Suppose in Figure 1, you are given two vectors: $A$ with a magnitude of $3$ units pointing east, and $B$ with a magnitude of $4$ units pointing north. Which of the following combinations of these vectors results in the largest possible magnitude?

• A. The vector difference $A-B$
• B. The vector $B$ alone
• C. The vector $A$ alone
• D. The vector sum $A+B$

Answer: (D)

Verified step by step guidance

1. Identify the magnitudes and directions of the given vectors: vector \( \mathbf{A} \) has magnitude 3 units pointing east, and vector \( \mathbf{B} \) has magnitude 4 units pointing north.
2. Recall that when two vectors are perpendicular (east and north directions are at 90°), the magnitude of their sum \( \mathbf{A} + \mathbf{B} \) can be found using the Pythagorean theorem: \( |\mathbf{A} + \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2} \).
3. Similarly, the magnitude of their difference \( \mathbf{A} - \mathbf{B} \) is also found using the Pythagorean theorem because the vectors are perpendicular: \( |\mathbf{A} - \mathbf{B}| = \sqrt{|\mathbf{A}|^2 + |\mathbf{B}|^2} \).
4. Compare the magnitudes of the individual vectors \( |\mathbf{A}| = 3 \) and \( |\mathbf{B}| = 4 \) with the magnitudes of the sum and difference vectors calculated using the Pythagorean theorem.
5. Conclude that the vector sum \( \mathbf{A} + \mathbf{B} \) results in the largest magnitude because adding perpendicular vectors results in a vector whose magnitude is greater than either individual vector or their difference.