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Vectors: Angle between two vectors
For the three points A( –1, 6, 9), B(0, –1, 17) and C(4, 12, 6),
the cosine of the angle between
->
AB
 and
->
AC
 is
to 3 dps.

tickcross


Worked Solution:

The two vectors
->
AB
 and
->
AC
 are
->
AB
 =
(
1
–7
8
)
->
AC
 =
(
5
6
–3
)
The dot product (scalar product) of two vectors a and b is :
a.b = |a| |b| cos(α) where α is the angle between the vectors, and also
(p i + q j + r k).(t i + u j + v k) = p×t + q×u + r×v
=>cos(α) =
a.b
---
|a| |b|
The two vectors in this question have lengths
|
->
AB
| = 10.67708 and |
->
AC
| = 8.3666
The dot product of the two vectors is 1×5 + (–7)×6 + 8×(–3) = –61
=>cos(α) =
–61
---
10.67708×8.3666
 = –0.683 to 3 dps.
Since the angle and its cosine can be measured clockwise or anticlockwise, the answer 0.683 is also correct.

© MEI Produced by Dr Ron Knott, revised 15 May 2007
tom . button [AT] mei . org . uk 		Test reference: VectorsAng2.html?ref=79076,qu=2pts