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). What are the location of the vertices?what are the vectors representing the diagonals? what are...

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).

What are the location of the vertices?

what are the vectors representing the diagonals?

what are the length of the diagonals?

Asked on by bobby455

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labrat256 | (Level 2) Adjunct Educator

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a) Vertexes, in this situation, can be said to mean 'corners'. Thus, you have a parallelogram (a 4 sided shape with 2 pairs of parallel sides) with 4 corners where the parallelogram is described by two of those sides (PA and PB) and a starting point (P).

The first vertex must be at the starting point, P. The first vertex is therefore at (0,2,3). The second vertex will be at the corner A which occurs after the vector PA away from P, so P+PA is the second vertex. 

A=PA+P=(3,4,0)

Another vertex, B, will be at the vector PB away from P, so the third vertex will be at

B=PB+P=(4,3,8)

The fourth and final vertex can be found by applying vector PA to point B or equivalently vector PB to point A. Lets call this point C.

C=P+PA+PB=(3,4,0)+(4,1,5)=(4,3,8)+(3,2,-3)=(7,5,5)

Therefore, the vertexes are at (0,2,3), (3,4,0), (4,3,8) and (7,5,5).

b) The diagonals are the lines which go diagonally across the parallelogram. These are therefore the lines PC and AB.

We know what both P and C are, thus we make a vector going from one to the other.

P+PC=C

PC=C-P

PC=(7,5,5)-(0,2,3)=(7,3,2)

The same goes for AB. We know where A is, we know where B is, so to know AB we make a vector going from A to B.

A+AB=B

AB=B-A

AB=(4,3,8)-(3,4,0)=(1,1,8)

We know now the diagonal vectors AB and PC.

c) The length of these vectors, and hence the diagonals themselves, can be calculated using Pythagoras' Theorem. If the length of a 3d vector is L and the lengths of the x,y and z components are x,y and z, then Pythagoras' Theorem in 3d is

L^2=x^2+y^2+z^2

The length of AB is 

Length(AB)=sqrt(1^2+1^2+8^2)=sqrt(66) =sqrt(11*3*2)=sqrt(11)*sqrt(3)*sqrt(2)

The length of PC is

Length(PC)=sqrt(7^2+3^2+2^2)=sqrt(49+9+4)=sqrt(62)=sqrt(2*31)=sqrt(2)*sqrt(31)

 

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