Question

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+^(5i-4j+3k)
r2=5i+5j-k+u(i-3j+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

Answer 1

I don't like your use of ^ in this context.
^ is usually used to show exponents

so
r1 = (2i+3j+2k) + m(5i-4j+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 3-4m = 5-3u
5m - 3 = u and 3u = 2 + 4m
sub the first into the second
3(5m-3) = 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