Original post by Dm998

Question lifted from past edexcel paper .

OAN OMB and APB are straight lines. M is the midpoint of OB.

OA=a. Vector

OB=b. Vector.

AP(vector)=k x AB(vector) where k is a scalar quantity.

Given that MPN is a straight line find the value of k

Thanks

OAN OMB and APB are straight lines. M is the midpoint of OB.

OA=a. Vector

OB=b. Vector.

AP(vector)=k x AB(vector) where k is a scalar quantity.

Given that MPN is a straight line find the value of k

Thanks

I can't do this... Have you got the question down completely? Could you post a picture of it?

Also, you have to post what you've worked out/tried so far before people can help you.

Original post by Dm998

I'll add the question and working out if someone can tell me how to add an attachment

There are insert image and attachment buttons above the box you type in when you post.

Original post by ada_786

The answer is 2/5 but I keep getting -5/2

How are you tackling it?

Original post by Dm998

Not on my screen

https://imgur.com/b7PDxSM

This is the question? (you left out AN=2OA in your original post).

This questions has already been answered in a different thread https://www.thestudentroom.co.uk/showthread2.php?t=5262824

You can thank @ghostwalker

You can thank @ghostwalker

AB=b-a

MN=-0.5b+a+3a

MN=4a-0.5b

MP=-0.5b+a+k(b-a)

MP=-0.5b+a+kb-ka

4/1-k=-0.5/-0.5+k (number of As in MN divided by As in MP = no. Bs in MN/Bs in MP)

8/2-2k=-1/-1+2k

8(-1+2k)=-1(2-2k)

-8+16k=-2+2k

14k=6

k=6/14

k=3/7

Hope that helps

MN=-0.5b+a+3a

MN=4a-0.5b

MP=-0.5b+a+k(b-a)

MP=-0.5b+a+kb-ka

4/1-k=-0.5/-0.5+k (number of As in MN divided by As in MP = no. Bs in MN/Bs in MP)

8/2-2k=-1/-1+2k

8(-1+2k)=-1(2-2k)

-8+16k=-2+2k

14k=6

k=6/14

k=3/7

Hope that helps

(edited 5 years ago)

AB = AO+OB

AB = -a+b

MN = MO+OA+AN

MN = -0.5b+a+2a

MN = 3a-0.5b

MP = MO+OA+AP

MP = -0.5B+a-ka+kb

NM = -3a+0.5b

NP = -2a+k(-a+b)

NPM is a straight line, so,

x × NP = NM

k = 2/5

Little late but correct

AB = -a+b

MN = MO+OA+AN

MN = -0.5b+a+2a

MN = 3a-0.5b

MP = MO+OA+AP

MP = -0.5B+a-ka+kb

NM = -3a+0.5b

NP = -2a+k(-a+b)

NPM is a straight line, so,

x × NP = NM

k = 2/5

Little late but correct

(edited 4 years ago)

Original post by Pro student

AB = AO+OB

AB = -a+b

MN = MO+OA+AN

MN = -0.5b+a+2a

MN = 3a-0.5b

MP = MO+OA+AP

MP = -0.5B+a-ka+kb

NM = -3a+0.5b

NP = -2a+k(-a+b)

NPM is a straight line, so,

x × NP = NM

k = 2/5

Little late but correct

AB = -a+b

MN = MO+OA+AN

MN = -0.5b+a+2a

MN = 3a-0.5b

MP = MO+OA+AP

MP = -0.5B+a-ka+kb

NM = -3a+0.5b

NP = -2a+k(-a+b)

NPM is a straight line, so,

x × NP = NM

k = 2/5

Little late but correct

Can you please explain the last step ?Exactly how did you get the 2/5 . I do get other steps though

Original post by Dm998

Question lifted from past edexcel paper .

OAN OMB and APB are straight lines. M is the midpoint of OB.

OA=a. Vector

OB=b. Vector.

AP(vector)=k x AB(vector) where k is a scalar quantity.

Given that MPN is a straight line find the value of k

Thanks

OAN OMB and APB are straight lines. M is the midpoint of OB.

OA=a. Vector

OB=b. Vector.

AP(vector)=k x AB(vector) where k is a scalar quantity.

Given that MPN is a straight line find the value of k

Thanks

-3a+1/2b = x(-2+k(a-b))

-3a+1/2b = -2xa-kxa+kxb

inspect a and b:

2x+kx=3

2(1/2k)+k(1/2k)=3

1/k+1/2=3

1/k+1/2=3

k=2/5

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