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A vector is drawn on a grid

Please help A vector is drawn on a grid-example-1
User Gurhan
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Answer:

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Explanation:

I cant see the grid

User Nic Szerman
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3 votes

The vector on the grid, from (2, 1) to (8, 5), is expressed as
\( \mathbf{v} = 2\mathbf{a} + \mathbf{b} \) using vector decomposition with given vectors
\( \mathbf{a} \) and \( \mathbf{b} \).

To express the vector drawn on the grid in terms of vectors
\( \mathbf{a} \) and
\( \mathbf{b} \), we can utilize the given vectors
\( \mathbf{a} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} \) and
\( \mathbf{b} = \begin{bmatrix} 4 \\ -2 \end{bmatrix} \). The vector on the grid starts from (2, 1) and ends at (8, 5). The displacement between these points represents the desired vector.

The displacement vector
\( \mathbf{v} \) can be obtained by subtracting the initial point from the terminal point:


\[ \mathbf{v} = \begin{bmatrix} 8 - 2 \\ 5 - 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 4 \end{bmatrix} \]

Now, to express
\( \mathbf{v} \) in terms of
\( \mathbf{a} \) and \( \mathbf{b} \), we can use vector decomposition. We want to find scalars
\( \alpha \) and
\( \beta \) such that
\( \mathbf{v} = \alpha \mathbf{a} + \beta \mathbf{b} \).


\[ \begin{bmatrix} 6 \\ 4 \end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 3 \end{bmatrix} + \beta \begin{bmatrix} 4 \\ -2 \end{bmatrix} \]

Solving for
\( \alpha \) and
\( \beta \), we find
\( \alpha = 2 \) and \( \beta = 1 \). Therefore, the vector
\( \mathbf{v} \) can be expressed as:


\[ \mathbf{v} = 2\mathbf{a} + \mathbf{b} \]

In conclusion, the vector drawn on the grid can be expressed as twice vector
\( \mathbf{a} \) plus vector
\( \mathbf{b} \).

User Brian Rothstein
by
8.9k points

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