Multivariable Calculus Lecture Slides Part 5

Friday September, 11, 2009 Examples Problem Find a point on the line and a direction vector. 1 h3 + t, 1 − t, 3 + 2ti. 2 x−1 2 = y+1 −3 = z−3 2 . Proof. √1 6 1 Goes through the point (3, 1, 3) with direction 2 Goes through the point (1, −1, 3) with direction Math 12 h1, −1, 2i. √1 17 h2, −3, 2i. September 11, 2009 41 / 51 Friday September, 11, 2009 Parallel Lines Problem Find the symmetric and parametric equations for the line through the point (2, 5, 3) that is parallel to the line x −3 y −2 z −1 = = . −4 −3 2 Math 12 September 11, 2009 42 / 51 Friday September, 11, 2009 Parallel Lines Problem Find the symmetric and parametric equations for the line through the point (2, 5, 3) that is parallel to the line x −3 y −2 z −1 = = . −4 −3 2 Proof. The line must be in the direction of h−4, −3, 2i so we have the parametric equations h2 − 4t, 5 − 3t, 3 + 2ti and the symmetric equations x −2 y −5 z −3 = = . −4 −3 2t Math 12 September 11, 2009 42 / 51 Friday September, 11, 2009 Intersection Points To determine if two lines intersect, we must determine if their equations have a common solution. Problem Determine if these two lines intersect: L1 : h3 + 2t, −2 + 4t, −1 − 3ti L2 : h−1 + 3s, −2 + 2s, −2 − si Math 12 September 11, 2009 43 / 51 Friday September, 11, 2009 Solution Proof. We use the z coordinate to solve −1 − 3t = −2 − s and so s = −1 + 3t. Substituting into ont of the other two equations yields 3 + 2t = −1 + 3(−1 + 3t) which is 7t = 7 which is t = 1 and so s = 2. So we have the solution (5, 2, −4) for both lines. Math 12 September 11, 2009 44 / 51 Friday September, 11, 2009 Skew Lines Problem Determine if these two lines intersect: L1 : h−2 + t, 4 − 3t, −5 − ti L2 : h3 − s, 5 + 2s, 4 + si Math 12 September 11, 2009 45 / 51 Friday September, 11, 2009 Solution Proof. Attempting to solve we find from the x coordinates that t = 5 − s. Substituting into one of the other two equations yields 4 − (5 − s) = 5 + 2s and so s = −6 and then t = 11. So we have the point (9, −29, −16) and (9, −7, −2). So the lines do not intersection since the solution is inconsistent. Math 12 September 11, 2009 46 / 51 Friday September, 11, 2009 Another Example Problem For the three following lines determine for the three possible pairs of two lines whether they are parallel, intersecting, or skew. If intersecting, determine the point of intersection. L1 : −6t î + (1 + 9t)ĵ − 3t k̂ L2 : x = 5 + 2t, y = −2 − 3t, z = 3 + t x −3 −1 − y L3 : = =z −2 2 5 Math 12 September 11, 2009 47 / 51 Friday September, 11, 2009 Solution Proof. Lines L1 , L2 are parallel since their directions h−6, 9, −3i and h2, −3, 1i are parallel (they differ by the multiple −3). Lines L2 , L3 intersect at (1, 4, 1). Lines L1 , L3 have direction h−6, 9, −3i and h2, −5, 1i respectively, so are clearly not parallel. If they intersect we must have −6t = 2s + 3 −3t = 2 + s So we must have 2s + 3 = 2s + 2 which is not possible. So the lines do not intersect. Therefore, they must be skew. Math 12 September 11, 2009 48 / 51 Friday September, 11, 2009 Planes Minimum information needed to define a plane 1 a point and a (normal) vector 2 Two non-parallel vectors 3 3 (non-colinear) points Math 12 September 11, 2009 49 / 51 Friday September, 11, 2009 Planes Minimum information needed to define a plane 1 a point and a (normal) vector 2 Two non-parallel vectors 3 3 (non-colinear) points To determine the equation for a plane consider that the normal vector is perpendicular to every vector in the plane. If P0 is represented by the position vector r~0 and P is any other point, represented by the vector ~r , then the plane is determined buy (~r − r~0 ) · ~n = 0 where ~n is the normal vector. In particular we have hx − x0 , y − y0 , z − z0 i · ha, b, ci = 0 In particular we can represent any plane as ax + by + cz + d = 0. Math 12 September 11, 2009 49 / 51 Friday September, 11, 2009 A Simple Example Problem Find the plane through (1, 1, 1) with normal ~n = h0, 2, −1i. Math 12 September 11, 2009 50 / 51 Friday September, 11, 2009 A Simple Example Problem Find the plane through (1, 1, 1) with normal ~n = h0, 2, −1i. Proof. This is the plane 0(x − 1) + 2(y − 1) − (z − 1) = 0 which is 2y − z − 1 = 0. Math 12 September 11, 2009 50 / 51 Friday September, 11, 2009 Two vectors Problem Find the plane containing the vectors ~a = h2, −4, 4i and ~b = h4, −1, −2i and the point P = (1, 3, 2). Math 12 September 11, 2009 51 / 51 Friday September, 11, 2009 Two vectors Problem Find the plane containing the vectors ~a = h2, −4, 4i and ~b = h4, −1, −2i and the point P = (1, 3, 2). Proof. We compute ~n = ~a × ~b = h12, 20, 14i and given P = (1, 3, 2) on the plane we have 12(x − 1) + 20(y − 3) + 14(z − 2) = 0 which is 6x + 10y + 7z = 50. Math 12 September 11, 2009 51 / 51