Vectors Question.

Hey guys, I need help with this vectors question! Thank you so much!

OAN, OMB, APB and MPN are straight lines and AN=3OA.
M is the midpoint of OB.
OA=a and OB=b
AP=kAB where k is scalar quantity.
Question 1. Express AB and MN in terms of and b.
Question 2. MP in terms of a, b and k.
Question 3. Find the value of k.

I done 1, (AB=b-a and MN=4a-0.5a) and 2 MP=-0.5b+a+k(b-a)

But I'm struggling on finding k.
I attached a picture of the triangles.

Thank you very much.

From another thread, the hint is good so try and work through it without peeking ...

by ghostwalker

I’m sure it doesn’t, I’ll try to post it again anyway
Well it did when I tried to look at it. However your latest post shows it.

So, MPN is a straight line.

This means MP will be a multple of PN (parallel vectors), or of a multiple of MN (either one).

I.e. MP = mPN for some unknown scalar m.

If we can express each of those in terms of a,b,

For them to be equal they must have the same scalar for the vector a on each side, and the same scalar for the vector b on each side (not necessarily the same as a's), since a,b are not parallel.

You now have two equations in two unknowns, k,m. And solve

Reply 2

Towcestermaths

8

Tidy up your answer for MP. That is, write it as so many a's and so many b's. Then work out an expression for PN, written in the same way. You know that PN must be a multiple of MP as they are collinear/parallel. Thus you can write that PN =qMP (where q is just the unknown multiple. Now equate how many a's there are on each side of the equation and how many b's. This will give you two equations with two unknowns which you should be able to solve. Its a nice question.

Original post by NeedHe1pWithMath

Hey guys, I need help with this vectors question! Thank you so much!

OAN, OMB, APB and MPN are straight lines and AN=3OA.
M is the midpoint of OB.
OA=a and OB=b
AP=kAB where k is scalar quantity.
Question 1. Express AB and MN in terms of and b.
Question 2. MP in terms of a, b and k.
Question 3. Find the value of k.

I done 1, (AB=b-a and MN=4a-0.5a) and 2 MP=-0.5b+a+k(b-a)

But I'm struggling on finding k.
I attached a picture of the triangles.

Thank you very much.

Reply 3

NeedHe1pWithMath

OP

2

So i got PN=PA+AN. Hence got PN=-k(b-a)+3a
So then -k(b-a)+3a=q(-0.5b+a+k(b-a))
Then for a's, ak+3a=qa-qak
and for b's, -bk=-0.5qb+qkb
==> k+3=q-qk
==> -k=-0.5q+qk

I've got this so far but I feel like I'm not on the right lines

Thanks guys!

Reply 4

mqb2766

21

Just add those two equations to get a then sub back in to get k?

Original post by NeedHe1pWithMath

So i got PN=PA+AN. Hence got PN=-k(b-a)+3a
So then -k(b-a)+3a=q(-0.5b+a+k(b-a))
Then for a's, ak+3a=qa-qak
and for b's, -bk=-0.5qb+qkb
==> k+3=q-qk
==> -k=-0.5q+qk

I've got this so far but I feel like I'm not on the right lines

Thanks guys!

Reply 5

Towcestermaths

8

You are on the right lines. You now have 2 equations and 2 unknowns so you can solve them simultaneously

Original post by NeedHe1pWithMath

So i got PN=PA+AN. Hence got PN=-k(b-a)+3a
So then -k(b-a)+3a=q(-0.5b+a+k(b-a))
Then for a's, ak+3a=qa-qak
and for b's, -bk=-0.5qb+qkb
==> k+3=q-qk
==> -k=-0.5q+qk

I've got this so far but I feel like I'm not on the right lines

Thanks guys!

OP

Is k=3/7?

Thanks