2014Planes in R3(§12.5)Cross Product (§12.4)Determinant in 3
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