2.1 Scalars and vectors  (Page 4/29)

In general, when a vector $\overrightarrow{A}$ is multiplied by a positive scalar $\alpha$ , the result is a new vector $\overrightarrow{B}$ that is parallel to $\overrightarrow{A}$ :

The magnitude $\left|\overrightarrow{B}\right|$ of this new vector is obtained by multiplying the magnitude $\left|\overrightarrow{A}\right|$ of the original vector, as expressed by the scalar equation    :

In a scalar equation, both sides of the equation are numbers. [link] is a scalar equation because the magnitudes of vectors are scalar quantities (and positive numbers). If the scalar $\alpha$ is negative in the vector equation [link] , then the magnitude $\left|\overrightarrow{B}\right|$ of the new vector is still given by [link] , but the direction of the new vector $\overrightarrow{B}$ is antiparallel to the direction of $\overrightarrow{A}$ . These principles are illustrated in [link] (a) by two examples where the length of vector $\overrightarrow{A}$ is 1.5 units. When $\alpha =2$ , the new vector $\overrightarrow{B}=2\overrightarrow{A}$ has length $B=2A=3.0\text{units}$ (twice as long as the original vector) and is parallel to the original vector. When $\alpha =\mathrm{-2}$ , the new vector $\overrightarrow{C}=\mathrm{-2}\overrightarrow{A}$ has length $C=|-2|A=3.0\text{units}$ (twice as long as the original vector) and is antiparallel to the original vector.

Now suppose your fishing buddy departs from point A (the campsite), walking in the direction to point B (the fishing hole), but he realizes he lost his tackle box when he stopped to rest at point C (located three-quarters of the distance between A and B , beginning from point A ). So, he turns back and retraces his steps in the direction toward the campsite and finds the box lying on the path at some point D only 1.2 km away from point C (see [link] (b)). What is his displacement vector ${\overrightarrow{D}}_{AD}$ when he finds the box at point D ? What is his displacement vector ${\overrightarrow{D}}_{DB}$ from point D to the hole? We have already established that at rest point C his displacement vector is ${\overrightarrow{D}}_{AC}=0.75{\overrightarrow{D}}_{AB}$ . Starting at point C , he walks southwest (toward the campsite), which means his new displacement vector ${\overrightarrow{D}}_{CD}$ from point C to point D is antiparallel to ${\overrightarrow{D}}_{AB}$ . Its magnitude $\left|{\overrightarrow{D}}_{CD}\right|$ is $D_{CD}=1.2\text{km}=0.2D_{AB}$ , so his second displacement vector is ${\overrightarrow{D}}_{CD}=\mathrm{-0.2}{\overrightarrow{D}}_{AB}$ . His total displacement ${\overrightarrow{D}}_{AD}$ relative to the campsite is the vector sum    of the two displacement vectors: vector ${\overrightarrow{D}}_{AC}$ (from the campsite to the rest point) and vector ${\overrightarrow{D}}_{CD}$ (from the rest point to the point where he finds his box):

The vector sum of two (or more) vectors is called the resultant vector    or, for short, the resultant . When the vectors on the right-hand-side of [link] are known, we can find the resultant ${\overrightarrow{D}}_{AD}$ as follows:

When your friend finally reaches the pond at B , his displacement vector ${\overrightarrow{D}}_{AB}$ from point A is the vector sum of his displacement vector ${\overrightarrow{D}}_{AD}$ from point A to point D and his displacement vector ${\overrightarrow{D}}_{DB}$ from point D to the fishing hole: ${\overrightarrow{D}}_{AB}={\overrightarrow{D}}_{AD}+{\overrightarrow{D}}_{DB}$ (see [link] (c)). This means his displacement vector ${\overrightarrow{D}}_{DB}$ is the difference of two vectors    :

Notice that a difference of two vectors is nothing more than a vector sum of two vectors because the second term in [link] is vector $\text{−}{\overrightarrow{D}}_{AD}$ (which is antiparallel to ${\overrightarrow{D}}_{AD})$ . When we substitute [link] into [link] , we obtain the second displacement vector: