Question Type 2: Sketching a path between two points using two given vectors on a grid Exercises

Can the point $(1,1)$ be reached by integer combinations of $\mathbf{v_1}=(1,-2)$ and $\mathbf{v_2}=(6,-5)$? Justify your answer.

Determine all integer solutions $(a,b)$ such that $a\mathbf{v}_1+b\mathbf{v}_2=\begin{pmatrix} 6 \\ -5 \end{pmatrix}$ with $\mathbf{v}_1=\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\mathbf{v}_2=\begin{pmatrix} 6 \\ -5 \end{pmatrix}$.

Solve for integers $a,b$ such that $a\mathbf{v_1}+b\mathbf{v_2}=(2,3)$ where $\mathbf{v_1}=(1,-2)$ and $\mathbf{v_2}=(6,-5)$.

List all points that can be reached from the origin in exactly 3 forward moves using the vectors $\mathbf{v}_1 = (1, -2)$ and $\mathbf{v}_2 = (6, -5)$.

Find the coordinates of the endpoint after taking 3 steps of vector $\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and 2 steps of vector $\mathbf{v}_2 = \begin{pmatrix} 6 \\ -5 \end{pmatrix}$, starting from the origin.

Is the point $(3,0)$ reachable using forward or backward moves of $\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\mathbf{v}_2 = \begin{pmatrix} 6 \\ -5 \end{pmatrix}$? If yes, find one integer combination; if not, explain why.

Explain why $\mathbf{v}_1=\begin{pmatrix}1\\-2\end{pmatrix}$ and $\mathbf{v}_2=\begin{pmatrix}6\\-5\end{pmatrix}$ form a basis for $\mathbb{R}^2$ by computing the determinant of the matrix with columns $\mathbf{v}_1,\mathbf{v}_2$.

Determine whether it is possible to reach the point $(2, 3)$ using exactly 4 forward steps of $\mathbf{v}_1$ or $\mathbf{v}_2$. If a sequence of moves exists, find it; otherwise, explain why it is impossible.

Determine whether it is possible to reach the point $(2,3)$ using exactly $5$ moves. If it is possible, state the number of forward or backward moves along each vector.

What is the minimum number of moves (allowing forward or backward steps of $\mathbf{v}_1$ and $\mathbf{v}_2$) needed to reach the point $(2,3)$ from the origin? Provide the sequence of moves.

If you take 2 forward steps of $\mathbf{v_1} = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and then 3 backward steps of $\mathbf{v_2} = \begin{pmatrix} 6 \\ -5 \end{pmatrix}$, what point do you reach?

If you first take one step of $\mathbf{v}_2 = \begin{pmatrix} 6 \\ -5 \end{pmatrix}$, then backtrack along $\mathbf{v}_1 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$ $k$ times to reach the point $(2, 3)$, find the value of $k$.