Answer
\[
d_{2}=3 \text { , } d_{1}=\sqrt{5}
\]
Work Step by Step
Let us consider a parallelogram with adjacent sides
\[
\vec{v}=\hat{i}-2 \hat{\jmath}\quad \text { , } \quad \vec{u}=\hat{\imath}+\hat{\jmath}
\]
Notice that the endpoint of the vector $\vec{v}+\vec{u}$ is a vertex of the parallelogram, and it is the opposite vertex to the vertex at the origin $O(0,0)$. So one diagonal length is
\[
d_{1}=\|\vec{v}+\vec{u}\|=\|2 \hat{\imath}-\hat{\jmath}\|=\sqrt{(-1)^{2}+2^{2}}=\sqrt{5}
\]
The extremes of the second diagonal are the endpoints of the given vectors, so:
\[
d_{2}=\|\vec{u}-\vec{v}\|=\|\quad \hat{\imath}+\hat{\jmath}-(\hat{\imath}-2 \hat{\jmath})\|=\|3 \hat{\jmath}\|=3
\]
So, the length of the diagonals are:
\[
d_{2}=3 \text { , } d_{1}=\sqrt{5}
\]
