For the pair of vectors, find $\mathbf{u}+\mathbf{v}, \mathbf{u}-\mathbf{v}$, and $2 \mathbf{u}-3 \mathbf{v}$.
\[
\mathbf{u}=\langle 0,-2\rangle, \mathbf{v}=\langle-2,0\rangle
\]
Hint: Use "< 1, $2> $ " for $\langle 1,2\rangle$.
\[
\begin{array}{l}
u+v= \\
u-v= \\
2 u-3 v=
\end{array}
\]
Answer
Final Answer: \(\mathbf{u}+\mathbf{v}=\boxed{\langle -2,-2 \rangle}\), \(\mathbf{u}-\mathbf{v}=\boxed{\langle 2,-2 \rangle}\), \(2\mathbf{u}-3\mathbf{v}=\boxed{\langle 6,-4 \rangle}\)
Step 1 ：Given vectors \(\mathbf{u} = \langle 0,-2 \rangle\) and \(\mathbf{v} = \langle -2,0 \rangle\)
Step 2 ：To find the sum \(\mathbf{u}+\mathbf{v}\), we add the corresponding components of the vectors together, which gives us \(\langle 0+(-2), -2+0 \rangle = \langle -2,-2 \rangle\)
Step 3 ：To find the difference \(\mathbf{u}-\mathbf{v}\), we subtract the components of \(\mathbf{v}\) from the corresponding components of \(\mathbf{u}\), which gives us \(\langle 0-(-2), -2-0 \rangle = \langle 2,-2 \rangle\)
Step 4 ：To find the linear combination \(2\mathbf{u}-3\mathbf{v}\), we multiply each vector by its corresponding scalar, then subtract the second result from the first. This gives us \(\langle 2*0-3*(-2), 2*(-2)-3*0 \rangle = \langle 6,-4 \rangle\)
Step 5 ：Final Answer: \(\mathbf{u}+\mathbf{v}=\boxed{\langle -2,-2 \rangle}\), \(\mathbf{u}-\mathbf{v}=\boxed{\langle 2,-2 \rangle}\), \(2\mathbf{u}-3\mathbf{v}=\boxed{\langle 6,-4 \rangle}\)
