Let [tex]\vec u[/tex] and [tex]\vec v[/tex] be the vectors, and let [tex]x=\|\vec u\|=\|\vec v\|[/tex] be their common magnitude.
The resultant [tex]\vec u + \vec v[/tex] is [tex]\sqrt 2[/tex] times larger in magnitude than either vector alone, so [tex]\|\vec u+\vec v\| = \sqrt2\,x[/tex].
Recall the dot product identity
[tex]\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)[/tex]
where [tex]\theta[/tex] is the angle between the vectors [tex]\vec a[/tex] and [tex]\vec b[/tex]. In the special case of [tex]\vec a=\vec b[/tex], we get
[tex]\vec a \cdot \vec a = \|\vec a\|^2 \cos(0^\circ) \implies \|\vec a\| = \sqrt{\vec a\cdot\vec a}[/tex]
Now, to get the angle between [tex]\vec u[/tex] and [tex]\vec v[/tex], we have
[tex]\vec u \cdot \vec v = \|\vec u\| \|\vec v\| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec u \cdot \vec v}{x^2}[/tex]
To compute the dot product, we take the dot product of the resultant with itself.
[tex](\vec u+\vec v) \cdot (\vec u + \vec v) = \|\vec u + \vec v\|^2[/tex]
Solve for [tex]\vec u\cdot\vec v[/tex].
[tex](\vec u\cdot\vec u) + 2(\vec u\cdot\vec v) + (\vec v\cdot\vec v) = \|\vec u + \vec v\|^2[/tex]
[tex]\|\vec u\|^2 + 2(\vec u\cdot \vec v) + \|\vec v\|^2 = \|\vec u+\vec v\|^2[/tex]
[tex]x^2 + 2(\vec u \cdot \vec v) + x^2 = (\sqrt2\,x)^2[/tex]
[tex]2(\vec u\cdot\vec v) + 2x^2 = 2x^2[/tex]
[tex]2(\vec u\cdot\vec v) = 0[/tex]
[tex]\vec u\cdot\vec v = 0[/tex]
Since their dot product is zero, [tex]\vec u[/tex] and [tex]\vec v[/tex] are perpendicular, so [tex]\theta=90^\circ[/tex].
