What is the value of 'm' in "i+mj+k" to be unit vector ?

Question

What is the value of 'm' in "i+mj+k" to be unit vector ?

Answer 1

v = (1, m, 1)

for a unit vector you need

||v|| = 1 = √(1² + m² + 1²)

so

m = ±i.................where i = √(-1)

∴ v = (1, i, 1) or (1, -i, 1) are the unit vectors, it would be a bit confusing to write them using i, j, k notation though

.,.,.

Answer 2

?

It cannot be a unit vector because you would require

1 + m^2 + 1 = 1

(that is the vector you gave "dotted" with itself)

so m^2 = -1  and m would have to be imaginary.

ASIDE:  This might be going a bit beyond what you were trying to say, but you can't have negative "lengths" so your question doesn't make sense in that vector spaces with inner products don't produce the phenomenon you require for your "vector" to have a unit length.  

:P  or maybe your question is simply over my head or I'm missing something.

EDIT:  There are other ways of defining length, but you still can't use imaginary numbers like that using the 2-norm to define length (that just means using what you think of as length in everyday type circumstances).  The reason is that it doesn't satisfy the triangle inequality

|| x + y || <= ||x|| + ||y||

A simple way to see that is to take the vectors

v1 = {1,0}  &  v2 = {1,i}

v1.v1 = 1

v2.v2 = 0

(v1+v2).(v1+v2) = {2,i}^2 = 4 - 1 = 3

So if we use imaginary numbers in the way the other answerer is describing we fail to satisfy one of the fundamental properties of the term "length".

[there is also the fundamental property that length needs to be positive which this doesn't mesh with]