If `abs(3veca+vecb)=abs(2veca+3vecb)` and angle between `veca and vecb` is `60^@` and `veca/8` is a unit vector then find the magnitude of `vecb`

Options

A. 5
B. 6
C. 8
D. 10

Answer

To solve the problem, we start with the given equation and the conditions provided:

1. We have the equation:
\[ |\vec{3a} + \vec{b}| = |\vec{2a} + \vec{3b}| \]

2. The angle between \(\vec{a}\) and \(\vec{b}\) is \(60^\circ\).

3. The vector \(\frac{\vec{a}}{8}\) is a unit vector, which implies:
\[ |\vec{a}| = 8 \]

### Step 1: Squaring Both Sides
We square both sides of the equation:
\[ |\vec{3a} + \vec{b}|^2 = |\vec{2a} + \vec{3b}|^2 \]

### Step 2: Expanding Both Sides
Using the formula \( |\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2|\vec{x}||\vec{y}|\cos\theta \):
- Left Side:
\[ |\vec{3a} + \vec{b}|^2 = |\vec{3a}|^2 + |\vec{b}|^2 + 2|\vec{3a}||\vec{b}|\cos(60^\circ) \]
\[ = 9|\vec{a}|^2 + |\vec{b}|^2 + 3|\vec{b}| \]
- Right Side:
\[ |\vec{2a} + \vec{3b}|^2 = |\vec{2a}|^2 + |\vec{3b}|^2 + 2|\vec{2a}||\vec{3b}|\cos(60^\circ) \]
\[ = 4|\vec{a}|^2 + 9|\vec{b}|^2 + 6|\vec{b}| \]

### Step 3: Substituting the Magnitudes
Since \( |\vec{a}| = 8 \):
- Left Side:
\[ = 9(8^2) + |\vec{b}|^2 + 3|\vec{b}| = 576 + |\vec{b}|^2 + 3|\vec{b}| \]
- Right Side:
\[ = 4(8^2) + 9|\vec{b}|^2 + 6|\vec{b}| = 256 + 9|\vec{b}|^2 + 6|\vec{b}| \]

### Step 4: Setting the Equations Equal
Now we equate both sides:
\[ 576 + |\vec{b}|^2 + 3|\vec{b}| = 256 + 9|\vec{b}|^2 + 6|\vec{b}| \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 576 - 256 + |\vec{b}|^2 - 9|\vec{b}|^2 + 3|\vec{b}| - 6|\vec{b}| = 0 \]
\[ 320 - 8|\vec{b}|^2 - 3|\vec{b}| = 0 \]

### Step 6: Forming a Quadratic Equation
Rearranging the equation gives:
\[ 8|\vec{b}|^2 + 3|\vec{b}| - 320 = 0 \]

### Step 7: Solving the Quadratic Equation
Using the quadratic formula \( |\vec{b}| = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 8 \), \( b = 3 \), and \( c = -320 \):
\[ |\vec{b}| = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 8 \cdot (-320)}}{2 \cdot 8} \]
\[ = \frac{-3 \pm \sqrt{9 + 10240}}{16} \]
\[ = \frac{-3 \pm \sqrt{10249}}{16} \]
\[ = \frac{-3 \pm 101.24}{16} \]

Calculating the two possible values:
1. \( |\vec{b}| = \frac{98.24}{16} \approx 6.14 \)
2. \( |\vec{b}| = \frac{-104.24}{16} \) (not valid as magnitude cannot be negative)

Thus, the magnitude of \(\vec{b}\) is approximately:
\[ |\vec{b}| \approx 6.14 \]