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

Determine the x-component of a vector with the same magnitude as \( \vec{A} = -3\hat{i} + 4\hat{j} \) and parallel to \( \vec{B} = 4\hat{i} + 3\hat{j} \). The vector is located in the first quadrant.

Concepts Utilized:

Vector algebra principles applied include:

1. Vector Magnitude: For \( \vec{V} = V_x\hat{i} + V_y\hat{j} \), the magnitude is \( |\vec{V}| = \sqrt{V_x^2 + V_y^2} \).

2. Unit Vector: A unit vector in the direction of \( \vec{V} \) is \( \hat{V} = \frac{\vec{V}}{|\vec{V}|} \).

3. Vector Representation: A vector is its magnitude multiplied by its unit vector: \( \vec{V} = |\vec{V}| \hat{V} \).

4. Parallel Vectors: Parallel vectors share the same unit vector.

Solution Steps:

Step 1: Calculate \( |\vec{A}| \).

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

Let the target vector be \( \vec{V} \). Given \( |\vec{V}| = |\vec{A}| \), then \( |\vec{V}| = 5 \).

Step 2: Determine the unit vector for \( \vec{B} \).

First, find \( |\vec{B}| \):

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

The unit vector \( \hat{B} \) is:

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

Step 3: Formulate \( \vec{V} \).

Since \( \vec{V} \) is parallel to \( \vec{B} \), \( \vec{V} \) has the unit vector \( \hat{B} \). Thus:

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

Step 4: Simplify \( \vec{V} \).

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

Step 5: Extract components of \( \vec{V} \).

For \( \vec{V} = x\hat{i} + y\hat{j} = 4\hat{i} + 3\hat{j} \), the x-component is \( x = 4 \) and the y-component is \( y = 3 \). Both are positive, confirming first quadrant location.

The x-component is 4.