1. The problem statement, all variables and given/known data Using vectors, demonstrate that these points are collinear. a) P(15 , 10) , Q(6 , 4) , R(-12 , -8) b) D(33, -5, 20) , E(6, 4, -16) , F(9, 3, -12) 2. Relevant equations [tex]\frac{x_{1}}{x_{2}}[/tex] = [tex]\frac{y_{1}}{y_{2}}[/tex] [tex]\frac{x_{1}}{x_{2}}[/tex] = [tex]\frac{y_{1}}{y_{2}}[/tex] = [tex]\frac{z_{1}}{z_{2}}[/tex] 3. The attempt at a solution a) Vector PQ = (-9 , -6) Vector QR = (-18 , -12) Vector RP = (27 , -18) (-9 / 27 / -18) = (-6 / -18 / -12) Therefore, not collinear. b) Vector DE = (-27, 9, -36) Vector EF = (3, -1, 4) Vector FD = (24, -8, 32) (-27 / 3 / 24) = (9 / - 1 / -8) = (-36 / 4 / 32) Am I correct or did I do something wrong? If I did, can you please point it out and tell me on how to fix it?
Since you don't say what x_{1}, x_{2}, y_{1}, and y_{2} have to do with this problem you "relevant equation" really doesn't make sense. The problem said "Usining vectors". Okay, what is the vector from P(15, 10) to Q(6,4)? What is the vector from P(15, 10) to R(-12, -8)?
well what u need to do is form two vectors like PQ as a vector, and QR, also as a vector. Where P is the starting point wheras Q is the end of the vector for the first one, and similarly for the second. I assume you are working on a cartesian system of coordinates. so now if you manage to show something similar to PQ=k*QR, where k is a constant, than i guess you also have managed to show that those four points are collinear, since they all lie in a line!
I don't follow. . . The question does mention what are x1, x2, y1, y2, (if relevant) z1 and z2. . . And from my attempted solution, I did find the vectors. . .but I don't know what to do after finding the vectors. . .
why don't u just follow the suggestions! for the first part you have PQ(-9,-6) QR(-18,-12) ,now you see that their coordinates are proportional, that is -9/-18=-6/-12=1/2. Just do the same thing with the other!!! Also you can proceede with PR like halls suggested, you will get the same thing.