Exercise 1.1.14

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Suppose the twelve vectors start from $6: 00$ at the bottom instead of $(0,0)$ at the center. The vector to $12: 00$ is doubled to $(0,2) .$ The new twelve vectors add to Problems 15-19 go further with linear combinations of $v$ and $w$ (Figure 1.5a).

Prakash Anand · Accepted Answer

Moving the origin to $6: 00$ adds $j=(0,1)$ to every vector. So the sum of twelve vectors changes from $\mathbf{0}$ to $12 \boldsymbol{j}=(0,12)$.

Exercise 1.1.14

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Exercise 1.1.14

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