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P3 vectors question

Relative to 0, OA = i + 3j + 4k
OB = 2i + j + 6k

The point C has coordinates (2,2,2)
The point D has coordinates (4,6,6)

Find to the nearest degree, the acute angle between the directions of the lines AB and CD.

Thanks

Reply 1

Gaz031

sequence123

The line AB has direction b-a=i-2j+2k
The line CD has direction d-c=2i+4j+4k
To find the angle we use the scalar product so AB.CD=[AB][CD]cost where i've let [x] represent the modulus of x.
(i-2j+2k).(2i+4j+4k)=[3][6]cost
cost=(2-8+8)/18=2/18=1/9.