Dr. Z’s Math251 Handout #12.1 [Three Dimensional Coordinate Systems]By Doron ZeilbergerProblem Type 12.1a: Show that the triangle with vertices P = (p1, p2, p3), Q = (q1, q2, q3),R = (r1, r2, r3) is an equilateral triangle.Example Problem 12.1a: Show that the triangle with vertices P = (−4, 8, 0), Q = (2, 4, −2),R = (−2, 2, 4) is an equilateral triangle.Steps Example1. Use the distance formula|P1P2| =p(x2− x1)2+ (y2− y1)2+ (z2− z1)2,for the distance between two points P1(x1, y1, z1)and P2(x2, y2, z2), to find the three dis-tances |P Q|, |P R|, |QR|.1. Here P = (−4, 8, 0), Q = (2, 4, −2),R = (−2, 2, 4), so|P Q| =p(2 − (−4))2+ (4 − 8)2+ ((−2) − 0)2=√36 + 16 + 4=√56 .|P R| =p((−2) − 4)2+ (2 − 8)2+ (4 − 0)2=√4 + 36 + 16=√56 .|QR| =p((−2) − 2)2+ (2 − 4)2+ (−4 − (−2))2=√16 + 4 + 36=√56 .2. Check whether theese three distancesare all the same. It there are, it is anequilateral triangle, otherwise not.2. All the distances are the same (√56).Ans.: It is an equilateral triangle sinceall the sides have equal length , namely:√56.1Problem Type 12.1b: Find an equation of the sphere with center C(h, k, l) and radius r.Example Problem 12.1b: Find an equation of the sphere with center (1, 2, −1) and radius 2.Steps Example1. Implement the formula(x − h)2+ (y − k)2+ (z − l)2= r21. In this problem (h, k, l) = (1, 2, −1)and r = 2 so the equation is:(x − 1)2+ (y − 2)2+ (z − (−1))2= 22,which is the same as(x − 1)2+ (y − 2)2+ (z + 1)2= 22.2. Expand everything and move every-thing to the left leaving 0 at the right side.Also rearange terms so that the quadraticterms come before the linear terms.2.x2−2x+1+y2−4y+4+z2+2z+1 = 4 ,Cleaning up:x2+ y2+ z2− 2x − 4y + 2z + 2 = 0 .Ans.: x2+ y2+ z2−2x −4y + 2z + 2 = 0.2Problem Type 12.1c: Show that the equation represents a sphere, and find the center and radius.x2+ y2+ z2+ ax + by + cz + d = 0 .Example Problem 12.1c: Show that the equation represents a sphere, and find the center andradius.x2+ y2+ z2− 2x − 4y + 2z + 2 = 0 .Steps Example1. The coefficients of x2, y2, z2should allbe the same! If they are not, for example,if the equation is x2+ y2+ 3z2+ 2x +6y −5 + 11 = 0 where the coefficients arenot all the same, then it is not a sphere.Usually they are all 1, If the coefficient ofx2(=coeff. of y2=coeff. of z2) is not 1,divide the whole equation by that coeffi-cient, making it 1. The coeffs. of x2, y2,z2should now be all 1. Now group theterms so that x2is next to the x term, y2is next to the y term, and z2is next tothe z term.1. In this problem, the coeffs. of x2is al-ready 1, as are those of y2and z2. Group-ing the x-terms, y-terms and z-terms, weget:x2− 2x + y2− 4y + z2+ 2z + 2 = 02. For each part separately, complete thesquare, using X2+ aX = (X + a/2)2−(a/2)22.x2− 2x = (x − 1)2− 1y2− 4y = (y − 2)2− 4z2+ 2z = (z + 1)2− 1Sox2− 2x + y2− 4y + z2+ 2z + 2 = 0becomes(x−1)2−1+(y−2)2−4+(z+1)2−1+2 = 033. Move all the numbers to the right andexpress the resulting number on the rightas r2. Compare with the equation of thesphere(x − h)2+ (y − k)2+ (z − l)2= r2,and read-off the center (h, k, l) and theradious, r.3.(x−1)2−1+(y−2)2−4+(z+1)2−1+2 = 0is the same as(x − 1)2+ (y − 2)2+ (z + 1)2= 4which, in turn, is the same as(x − 1)2+ (y − 2)2+ (z − (−1))2= 22,which is an equation of a sphere with cen-ter (1, 2, −1) and radius