## 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.

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### List of participants with their entries:

Name | solution found | comment |
---|---|---|

@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 |

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## Winner draw:

Not necessary since exactly 2 successful participant:

Congratulations @jeffjagoe and @lallo , you won 1 SBI each!

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