Can a vector have direction angles `45^0,60^0,120^0`

Explanation

We know that, If `l, m, n` are the direction cosine of a vector and `alpha, beta, gamma` are the direction angle, then

`-l=cos alpha, m=cos beta, n=cos gamma`

And, `l^{2}+m^{2}+n^{2}=1 ldots ldots (i)`

therefore `l=cos 45^{circ}, {m}=cos 60^{circ}, {n}=cos 120^{circ}`

`l=frac{1}{sqrt{2}}, m=frac{1}{2}, n=-frac{1}{2}`

Now, substituting `l, m, n` in equation `(i)`, we get -
`(frac{1}{sqrt{2}})^{2}+(frac{1}{2})^{2}+(-frac{1}{2})^{2}=1`

`Rightarrow frac{1}{2}+frac{1}{4}+frac{1}{4}=1`

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