A pt P is between R=(-1,3,2) & Q=(5,2,5) & is twice as far from R as Q. Find the coordinates of P.?

Question

How do u solve this using vector equation? My answer is (11,1,8), but the actual answer is (3,7/3,4). How do u get the actual answer?

Answer 1

Since P is twice as close to Q as R, Q gets twice the weighting of R.

P = (2Q + R)/3 = [2Q(5,2,5) + R(-1,3,2)] / 3

= (2*5-1, 2*2+3, 2*5+2) / 3

= (9, 7, 12) / 3 = (3, 7/3, 4)

Answer 2

yeah...well the solution is pretty simple...
and u do get the result (3, 3/7, 4)
the formula used is called section formula...the formula is
( X2*m + X1*n)/m+n
where X1 is the X co-ordinate of the first point(in this case R), X2 is the x co-ordinate of the scond point(in this case Q)
m, n is the ratio of the dist between the point P from R and Q..(in this case m=2 and n= 1, since point P is twice as far frm R as it is frm Q)
u do the same thing for the Y and Z co-ordinates, m and n will be the same....
let me solve the C point for u so that u understand better
point R( -1,3,2)
Q (5,2,5)
m=2
n=1
therefore
(5*2 + -1*1)/2+1
ull get the point as 3
try it with the other points, ull definitely get the answer

Answer 3

2(5 - (-1))/3 - 1 = 3
2(2 - 3)/3 + 3 = 7/3
2(5 - 2)/3 + 2 = 4
P = (3, 7/3, 4)

Answer 4

wud it not b easier 2 ask yur maths teacher instead of askin a bunch of ppl youve never met dat dnt no de answer????