Vectors q (1 Viewer)

[Bingxilin John Xena]

How do I do this one?

 

[Hughmaster]

View attachment 42928
How do I do this one?

We have that OP = 2/3 OA and suppose that PQ = k PC.

Then OQ = OP + PQ = 2/3 OA + k PC = 2/3 OA + k (OC - OP) = 2/3 OA + k (OC - 2/3 OA).

But OQ =lambda OB = lambda (OA + OC).

So you can now equate the coefficients of OA and OC in :

lambda (OA + OC) = 2/3 OA + k (OC - 2/3 OA).

Can supply some proper working if needed, but that should be enough for you to digest and have another crack at solving?

 

[Bingxilin John Xena]

We have that OP = 2/3 OA and suppose that PQ = k PC.

Then OQ = OP + PQ = 2/3 OA + k PC = 2/3 OA + k (OC - OP) = 2/3 OA + k (OC - 2/3 OA).

But OQ =lambda OB = lambda (OA + OC).

So you can now equate the coefficients of OA and OC in :

lambda (OA + OC) = 2/3 OA + k (OC - 2/3 OA).

Can supply some proper working if needed, but that should be enough for you to digest and have another crack at solving?

ty ty

 

[Bingxilin John Xena]

View attachment 42928
How do I do this one?

Actually can someone finish this?

 

[KloppsAndRobbers]

Actually can someone finish this?

Finished it. Just making sure is the answer lambda = 2/5? If it is I'll send you the working.

 

[jimmysmith560]

Finished it. Just making sure is the answer lambda = 2/5? If it is I'll send you the working.

Your answer is correct.

 

[KloppsAndRobbers]

Actually can someone finish this?

I just found 2 ways to express OQ & then equated it to find Lambda. My working is a bit different to Hughmaster.

PQ = kPC
OQ - OP = k(OC - OP) [Note: Head minus tail - both sides]
= kOC - kOP + OP
= kOC - OP(k-1)
Therefore, OQ = kOC - 2/3OA(k-1) [Note: OP = 2/3OA, from |OA|:|PA| = 2:1]

OQ = lambda OB
= lambda (OA+AB)
= lambda (OA+OC) [Note: AB = OC]
Therefore, OQ = lambdaOA + lambdaOC

OQ = OQ
lambdaOA + lambdaOC = kOC - 2/3OA(k-1)
lambdaOA + lambdaOC = kOC + OA (-2/3k+2/3)

Equating:
1)
lambdaOC = kOC
Therefore, lambda = k.

2)
lambdaOA = OA(-2/3k+2/3)
lambda = -2/3k + 2/3
lambda = -2/3lambda + 2/3 [Note: lambda = k from previous equation]
5/3lambda = 2/3
Therefore, lambda = 2/5

 

[ivanradoszyce]

I just found 2 ways to express OQ & then equated it to find Lambda. My working is a bit different to Hughmaster.

PQ = kPC
OQ - OP = k(OC - OP) [Note: Head minus tail - both sides]
= kOC - kOP + OP
= kOC - OP(k-1)
Therefore, OQ = kOC - 2/3OA(k-1) [Note: OP = 2/3OA, from |OA|:|PA| = 2:1]

OQ = lambda OB
= lambda (OA+AB)
= lambda (OA+OC) [Note: AB = OC]
Therefore, OQ = lambdaOA + lambdaOC

OQ = OQ
lambdaOA + lambdaOC = kOC - 2/3OA(k-1)
lambdaOA + lambdaOC = kOC + OA (-2/3k+2/3)

Equating:
1)
lambdaOC = kOC
Therefore, lambda = k.

2)
lambdaOA = OA(-2/3k+2/3)
lambda = -2/3k + 2/3
lambda = -2/3lambda + 2/3 [Note: lambda = k from previous equation]
5/3lambda = 2/3
Therefore, lambda = 2/5

Yes, that's the way I did it.

 

[Bingxilin John Xena]

I just found 2 ways to express OQ & then equated it to find Lambda. My working is a bit different to Hughmaster.

PQ = kPC
OQ - OP = k(OC - OP) [Note: Head minus tail - both sides]
= kOC - kOP + OP
= kOC - OP(k-1)
Therefore, OQ = kOC - 2/3OA(k-1) [Note: OP = 2/3OA, from |OA|:|PA| = 2:1]

OQ = lambda OB
= lambda (OA+AB)
= lambda (OA+OC) [Note: AB = OC]
Therefore, OQ = lambdaOA + lambdaOC

OQ = OQ
lambdaOA + lambdaOC = kOC - 2/3OA(k-1)
lambdaOA + lambdaOC = kOC + OA (-2/3k+2/3)

Equating:
1)
lambdaOC = kOC
Therefore, lambda = k.

2)
lambdaOA = OA(-2/3k+2/3)
lambda = -2/3k + 2/3
lambda = -2/3lambda + 2/3 [Note: lambda = k from previous equation]
5/3lambda = 2/3
Therefore, lambda = 2/5

tysm!!

 

[ivanradoszyce]

Here is the solution written in Latex.

 

[ivanradoszyce]

View attachment 42928
How do I do this one?

As a matter of interest, where does this question come from?

 

[Bingxilin John Xena]

As a matter of interest, where does this question come from?

My school’s question bank

 

[jimmysmith560]

As a matter of interest, where does this question come from?

This question also appeared in the 2023 Baulkham Hills High School Mathematics Extension 1 trial paper.