Problem 43 - Master Resource Book in JEE Main Physics

Question

The component of vector \(\mathbf{A}=a_{x} \hat{\mathbf{i}}+a_{y} \hat{\mathbf{j}}+a_{z} \mathbf{\mathbf { k }}\) along the direction of \((\hat{\mathbf{i}}-\hat{\mathbf{j}})\) is \(\quad\) [EAMCET 2008] (a) \(\left(a_{x}-a_{y}+a_{z}\right)\) (b) \(\left(a_{x}+a_{y}\right)\) (c) \(\left(a_{x}-a_{y}\right) / \sqrt{2}\) (d) \(\left(a_{x}-a_{y}+a_{z}\right)\)

Step-by-Step Solution

Answer

The correct answer is (c) \((a_x - a_y)/\sqrt{2}\).

Step 1: Understanding the Problem

We need to find the component of the vector \(\mathbf{A} = a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}} + a_z \mathbf{k}\) in the direction of the vector \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\). Essentially, this requires us to project vector \(\mathbf{A}\) onto the direction of the vector \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\).

Step 2: Find the Unit Vector

The direction vector is given as \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\). We need to convert this to a unit vector. The magnitude of the vector \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\) is calculated as: \[\sqrt{(1)^2 + (-1)^2} = \sqrt{2}\]. Thus, the unit vector in this direction is \[\frac{1}{\sqrt{2}}( \hat{\mathbf{i}} - \hat{\mathbf{j}} )\].

Step 3: Calculate the Dot Product

To find the component of \(\mathbf{A}\) along the unit vector \(\frac{1}{\sqrt{2}}( \hat{\mathbf{i}} - \hat{\mathbf{j}} )\), calculate the dot product: \[\mathbf{A} \cdot \frac{1}{\sqrt{2}}( \hat{\mathbf{i}} - \hat{\mathbf{j}} ) = \frac{1}{\sqrt{2}} ( a_x \hat{\mathbf{i}} + a_y \hat{\mathbf{j}} + a_z \mathbf{k} ) \cdot ( \hat{\mathbf{i}} - \hat{\mathbf{j}} )\].

Step 4: Solve the Dot Product

Compute the dot product: \[\mathbf{A} \cdot \frac{1}{\sqrt{2}}( \hat{\mathbf{i}} - \hat{\mathbf{j}} ) = \frac{1}{\sqrt{2}} ( a_x \times 1 + a_y \times (-1) ) = \frac{1}{\sqrt{2}}(a_x - a_y)\]. There is no contribution from \(a_z\) as there is no component in the \(\mathbf{k}\) direction in the second vector.

Step 5: Choosing the Correct Answer

The final expression for the component of \(\mathbf{A}\) in the required direction is \(\frac{1}{\sqrt{2}}(a_x - a_y)\). This matches option (c).

Key Concepts

Dot Product

The dot product is a fundamental concept in vector mathematics that is used to calculate the magnitude of one vector in the direction of another. Simply put, it allows us to assess how much of one vector is pointing in the same direction as another vector. The dot product of two vectors, \(\mathbf{A}\) and \(\mathbf{B}\), is computed as follows:\[\mathbf{A} \cdot \mathbf{B} = a_x b_x + a_y b_y + a_z b_z\]where \(a_x, a_y, a_z\) and \(b_x, b_y, b_z\) are the components of vectors \(\mathbf{A}\) and \(\mathbf{B}\), respectively.

• To calculate the dot product, multiply the corresponding components of the vectors.
• Add up these multiplications to get a scalar value.
• The result is not a vector but a scalar, signifying magnitude and direction relations.

Hence, it's an efficient way to find the angle between vectors or for projecting vectors onto one another.

Unit Vector

A unit vector is a vector of length one that indicates direction but not magnitude. It is particularly useful when you need to transform a vector into a specific or normalized direction. For instance, when given a vector \(\mathbf{V}\), its unit vector \(\hat{\mathbf{V}}\) is computed by dividing each component by the vector's magnitude:
\[\hat{\mathbf{V}} = \frac{1}{\|\mathbf{V}\|} \cdot \mathbf{V}\]where \(\|\mathbf{V}\|\) is the magnitude, calculated as \(\sqrt{(v_x)^2 + (v_y)^2 + (v_z)^2}\).

• Unit vectors preserve direction and discard magnitude.
• They are denoted typically with a caret (e.g., \(\hat{\mathbf{i}}\)).
• Used for aligning vectors and calculating directional components such as projection.

In our problem, turning the vector \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\) into a unit vector involved dividing by its magnitude \(\sqrt{2}\) to normalize it.

Projection of Vectors

The projection of one vector onto another involves creating a vector that aligns entirely in the direction of the unit vector, while having a magnitude representative of the original vector's influence in that direction. You can visualize it as the shadow or footprint of one vector onto another.
To find the projection of vector \(\mathbf{A}\) onto vector \(\mathbf{B}\), follow these steps:

• Calculate the unit vector of \(\mathbf{B}\), \(\hat{\mathbf{B}}\).
• Compute the scalar projection using the dot product: \(\mathbf{A} \cdot \hat{\mathbf{B}}\).
• This scalar tells you the magnitude of \(\mathbf{A}\) in the \(\mathbf{B}\) direction.

This results in a new vector which is in the direction of \(\mathbf{B}\) with a magnitude noted by the scalar. In our exercise, projecting \(\mathbf{A}\) along \(\hat{\mathbf{i}} - \hat{\mathbf{j}}\) required us to calculate \(\mathbf{A}\)'s dot product with its unit vector. Hence, proving the right answer of \(\frac{1}{\sqrt{2}}(a_x - a_y)\). This helps us see precisely how much of \(\mathbf{A}\) is in that specific direction.