Instructions: On occasion, the notation \vec{A} = [A, \theta] will be a shorthand notation for \vec{A} = A \cos \theta \hat{i} + A \sin \theta \hat{j} Given that \vec{A} + \vec{B} = x_1 \hat{i} + y_1 \hat{j} and \vec{A} - \vec{B} = x_2 \hat{i} + y_2 \hat{j}, what is \vec{A}? \vec{A} = \frac{1}{2} (x_1 + x_2) \hat{i} + \frac{1}{2} (y_1 + y_2) \hat{j} \vec{A} = \frac{1}{2} (x_1 - x_2) \hat{i} + \frac{1}{2} (y_1 + y_2) \hat{j} \vec{A} = \frac{1}{2} (x_1 + x_2) \hat{i} + \frac{1}{2} (y_1 - y_2) \hat{j} \vec{A} = \frac{1}{2} (x_1 + x_2) \hat{i} \vec{A} = \frac{1}{2} (x_1 - x_2) \hat{i} + \frac{1}{2} (y_1 - y_2) \hat{j}
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Step 1: Given that A+B = x,i+yi and A-B = x,i+yj, we can write the equations as follows: A + B = x + xi + yi A - B = x + xi + yj Show more…
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