A-level maths question help (vectors)?
The points A, B and C have position vectors i + 3j + k, 2i + 7j - 3k and 4i - 5j +2k respectively.
a) Find, as surds the lengths of AB and BC.
b) Calculate, in degrees, to 1 decimal place, the size of angle ABC.
So, I could do part a) easily, worked out AB to be (root)33 and BC to be (root)173. These answers are both correct. I just can't seem to do part b), I tried various methods but the answers I got were never correct.
thanks
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Well,
we have:
i + 3j + k, 2i + 7j - 3k and 4i - 5j +2k
so the coordinates of the points are :
A(1,3,1) , B(2,7, -3) , C(4, -5, 2)
then :
AB = <1 , 4, -4> ----> ||AB||^2 = 1+4^2+(-4)^2 = 33 ---> ||AB|| = sqrt33
BC = <2, -12, 5> ---> ||BC||^2 = 2^2+(-12)^2+5^2=173 ---> ||BC|| = sqrt173
therefore : the angle is given by its sine using the vectorial product:
u = BA x BC = - AB x BC = < -28, -13, -20>
||u||^2 = 1353 # 36.783^2
and so
sin(ABC) # 36.783 / (sqrt33 . sqrt173) =# 0.486811
finally :
angle(ABC) # 29.13°
hope it' ll help !!
Yes, your answers to part a) are correct.
b) cos(angle ABC) = (vector BA) dot (vector BC) / (BA*BC)
= (-i - 4j + 4k) dot (2i - 12j + 5k) / [sqrt(33)*sqrt(173)]
= [(-1)(2) + (-4)(-12) + (4)(5)] / [sqrt(33)*sqrt(173)]
= 66 / [sqrt(33)*sqrt(173)].
m(angle ABC) = cos^-1 {66 / [sqrt(33)*sqrt(173)]}
= 29.1 degrees.
Alternatively, we can use the law of cosines instead of dot products.
AB = sqrt(33), BC = sqrt(173), AC = sqrt(74).
(AC)^2 = (AB)^2 + (BC)^2 - 2AB*BC cos(angle ABC)
cos(angle ABC) = [(AB)^2 + (BC)^2 - (AC)^2]/(2AB*BC)
= (33 + 173 - 74)/[2*sqrt(33)*sqrt(173)]
= 66 / [sqrt(33)*sqrt(173)].
Once again,
m(angle ABC) = cos^-1 {66 / [sqrt(33)*sqrt(173)]}
= 29.1 degrees.
Have a blessed, wonderful day!