How much work does a force \(\vec{F}=67 \hat{\imath}+23 \hat{\jmath}+55 \hat{k} \mathrm{N}\) do as it acts on a body moving in a straight line from \(\vec{r}_{1}=16 \hat{\imath}+31 \hat{\jmath} \mathrm{m}\) to \(\vec{r}_{2}=21 \hat{\imath}+10 \hat{\jmath}+14 \hat{k} \mathrm{m} ?\)

Understanding the concept of 'displacement vector' is critical in physics, especially when analyzing motion. A displacement vector represents the change in position of an object, pointing directly from the initial position to the final position. It’s a vector quantity, which means it has both magnitude and direction.

In the problem provided, the displacement vector was determined by subtracting the initial position vector \( \vec{r}_{1} \) from the final position vector \( \vec{r}_{2} \) to reveal the object’s overall change in position. The question involved a change in position from \( \vec{r}_{1} = 16 \hat{\imath}+31 \hat{\jmath} \mathrm{m} \) to \( \vec{r}_{2} = 21 \hat{\imath}+10 \hat{\jmath}+14 \hat{k} \mathrm{m} \). The formula applied was:\
\[ \vec{r} = \vec{r}_{2} - \vec{r}_{1} \]
As emphasized in the solution, these calculations result in the displacement vector \( \vec{r} = 5\hat{\imath} -21 \hat{\jmath} + 14 \hat{k} \mathrm{m} \).

The student should visualize this as a straight line drawn from the starting to the ending point in a 3-dimensional space. The concept helps in understanding how objects move from one point to another and forms the basis for further calculations in work and energy studies.

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is key when you want to measure the work done by a force on an object when the force applied is constant and the object moves in a straight line.

In mathematical terms, if you have two vectors \( \vec{A} \) and \( \vec{B} \), their dot product is denoted and calculated as follows:
\[ \vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z \]
Where \( A_x, A_y, A_z \) and \( B_x, B_y, B_z \) are the vector components along the x, y, and z axes, respectively. In the context of work, the dot product will only result in non-zero work if there is a component of the force in the direction of the displacement.

The knack to understanding the dot product lies in recognizing that it's a measure of how much one vector extends in the direction of another. The problem given requires computing the dot product to determine the work done, which is a real-world application of the concept.

In physics, the term vector force is significant because it not just specifies the magnitude of the force but also the direction in which the force is applied. Forces can cause objects to start moving, stop moving, or change direction. A force is a vector quantity described by its magnitude, direction, and point of application.

The problem presents the force vector as \( \vec{F}=67 \hat{\imath}+23 \hat{\jmath}+55 \hat{k} \mathrm{N} \). This tells us the strength of the force along each axis: 67 N in the x-direction, 23 N in the y-direction, and 55 N in the z-direction. When an object is moved by a force, as it happens with the displacement in this problem, you can determine the work done using this force vector.

It’s essential to correctly interpret the vector components, to apply other vector operations successfully, like when calculating the work done where the dot product is required. For students, envisioning the force vector applied to an object along with its displacement helps to make sense of the work and energy concepts.