Points A and B have position vectors a and b, respectively, relative to an origin O, and are such that OAB is a triangle with OA = a and OB = b.

the point C, with position vector c, lies on the line through O that bisects the angle AOB.

1.) prove that the vector ba-ab is perpendicular to c.

the point D, with position vector d, lies on the line AB between A and B.

2.) explain why d can be expressed in the form d=(1-λ)a+λb for some scalar λ with 0<λ<1

3.) given that D is also on the line OC, find an expression for λ in terms of a and b only and hence show that

DAB = OA:OB

the point C, with position vector c, lies on the line through O that bisects the angle AOB.

1.) prove that the vector ba-ab is perpendicular to c.

the point D, with position vector d, lies on the line AB between A and B.

2.) explain why d can be expressed in the form d=(1-λ)a+λb for some scalar λ with 0<λ<1

3.) given that D is also on the line OC, find an expression for λ in terms of a and b only and hence show that

DAB = OA:OB

Original post by shobby_999

Points A and B have position vectors a and b, respectively, relative to an origin O, and are such that OAB is a triangle with OA = a and OB = b.

the point C, with position vector c, lies on the line through O that bisects the angle AOB.

1.) prove that the vector ba-ab is perpendicular to c.

the point D, with position vector d, lies on the line AB between A and B.

2.) explain why d can be expressed in the form d=(1-λ)a+λb for some scalar λ with 0<λ<1

3.) given that D is also on the line OC, find an expression for λ in terms of a and b only and hence show that

DAB = OA:OB

the point C, with position vector c, lies on the line through O that bisects the angle AOB.

1.) prove that the vector ba-ab is perpendicular to c.

the point D, with position vector d, lies on the line AB between A and B.

2.) explain why d can be expressed in the form d=(1-λ)a+λb for some scalar λ with 0<λ<1

3.) given that D is also on the line OC, find an expression for λ in terms of a and b only and hence show that

DAB = OA:OB

What have you tried and can you upload a pic of the question? Not sure about the ba-ab in part 1)?

Edit - its question 5 in

https://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2018/exam-materials/9811_01_que_20190815.pdf

So what have you tried, the first two parts should be relatively straightforward?

(edited 2 months ago)

Original post by mqb2766

What have you tried and can you upload a pic of the question? Not sure about the ba-ab in part 1)?

Edit - its question 5 in

https://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2018/exam-materials/9811_01_que_20190815.pdf

So what have you tried, the first two parts should be relatively straightforward?

Edit - its question 5 in

https://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2018/exam-materials/9811_01_que_20190815.pdf

So what have you tried, the first two parts should be relatively straightforward?

thank you i was able to solve it, i just had a bit of difficulty recognising that it would be an isosceles triangle, and if you add 2 points such that the position of the vector from the origin was ba for one and ab for the other it is quite simple. i was also told it can be solved using dot products but not entirely sure on the method for that

Original post by mqb2766

What have you tried and can you upload a pic of the question? Not sure about the ba-ab in part 1)?

Edit - its question 5 in

https://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2018/exam-materials/9811_01_que_20190815.pdf

So what have you tried, the first two parts should be relatively straightforward?

Edit - its question 5 in

https://qualifications.pearson.com/content/dam/pdf/Advanced%20Extension%20Award/Mathematics/2018/exam-materials/9811_01_que_20190815.pdf

So what have you tried, the first two parts should be relatively straightforward?

thanks

Original post by shobby_999

thank you i was able to solve it, i just had a bit of difficulty recognising that it would be an isosceles triangle, and if you add 2 points such that the position of the vector from the origin was ba for one and ab for the other it is quite simple. i was also told it can be solved using dot products but not entirely sure on the method for that

The isosceles solution is the geometric way to do it, but I suspect they would expect the dot product. So two vectors are orthogonal if their dot product is zero so

c.(ab-ba) = ac.b - bc.a = abc*cos(theta) - abc*cos(theta) = 0

so its a "write down" solution rather than having to construct the isosceles triangle, though the latter is a more elementary method (gcse). I suspect person setting the question adapted the basic geometry to vector lenguage.

For the last part about the ratio, you "can" also do a geometic argument as its the angle bisector theorem

https://en.wikipedia.org/wiki/Angle_bisector_theorem

and there are several proofs, but I like the one where you construct a(nother) isosceles triangle. Again though, theyd want a vector solution.

(edited 2 months ago)

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