# C4 vectorsWatch

#1
can anyone help me???

the points A, B and C have position vectors i + 3j + k , 2i + 7j - 3k and 4i - 5j + 2k respectively.

lengths of AB is √33 and length of BC is √173

calculate, in degrees to 1dp, the size of angle ABC.

i know the answer is 29.1 but i cant get to it-any ideas?
0
13 years ago
#2
Are you sure thats the right answer?

I may be wrong but I think you need to use the cosine rule a2 =b2 +c2 -2bccosA
I've done it and worked the angle out as being 24.7 to 1 dp using this
0
13 years ago
#3
AB = B-A = (1, 4, -4)
CB = B-C = (-2, 12, -5)
|AB| = sqrt(33)
|CB| = sqrt(173)

AB . CB = |AB||CB| cos(ABC)
-2 + 48 + 20 = sqrt(5709) cos(ABC)
=> cos(ABC) = 66/sqrt(5709)
=> ABC =~ 29.1 (1dp)
0
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