2.2 Coordinate systems and components of a vector  (Page 2/15)

Displacement of a mouse pointer

Strategy

Solution

The vector component form of the displacement vector is

This solution is shown in [link] .

Significance

The vector x -component ${\overrightarrow{D}}_x=\mathrm{-4.0}\widehat{i}=4.0(\text{−}\widehat{i})$ of the displacement vector has the magnitude $\left|{\overrightarrow{D}}_x\right|=|-4.0|\left|\widehat{i}\right|=4.0$ because the magnitude of the unit vector is $\left|\widehat{i}\right|=1$ . Notice, too, that the direction of the x -component is $\text{−}\widehat{i}$ , which is antiparallel to the direction of the + x -axis; hence, the x -component vector ${\overrightarrow{D}}_x$ points to the left, as shown in [link] . The scalar x -component of vector $\overrightarrow{D}$ is $D_x=\mathrm{-4.0}$ .

Similarly, the vector y -component ${\overrightarrow{D}}_y=+2.9\widehat{j}$ of the displacement vector has magnitude $\left|{\overrightarrow{D}}_y\right|=\left|2.9\right|\left|\widehat{j}\right|=2.9$ because the magnitude of the unit vector is $\left|\widehat{j}\right|=1$ . The direction of the y -component is $+\widehat{j}$ , which is parallel to the direction of the + y -axis. Therefore, the y -component vector ${\overrightarrow{D}}_y$ points up, as seen in [link] . The scalar y -component of vector $\overrightarrow{D}$ is $D_y=+2.9$ . The displacement vector $\overrightarrow{D}$ is the resultant of its two vector components.

The vector component form of the displacement vector [link] tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.