Dr. Z’s Math251 Handout #12.5 [Equations of Lines and Planes]By Doron ZeilbergerProblem Type 12.5a: Find an equation of the plane that passes through three given pointsExample Problem 12.5a: Find an equation of the plane that passes through the points (1, 1, 1),(2, 0, 1), (2, 1, 0).Steps Example1. Calling these points P, Q, R find thevectors PQ and PR by doing Q − P andR − P .1. P = (1, 1, 1), Q = (2, 0, 1), R = (2, 1, 0),PQ = Q−P = h2−1, 0−1, 1−1i = h1, −1, 0i ,PR = R−P = h2−1, 1−1, 0−1i = h1, 0, −1i .2. Find a vector normal to the plane bycomputing the cross-product PQ × PR.2.PQ × PR = h1, −1, 0i × h1, 0, −1i =i j k1 −1 01 0 −1=i−1 00 −1− j1 01 −1+ k1 −11 0= i + j + k = h1, 1, 1i .This is the normal vector n = ha, b, ci.So n = ha, b, ci = h1, 1, 1i13. Pick any of the three points (it doesnot matter which) as the refence point(x0, y0, z0) and use the formulaa(x − x0) + b(y − y0) + c(z − z0) = 0 .3. Picking P we get (x0, y0, z0) = (1, 1, 1),since a = 1, b = 1, c = 1, we get that anequation is1 · (x − 1) + 1 · (y − 1) + 1 · (z − 1) = 0and expanding we getx + y + z = 3 .Ans.: An equation for the plane passingthrough P, Q and R is x + y + z = 3.Check: Plug-in all the three points intothe equation and make sure that they agree.Problem Type 12.5b: Find direction numbers for the line of intersections of the planes a1x +b1y + c1z = d1and a2x + b2y + c2z = d2.Example Problem 12.5b: Find direction numbers for the line of intersections of the planes2x + 3y + 4z = 2 and −3x + 2y + 3z = 1.Steps Example1. By looking at the coeffs. of x, y, z ex-tract the normal vectors n1= ha1, b1, c1iand n2= ha2, b2, c2i.Note: the numbers on the right sides(d1, d2) are not needed.1. n1= h2, 3, 4i and n2= h−3, 2, 3i.2. Take the cross-product n1× n2. Thecomponents are the direction numbersof the line of intersection.2.n1×n2= h2, 3, 4i×h−3, 2, 3i = h1, −18, 13i .(You do it!)Ans.: The direction numbers are h1, −18,
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