The resultant of two vectors at right angles is 5 N. If the angle between them is `120^(@)` and the resultant is `sqrt13` then the magnitude of vectors are

Options

A. `sqrt(12) N, sqrt(13) N`
B. `sqrt(20) N, sqrt(5) N`
C. `3 N, 4 N`
D. `sqrt(40) N, sqrt(15) N`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We are given two vectors \( A \) and \( B \) with a resultant of \( 5 \, \text{N} \) when they are at right angles (90 degrees). We also know that when the angle between them is \( 120^\circ \), the resultant is \( \sqrt{13} \).

### Step 2: Use the Formula for Resultant Vectors
The formula for the resultant \( R \) of two vectors \( A \) and \( B \) at an angle \( \theta \) is given by:
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(\theta)} \]

### Step 3: Set Up the First Equation
When the vectors are at right angles (\( \theta = 90^\circ \)):
\[ R = \sqrt{A^2 + B^2} \]
Given \( R = 5 \, \text{N} \):
\[ 5 = \sqrt{A^2 + B^2} \]
Squaring both sides:
\[ A^2 + B^2 = 25 \quad \text{(Equation 1)} \]

### Step 4: Set Up the Second Equation
When the angle between the vectors is \( 120^\circ \):
\[ R = \sqrt{A^2 + B^2 + 2AB \cos(120^\circ)} \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ R = \sqrt{A^2 + B^2 - AB} \]
Given \( R = \sqrt{13} \):
\[ \sqrt{13} = \sqrt{A^2 + B^2 - AB} \]
Squaring both sides:
\[ 13 = A^2 + B^2 - AB \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 1 into Equation 2
From Equation 1, we know \( A^2 + B^2 = 25 \). Substitute this into Equation 2:
\[ 13 = 25 - AB \]
Rearranging gives:
\[ AB = 25 - 13 = 12 \quad \text{(Equation 3)} \]

### Step 6: Solve the System of Equations
Now we have two equations:
1. \( A^2 + B^2 = 25 \)
2. \( AB = 12 \)

These can be treated as a system of equations. Let \( A \) and \( B \) be the roots of the quadratic equation:
\[ x^2 - (A+B)x + AB = 0 \]
Let \( S = A + B \) (sum of the roots), then we can use the identity:
\[ (A+B)^2 = A^2 + B^2 + 2AB \]
Substituting the known values:
\[ S^2 = 25 + 2(12) = 25 + 24 = 49 \]
Thus,
\[ S = \sqrt{49} = 7 \]

### Step 7: Find the Values of A and B
Now we have:
1. \( A + B = 7 \)
2. \( AB = 12 \)

The quadratic equation becomes:
\[ x^2 - 7x + 12 = 0 \]
Factoring gives:
\[ (x - 3)(x - 4) = 0 \]
Thus, the solutions are:
\[ A = 3, \quad B = 4 \quad \text{or vice versa.} \]

### Conclusion
The magnitudes of the vectors are \( 3 \, \text{N} \) and \( 4 \, \text{N} \).