Math Contest #25 Results and Solution
The problem of this contest was to find a and x so that these vectors would be orthogonal:
(a, 5a, 2a/x) and (3a, 4a, x)
being orthogonal means that the scalar product gives 0:
a*3a + 5a*4a + 2a/x *x = 0
3a² + 20a² + 2a = 0
(23a + 2)a = 0
a = 0 or
23a+2 = 0 →
a = -2/23
Is a=0 really a solution?
After all it gives the 0-vector which should have no direction and therefor shouldn't be orthogonal to anything?
It depends on how you define orthogonality.
I didn't define it, so I guess I have to count a=0 as a solution.
One important thing is still missing: the values of x:
x can be anything, because the solution does not depend on it, but it mustn't be 0 because 2a/0 is not defined!
So x != 0.
List of participants with their entries:
|@jeffjagoe||a = 0||You forgot to define x and you didn't find the second solution :(|
|@lallo||a = 0 or a = -2/23, x != 0|