I’m doing question 16

I’m alright with it until part c. I can do it but I’m confused with the solution given

I got the values of lambda as per solution but what is Lamda = 1 for C and -1 for D. Could it have been the other way round. That’s the only bit I’m struggling to grasp

Thanks

I’m alright with it until part c. I can do it but I’m confused with the solution given

I got the values of lambda as per solution but what is Lamda = 1 for C and -1 for D. Could it have been the other way round. That’s the only bit I’m struggling to grasp

Thanks

Original post by maggiehodgson

I’m doing question 16

I’m alright with it until part c. I can do it but I’m confused with the solution given

I got the values of lambda as per solution but what is Lamda = 1 for C and -1 for D. Could it have been the other way round. That’s the only bit I’m struggling to grasp

Thanks

I’m alright with it until part c. I can do it but I’m confused with the solution given

I got the values of lambda as per solution but what is Lamda = 1 for C and -1 for D. Could it have been the other way round. That’s the only bit I’m struggling to grasp

Thanks

Whichever way round it is, the midpoint is the same, so just P (lambda=0)

Original post by mqb2766

Whichever way round it is, the midpoint is the same, so just P (lambda=0)

Yes, thanks. It's just that the solution appeared to be specific that Lamda = 1 for C. There was no "it doesn't matter" indicated.

I'm going away from the marking scheme here.

Alternatively, there is a pretty neat geometric argument without needing to find what the positional vectors C and D are.

We look at the triangle ACD - it is isosceles which is given (AC=AD). In order for P to be the mid point of CD, AP should be perpendicular to CD by properties of isosceles triangles. So if we can show AP is indeed perpendicular to CD (how?), we are done.

Alternatively, there is a pretty neat geometric argument without needing to find what the positional vectors C and D are.

We look at the triangle ACD - it is isosceles which is given (AC=AD). In order for P to be the mid point of CD, AP should be perpendicular to CD by properties of isosceles triangles. So if we can show AP is indeed perpendicular to CD (how?), we are done.

(edited 11 months ago)

Original post by vc94

I would have used Mu as the parameter for line 2 since lambda is used for line 1.

So I end up with Mu=1 or Mu=-1 ... these will correspond to the 2 points C and D on line 2 which satisfy AB=AC=AD.

Could write it the other way round.

So I end up with Mu=1 or Mu=-1 ... these will correspond to the 2 points C and D on line 2 which satisfy AB=AC=AD.

Could write it the other way round.

Thanks. I'm happy with that.

Original post by maggiehodgson

Yes, thanks. It's just that the solution appeared to be specific that Lamda = 1 for C. There was no "it doesn't matter" indicated.

Its irrelevant which is chosen to be C or D and you dont need to calculate which is which. Like the other question, the solution method isnt that great. As PA is perpendicular to the direciton vector 1,-1,1, the midpoint of two points will be P irrespective of the distance from A (assuming its large enough).

(edited 11 months ago)

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