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Vectors Help (1 Viewer)

rickypontingdagoat

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If M is the midpoint of the side BC of triangle ABC. Let vector AB =a and vector AC = b

a ) Find vectors AM and CM in terms of a and b

i got
AM = (3b+a)/2 , CM=(a+b)/2

part b)

Show that |AM|^2 +|CM|^2 = 1/2 (|AB|^2 + |AC|^2)

Need help with part b, I tried to sub in the vectors into the expression but I couldn't get it.
 

liamkk112

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i would say to use the fact that for a vector v, |v|^2 is the dot product of v with itself, should fall apart then
 

Luukas.2

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I agree with @liamkk112, use the fact that the magnitude of a vector, squared, is equal to its dot product.

This question is a variation on a standard question covered in the syllabus, that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its four sides. The syllabus result is one that it is wise for students to be familiar with, as that result and variants on it (like this one) come up reasonably often.

Another such result is that the midpoint of the hypotenuse of a right angled triangle is equally distance from all three vertices, FYI. A corollary of this result is that the centre of the circumcircle of a right-angled triangle is the midpoint of the hypotenuse. Vectors can be used to prove the converse, that if the longest side of a triangle is also the diameter of its circumcircle, then the triangle is right-angled.
 
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rickypontingdagoat

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Thanks for the replies, however i still cant seem to figure it out, after replacing it with the dot product and expanding I am left with 1/2(5b^2+a^2+4ab) at which point I am not sure where else to go from there
 

liamkk112

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Thanks for the replies, however i still cant seem to figure it out, after replacing it with the dot product and expanding I am left with 1/2(5b^2+a^2+4ab) at which point I am not sure where else to go from there
i just realised u got the result for AM and CM wrong;

notice CB = AB - AC = a-b
then CM = 1/2 CB = (a-b)/2
and AM = AC + CM = b + (a-b)/2 = (a+b)/2

now the dot product property should make sense
 

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