Answer
Not orthogonal vectors
Work Step by Step
If $ u \cdot v = 0 $ for two nonzero vectors $u=\left\langle a,b\right\rangle$ and $v=\left\langle c,d\right\rangle$, then $\cos \theta = 0 $ means $\theta= 90°$ and the vectors $u$ and $v$ are called orthogonal vectors.
The dot product is given by:
$u\cdot v=\left\langle a,b\right\rangle\cdot\left\langle c,d\right\rangle=ac+bd$
Now for $\sqrt5i-2j=\left\langle \sqrt 5,-2\right\rangle$ and $-5i+2\sqrt 5j=\left\langle -5,2\sqrt 5\right\rangle$ we have
$\left\langle \sqrt 5,-2\right\rangle\cdot\left\langle -5,2\sqrt 5\right\rangle=(\sqrt 5)(-5)+(-2)(2\sqrt5)=-5\sqrt5-4\sqrt5=-9\sqrt5\neq0$.
Since $\left\langle \sqrt 5,-2\right\rangle\cdot\left\langle -5,2\sqrt 5\right\rangle\neq0$, therefore the vectors$\sqrt5i-2j$ and $-5i+2\sqrt 5j$ are not orthogonal vectors (to each other).
