Question

A vector has magnitude same as that of A = \(-3\hat{i} + 4\hat{j}\) and is parallel to B = \(4\hat{i} + 3\hat{j}\). The x and y components of this vector in the first quadrant are x and y respectively where:

\(x = \_\_\_\_\).

Answer 1

Correct Answer: 4

The problem asks to find the x-component of a vector that has the same magnitude as vector \( \vec{A} = -3\hat{i} + 4\hat{j} \) and is parallel to vector \( \vec{B} = 4\hat{i} + 3\hat{j} \). This new vector is stated to be in the first quadrant.

Concept Used:

To solve this problem, we will use the following concepts from vector algebra:

1. Magnitude of a Vector: The magnitude of a vector \( \vec{V} = V_x\hat{i} + V_y\hat{j} \) is given by:

\[ |\vec{V}| = \sqrt{V_x^2 + V_y^2} \]

2. Unit Vector: A unit vector in the direction of a non-zero vector \( \vec{V} \) is given by:

\[ \hat{V} = \frac{\vec{V}}{|\vec{V}|} \]

3. Representation of a Vector: Any vector can be represented as the product of its magnitude and its unit vector (direction):

\[ \vec{V} = |\vec{V}| \hat{V} \]

4. Parallel Vectors: If two vectors are parallel, they have the same direction, which means they share the same unit vector.

Step-by-Step Solution:

Step 1: Calculate the magnitude of vector \( \vec{A} = -3\hat{i} + 4\hat{j} \).

\[ |\vec{A}| = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

Let the required vector be \( \vec{V} \). The problem states that the magnitude of \( \vec{V} \) is the same as the magnitude of \( \vec{A} \). Therefore:

\[ |\vec{V}| = |\vec{A}| = 5 \]

Step 2: Find the unit vector in the direction of vector \( \vec{B} = 4\hat{i} + 3\hat{j} \).

First, we find the magnitude of \( \vec{B} \):

\[ |\vec{B}| = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \]

Now, we find the unit vector \( \hat{B} \):

\[ \hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{5} = \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j} \]

Step 3: Construct the required vector \( \vec{V} \).

Since \( \vec{V} \) is parallel to \( \vec{B} \), it must have the same direction as \( \vec{B} \). So, the unit vector for \( \vec{V} \) is \( \hat{B} \). The vector \( \vec{V} \) is the product of its magnitude and its unit vector:

\[ \vec{V} = |\vec{V}| \hat{B} \]

Substituting the values we found:

\[ \vec{V} = 5 \left( \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j} \right) \]

Step 4: Simplify the expression for \( \vec{V} \) to find its components.

\[ \vec{V} = 5 \times \frac{4}{5}\hat{i} + 5 \times \frac{3}{5}\hat{j} \] \[ \vec{V} = 4\hat{i} + 3\hat{j} \]

Step 5: Identify the x and y components of the vector \( \vec{V} \).

The vector is \( \vec{V} = x\hat{i} + y\hat{j} = 4\hat{i} + 3\hat{j} \). Both components are positive, which confirms the vector is in the first quadrant.

The x-component is \( x = 4 \).

The y-component is \( y = 3 \).

The question asks for the value of the x-component. The x-component of this vector is 4.