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

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

To prove that two lines intercept at a point in space, we need to check if there exists a common point that satisfies both vector equations.

a) To determine whether these lines intersect, we can set the two equations equal to each other and isolate the scalar parameters.

r1 = r2
2i + 3j + 2k + ^(5i - 4j + 3k) = 5i + 5j - k + u(i - 3j + 3k)

By comparing the components of the vectors on both sides of the equation, we can set up a system of equations:

2 = 5 + u (equating the coefficients of i)
3 = 5 - 4u (equating the coefficients of j)
2 = -1 + 3u (equating the coefficients of k)

Solving this system of equations will give us the values of u that satisfy the equations.

From the first equation: u = -3
Substituting u = -3 into the second equation: -4(-3) = 5 - 12 => 12 = -7 (not true)
Hence, there is no value of u that satisfies all three equations simultaneously.

Since no solution exists for the system of equations, the lines defined by the vector equations do not intersect at a point in space.

b) Since the lines do not intersect, we cannot find a single position vector for the point of intersection.

c) As the lines do not intersect, the concept of an acute angle between them does not apply.

In conclusion, the lines defined by the vector equations r1 and r2 do not intersect, and therefore, there is no position vector for the point of intersection, nor is there an acute angle between the lines.