Given \(\mathbf{A}=\mathbf{i}+\mathbf{j}-2 \mathbf{k}, \mathbf{B}=2 \mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{C}=\mathbf{j}-5 \mathbf{k}\): Find the total work done by forces \(\mathbf{A}\) and \(\mathbf{B}\) if the object undergoes the displacement \(\mathbf{C}\). Hint: Can you add the two forces first?

In vector calculus, understanding the concept of resultant forces is key to solving many problems, including the one at hand. A resultant force is essentially the single force that represents the vector sum of all individual forces acting on an object. When you add multiple vectors together, you are combining their effects to understand how an object will move under the influence of all these forces combined.
To calculate the resultant force of vectors \(\textbf{A}\) and \(\textbf{B}\), you simply add them component-wise:
\[\textbf{A} = \textbf{i} + \textbf{j} - 2\textbf{k}\]
\[\textbf{B} = 2\textbf{i} - \textbf{j} + 3\textbf{k}\]
Adding these:
\[\textbf{A} + \textbf{B} = (\textbf{i} + \textbf{j} - 2\textbf{k}) + (2\textbf{i} - \textbf{j} + 3\textbf{k}) = 3\textbf{i} + \textbf{k}\]
Notice how you add the \(\textbf{i}\), \(\textbf{j}\), and \(\textbf{k}\) components separately. This gives us the resultant force \(\textbf{R} = 3\textbf{i} + \textbf{k}\). By mastering this concept of adding vectors to find the resultant, you can better understand how different forces interact in physical space.

The dot product, also known as the scalar product, is a crucial concept in vector calculus used to find the multiplication of two vectors, which results in a scalar. Formally, the dot product of two vectors \(\textbf{A}\) and \(\textbf{B}\) is given by:
\[\textbf{A} \bullet \textbf{B} = A_x B_x + A_y B_y + A_z B_z\]
The dot product tells us about the directional relationship between two vectors and is essential in calculating quantities like work done. For our problem, we calculate the dot product of the resultant force and the displacement vector \(\textbf{C}\):
\[\textbf{R} \bullet \textbf{C} = (3\textbf{i} + \textbf{k}) \bullet (\textbf{j} - 5\textbf{k})\]
Breaking this down into components, we get:

• \(3\textbf{i} \bullet \textbf{j} = 0\) (since \(\textbf{i} \bullet \textbf{j} = 0\))
• \(3\textbf{i} \bullet -5\textbf{k} = 0\) (since \(\textbf{i} \bullet \textbf{k} = 0\))
• \(\textbf{k} \bullet \textbf{j} = 0\) (again, \(\textbf{k} \bullet \textbf{j} = 0\))
• \(\textbf{k} \bullet -5\textbf{k} = -5\) (since \(\textbf{k} \bullet \textbf{k} = 1\))

Summing these products gives us:
\[0 + 0 + 0 - 5 = -5\]
So, the dot product of \(\textbf{R}\) and \(\textbf{C}\) is \(-5\).

Work done by a force in physics is a measure of energy transfer when the force causes a displacement. Mathematically, the work done \(\textbf{W}\) by a force \(\textbf{F}\) over a displacement \(\textbf{d}\) is given by the dot product of the two vectors:
\[\textbf{W} = \textbf{F} \bullet \textbf{d}\]
In simple terms, this formula means that to find the work done, you multiply the magnitude of the force in the direction of displacement by the magnitude of that displacement. From our problem, we already calculated the resultant force \(\textbf{R} = 3\textbf{i} + \textbf{k}\) and the displacement vector \(\textbf{C} = \textbf{j} - 5\textbf{k}\). Using the dot product, we found that the work done is: #\[-5\]
The negative sign here indicates that the force is acting in the opposite direction to the displacement, which often implies that the force is working to slow down or resist the displacement rather than aiding it. In essence, understanding work done helps us grasp how forces transfer energy to or away from objects based on their directions and magnitudes.