In contrast to the proof in Exercise 81, we now use coordinates and position vectors to prove
Question:
In contrast to the proof in Exercise 81, we now use coordinates and position vectors to prove the same result. Without loss of generality, let P(x1, y1, 0) and Q(x2, y2, 0) be two points in the xy-plane and let R(x3, y3, z3) be a third point, such that P, Q, and R do not lie on a line. Consider ΔPQR.
a. Let M1 be the midpoint of the side PQ. Find the coordinates of M1 and the components of the vector
b. Find the vector 
from the origin to the point Z1 two-thirds of the way along 
c. Repeat the calculation of part (b) with the midpoint M2 of RQ and the vector
to obtain the vector 
d. Repeat the calculation of part (b) with the midpoint M3 of PR and the vector 
to obtain the vector 
e. Conclude that the medians of ΔPQR intersect at a point. Give the coordinates of the point.
f. With P(2, 4, 0), Q(4, 1, 0), and R(6, 3, 4), find the point at which the medians of ΔPQR intersect.
Step by Step Answer:
a The coordinates of M are 2 0 and thus RM 2 13 2 Y3 23 b O O RM 3 Y3 23 1232 YY2y 22 12373 9 ...View the full answer
Calculus Early Transcendentals
ISBN: 978-0321947345
2nd edition
Authors: William L. Briggs, Lyle Cochran, Bernard Gillett
