Posted by jon smith on .
The vector equations of 2 beams of light r1 and r2 in 3 dimensional spaces referred to a fixed origin 0 are
r1=2i+3j+2k+^(5i4j+3k)
r2=5i+5jk+u(i3j+3k)
where ^ and u are scalar parameters
a)prove that these lines intercept at a point in space
b)Find the position vector for the point at which these beams intercept
c)Find the acute angle between the beams
please help i am struggling with this question

maths 
Reiny,
I don't like your use of ^ in this context.
^ is usually used to show exponents
so
r1 = (2i+3j+2k) + m(5i4j+3k)
or
r1 = (2 + 5m , 3  4m , 2 + 3m)
in the same way
r2 = (5 +u , 5  3u , 1 + 3u)
at their intersection:
2+5m = 5+u and 34m = 53u
5m  3 = u and 3u = 2 + 4m
sub the first into the second
3(5m3) = 2 + 4m
15m  4m = 2 + 9
m = 1
u = 2
check for the 3rd component, that is, is 2+3m = 1 + 3u ?
LS = 2+3m = 2+3=5
RS = 1 + 3u = 1 +6 = 5
yes, they intersect
b) at what point?
r1 = (2,3,2) + 1(5,4,3) = (7,1, 5)
so the position vector to reach that point is
r = 7i  j + 5k
c)the direction vector of the first line is (5,4,3)
and of the second line is (1,3,3)
let Ø be the angle between them, you should know the definition of the dot product .....
(5,4,3)•(1,3,3) = (5,4,3) (1,3,3) cosØ
5 + 12 + 9 = √50√19 cos Ø
cos Ø = 26/√950 = .....
I got Ø = appr. 32.5° or .567 radians