Question

If resultant of \(\mathrm{A}^{\rightarrow}=2 \hat{1}+\hat{\jmath}-\mathrm{k}^{\wedge}, \mathrm{B}^{\rightarrow}=\hat{\imath}-2 \hat{\mathrm{j}}+3 \mathrm{k}^{\text {and }} \mathrm{C}^{\rightarrow}\) is unit vector in y direction, then \(\mathrm{C}^{\rightarrow}\) is (A) \(-\hat{j}\) (B) \(3 \hat{i}-2 \hat{j}+2 \mathrm{k}^{\wedge}\) (C) (D) \(2 \hat{i}+3 \mathrm{k}^{\wedge}\)

Short answer

Therefore, the short answer is: \(\mathrm{C}^{\rightarrow} = -\hat{j}\).

Step-by-step solution

Find the resultant vector#\tag_content#Add vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) to find the resultant vector \(\mathrm{R}^{\rightarrow}\): \[\mathrm{R}^{\rightarrow} = \mathrm{A}^{\rightarrow} + \mathrm{B}^{\rightarrow}\]

Step 2: Calculate \(\mathrm{R}^{\rightarrow}\)#\tag_content#Now, substitute the given values of \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) into the equation and calculate the resultant vector: \[\mathrm{R}^{\rightarrow} = (2\hat{\imath} + \hat{\jmath} - \hat{k}) + (\hat{\imath} - 2\hat{\jmath} + 3\hat{k})\]

Simplify the resultant vector#\tag_content#Combine the like terms (i.e. components in the same direction) to simplify the resultant vector: \[\mathrm{R}^{\rightarrow} = (2\hat{\imath} + \hat{\imath}) + (\hat{\jmath} - 2\hat{\jmath}) + (-\hat{k} + 3\hat{k})\] \[\mathrm{R}^{\rightarrow} = 3\hat{\imath} - \hat{\jmath} + 2\hat{k}\]

Step 4: Identify \(\mathrm{C}^{\rightarrow}\) as a unit vector in the y direction#\tag_content#According to the given information, \(\mathrm{C}^{\rightarrow}\) is a unit vector in the y direction and also the resultant of the two given vectors, meaning it is equal to the previously calculated \(\mathrm{R}^{\rightarrow}\). The only option among (A), (B), (C), and (D) which is a unit vector in the y direction, is: (A) \(-\hat{j}\) Thus, the correct option is (A), and \(\mathrm{C}^{\rightarrow} = -\hat{j}\).

Key concepts

Unit Vectors

When we talk about vectors, they can have different lengths, directions, and sometimes they can mix the two. But a unit vector is super special! It's created to point in a specific direction with a length of exactly one. No more, no less. This makes unit vectors incredibly useful when you want to specify direction only without worrying about the size or "magnitude" of the vector.

Why do we need unit vectors? Well, they become really handy when you're dealing with vector equations. Because they don't change the direction of a vector and only cancel out any scaling it might have. They're a classic example of keeping things simple when exploring complex problems.

• A unit vector in the x-direction is denoted as \(\hat{\imath}\).
• A unit vector in the y-direction is denoted as \(\hat{\jmath}\).
• A unit vector in the z-direction is denoted as \(\hat{\mathrm{k}}\).

In our exercise, the unit vector was found to be \(-\hat{\jmath}\) by the process of evaluating other vectors that share the same direction.

Vector Components

Vectors can be pretty complex, moving around in three-dimensional space. But to understand them better, we break them down into components. These components are essentially projections of the vector onto the x, y, and z axes. Imagine it as splitting a single vector into simpler parts aligned with each axis.

Thinking in components helps us perform vector operations more easily. For instance, to add vectors, we just add the respective components separately:

• The x-components combine horizontally.
• The y-components combine vertically.
• The z-components combine perpendicularly.

In the exercise you're examining, vector \(\mathrm{R}^{\rightarrow}\) was calculated by simplifying the vector components of \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\). By adding \(2\hat{\imath} + \hat{\jmath} - \hat{k}\) from \(\mathrm{A}^{\rightarrow}\) to \(\hat{\imath} - 2\hat{\jmath} + 3\hat{k}\) from \(\mathrm{B}^{\rightarrow}\), the resulting vector becomes \(3\hat{\imath} - \hat{\jmath} + 2\hat{k}\). This separation into components lets us handle and manipulate vectors more effectively.

Resultant Vector

Whenever we have multiple vectors, there’s a natural question—how do they sum up or cancel out? The answer is the resultant vector. It is essentially the vector sum of all the vectors being considered. It represents the combined effect of all the vectors.

Learning about resultant vectors helps us understand how different forces or motions combine together. It’s like solving a puzzle of finding the net effect when multiple influences act simultaneously.

• In most cases, to find a resultant vector, we add each individual component of the vectors separately: sum up all the x-components, likewise for y and z-components.

In the provided problem, the resultant vector \(\mathrm{R}^{\rightarrow}\) was determined by adding vectors \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) together to give \(3\hat{\imath} - \hat{\jmath} + 2\hat{k}\). Interestingly, the exercise also specifies a requirement that \(\mathrm{C}^{\rightarrow}\) should reflect this resultant vector, specifically maintaining the nature of a unit vector in the y-direction, which it achieves through \(-\hat{\jmath}\). This shows how we can use superposition calculus in vectors to arrive at simple conclusions from complex setups.