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Vectors: Angle between two vectors
For the three points A(3, 0, 9), B( –6, –1, 16) and C(10, –7, 1),
the cosine of the angle between
->
AB
 and
->
AC
 is
to 3 dps.

tickcross


Worked Solution:

The two vectors
->
AB
 and
->
AC
 are
->
AB
 =
(
–9
–1
7
)
->
AC
 =
(
7
–7
–8
)
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
| = 11.44552 and |
->
AC
| = 12.72792
The dot product of the two vectors is (–9)×7 + (–1)×(–7) + 7×(–8) = –112
=>cos(α) =
–112
---
11.44552×12.72792
 = –0.769 to 3 dps.
Since the angle and its cosine can be measured clockwise or anticlockwise, the answer 0.769 is also correct.

© MEI Produced by Dr Ron Knott, revised 15 May 2007
tom . button [AT] mei . org . uk 		Test reference: VectorsAng2.html?ref=8146