(a)
Parallelogram: PACB, where
P(0,2,3) is a vertex.
A=P+PA=(0,2,3)+(3,2,-3)=A(3,4,0)
B=P+PB=(0,2,3)+(4,1,5)=B(4,3,8)
C=A+PB=(3,4,0)+(4,1,5)=C(7,5,5)
(b)
PC=C-P=(7,5,5)-(0,2,3)=(7,3,2)
AB=B-A=(4,3,8)-(3,4,0)=(1,-1,8)
Lengths
PC=sqrt(7²+3²+2²)=sqrt(49+9+4)=√62
I leave you to find the length of AB in a similar way to finding length of PC.
A parallelogram is formed by R^3 by the vectors PA=(3,2,-3) and PB=(4,1,5). The point P=(0,2,3).
a) what are the location of the vertices?
b) what are the vectors representing the diagonals?
c) what are the length of the diagonals?
1 answer
1 answer