Question

Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that \(|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(60^\circ\). If \(\frac{1}{8} \vec{a}\) is a unit vector, then \(|\vec{b}|\) is equal to :

• A. \(4\)
• B. \(5\)
• C. \(6\)
• D. \(8\)

Hint:

Squaring both sides of a vector magnitude equality is the most effective way to produce an equation involving dot products.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The magnitude squared of a vector \(\vec{v}\) is given by \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\). We use the dot product properties and the given angle to find the unknown magnitude.
Step 2: Detailed Explanation:
1. Given \(\frac{1}{8} \vec{a}\) is a unit vector \(\implies |\frac{1}{8} \vec{a}| = 1 \implies \frac{1}{8} |\vec{a}| = 1 \implies |\vec{a}| = 8\).
2. Square both sides of the given magnitude equation:
\[ |2\vec{a} + 3\vec{b}|^2 = |3\vec{a} + \vec{b}|^2 \]
\[ 4|\vec{a}|^2 + 9|\vec{b}|^2 + 12(\vec{a} \cdot \vec{b}) = 9|\vec{a}|^2 + |\vec{b}|^2 + 6(\vec{a} \cdot \vec{b}) \]
\[ 8|\vec{b}|^2 + 6(\vec{a} \cdot \vec{b}) = 5|\vec{a}|^2 \]
3. Substitute \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 60^\circ\):
\[ 8|\vec{b}|^2 + 6(8|\vec{b}| \cdot \frac{1}{2}) = 5(8)^2 \]
\[ 8|\vec{b}|^2 + 24|\vec{b}| = 5 \cdot 64 = 320 \]
Divide by 8:
\[ |\vec{b}|^2 + 3|\vec{b}| - 40 = 0 \]
\[ (|\vec{b}| + 8)(|\vec{b}| - 5) = 0 \]
Since magnitude is always non-negative, \(|\vec{b}| = 5\).
Step 3: Final Answer:
The magnitude \(|\vec{b}|\) is \(5\).