91Ó°ÊÓ

To solve this problem, you should recall that in a parallelogram, opposite sides are parallel and equal in length. Therefore, as the sides of the parallelogram are determined by the given vectors \(\mathbf{u}\) and \(\mathbf{v}\), the opposite sides are also represented by the same vectors.

Let the vertices of the parallelogram be \(A\), \(B\), \(C\), and \(D\). Without loss of generality, let \(\mathbf{u}\) be the vector from vertex \(A\) to vertex \(B\) and let \(\mathbf{v}\) be the vector from vertex \(A\) to vertex \(C\). Then the opposite vertices are connected by the vectors \(\mathbf{AB}=\mathbf{u}\), \(\mathbf{BC}=\mathbf{v}\), \(\mathbf{CD}=\mathbf{u}\), and \(\mathbf{DA}=\mathbf{v}\).

The diagonals of the parallelogram can be now computed using vector addition. The diagonal \(\mathbf{AC}\) connects vertices \(A\) and \(C\). To find this diagonal, we start at vertex \(A\), follow the vector \(\mathbf{u}\) to get to vertex \(B\), and then follow the vector \(\mathbf{v}\) from vertex \(B\) to vertex \(C\). In terms of vector addition, this can be written as \(\mathbf{AC}=\mathbf{u}+\mathbf{v}\). Similarly, the diagonal \(\mathbf{BD}\) connects vertices \(B\) and \(D\). To find this diagonal, we start at vertex \(B\), follow the vector \(\mathbf{u}\) in the opposite direction (which gives us \(-\mathbf{u}\)) to get to vertex \(D\), and then follow the vector \(\mathbf{v}\) from vertex \(D\) to vertex \(B\). In terms of vector addition, this can be written as \(\mathbf{BD}=-\mathbf{u}+\mathbf{v}\) which is equivalent to \(\mathbf{BD}=\mathbf{v}-\mathbf{u}\).

Based on our computations, we have shown that the diagonals of the parallelogram are \(\mathbf{AC}=\mathbf{u}+\mathbf{v}\) and \(\mathbf{BD}=\mathbf{v}-\mathbf{u}\). Since this was the desired result, the proof is complete.