Answer
$\theta= 90^{\circ}$
The two vectors are orthogonal.
Work Step by Step
Here, we have $u=(1,-1); v=(1,1)$
$|\overrightarrow{u}|=\sqrt {(1)^2+(-1)^2}=\sqrt {2}$;
$|\overrightarrow{u}|=\sqrt {(1)^2+(1)^2}=\sqrt {2}$;
$\overrightarrow{u} \cdot \overrightarrow{v}=1(1)+(-1)(1)=0$
The angle $\theta$ can be calculated as:
$\cos \theta=\dfrac{\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|}$
This implies that
$\cos \theta=\dfrac{0}{|\sqrt {2}| |\sqrt {2}|}=0$
Thus,
$\theta=\cos^{-1} (0) = 90^{\circ}$
Hence, the two vectors are orthogonal.
