Question

Vector A has a magnitude of 63 units and points due west, while vector B has the same magnitude and points due south. Find the magnitude and direction of A+B and A-B

Answer 1

Answer on Question#64970, Physics / Molecular Physics | Thermodynamics

$A + B,$ $63\sqrt{2}$ units, south-west

$A - B,$ $63\sqrt{2}$ units, north-west

Question

Vector $A$ has a magnitude of 63 units and points due west, while vector $B$ has the same magnitude and points due south. Find the magnitude and direction of $A + B$ and $A - B$ .

Solution

$C = A - B,$ $D = A + B$

$C$ points due north-west and $D$ points due south-west, which can be clearly seen on the picture above. It follows from directions of $A$ and $B$ , and from equality of their magnitudes.

In order to find magnitudes of $C$ and $D$ we use Pythagorean theorem:

$$| C | = | D | = \sqrt {| A | ^ {2} + | \pm B | ^ {2}}$$ $$| A | = | + B | = | - B | = 6 3$$ $$| C | = | D | = \sqrt {6 3 ^ {2} + 6 3 ^ {2}} = 6 3 \sqrt {2} (u n i t s)$$

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