ZPQM is a parallelogram, express Vector OZ in terms of u and v

[Richie Smash]

Homework Statement

u and v are two vectors in the same plane.

Vector OM = u+2v
Vector OP = 6u+v
Vector OQ = 5u +2v
Vector OR =2(VecOM) +v

Given that ZPQM is a parallelogram, express Vector OZ in terms of u and v.

Homework Equations

The Attempt at a Solution

First they wanted me to find Vectors QP and PR in terms of u and v

I've done that so QP=(vector)OP -(vector)OQ= u-v

and vector PR = Vector OR- Vector OP = 4v-4u

SO now I'm completely stuck as to show OZ, I drew out a rough sketch guessing where the points may be in relation to the origin, and it sort of looks like OM is a continuation of OZ... but I am just guessing here I really need help with this difficult question.

I know OM is u+2v, so according to my rough sketch, it must be a multiple of OZ... if they lie on the same line.

I have also worked out vector QM to be Vector OM -Vector OQ= -4u

 

[andrewkirk]

Vectors QP and PR are not needed to solve this. What is needed is to express OZ as a sequence of vectors placed in sequence head to tail, that starts at O and ends at Z. The first vector can be one that goes from O to any of Z, P or Q. The remaining vectors in the sequence are edges of the parallelogram that lead from the head of that first vector, ultimately, to Z. You will need to use the fact that the final vector in the sequence, the one with head at Z, is parallel to one of the other parallelogram-edge vectors.

 

[Richie Smash]

Hmm I'm sort of following you sir but they don't give any value for OZ, so how can we start from O and go straight to Z? What you're trying to say it might something like Vec(OP)+Vec(PM)+Vec(QZ)+Vec(OZ)?

 

[andrewkirk]

Vec(OP)+Vec(PM)+Vec(QZ)+Vec(OZ)

That's the general idea. Your first vector gets you from O to the parallelogram. then you want to follow a path in the 'gram to Z. However

(1) since the parallelogram has been described as ZPQM, PM is a diagonal, not an edge of the shape. Go around the edges to get to Z
(2) write your sum as a head-to-tail sequence, so that the head of each vec is the tail of the next one.
(2) The last term OZ doesn't fit. The whole sum is supposed to be equivalent to OZ, ie a trail that leads from O to Z. So we don't want to put OZ in there. O should appear only in the first vec in the sequence, to get us away from the origin. We don't ever want to return to O.

 

[Richie Smash]

Oh ok, So I did OP + PQ+QM+MZ

Thats basically (6u+v)+(v-q)+(-4u)+MZ

= u+2v + Vec(MZ)

I'm getting closer but I don't have a value for MZ...

Wait a second, vector MZ is parallel to vector QP which is u-v so MZ= u-v

Now just add the two final products,(u+2v)+(u-v)= 2u+v

Therefore Vector OZ = 2u+v

 

[Richie Smash]

hello?

 

[andrewkirk]

Oh ok, So I did OP + PQ+QM+MZ

Thats basically (6u+v)+(v-q)+(-4u)+MZ

= u+2v + Vec(MZ)

I'm getting closer but I don't have a value for MZ...

Wait a second, vector MZ is parallel to vector QP which is u-v so MZ= u-v

Now just add the two final products,(u+2v)+(u-v)= 2u+v

Therefore Vector OZ = 2u+v

Yes that is correct. On reflection, it's simpler just to go direct from O to M and then from M to Z using the parallel fact you used above. But the answer is of course the same.