Question

If \(\vec{a}, \vec{b} \& \vec{c}\) are unit vectors such that \((\vec{a}-\vec{b})^2 + (\vec{b}-\vec{c})^2 + (\vec{c}-\vec{a})^2 = 9\). Find positive k if \(|2\vec{a} + k\vec{b} + k\vec{c}| = 3\) :

• A. 2
• B. 4
• C. 5
• D. 7

Hint:

If the sum of squared differences of three unit vectors is 9, it is a definitive signal that the vectors sum to zero, as this is the maximum possible value for that sum in 3D space.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem uses properties of dot products and magnitudes of unit vectors to find the relationship between three vectors and then determines a scalar parameter.
Step 2: Key Formula or Approach:
1. For unit vectors: \(|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1\).
2. Expansion: \(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}\).
3. Squared sum: \(|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\).
Step 3: Detailed Explanation:
Expanding the given condition:
\[ (2 - 2\vec{a}\cdot\vec{b}) + (2 - 2\vec{b}\cdot\vec{c}) + (2 - 2\vec{c}\cdot\vec{a}) = 9 \] \[ 6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9 \implies \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2} \] Now consider \(|\vec{a} + \vec{b} + \vec{c}|^2\):
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + 2\left( -\frac{3}{2} \right) = 3 - 3 = 0 \] This implies \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), so \(\vec{b} + \vec{c} = -\vec{a}\).
Substitute this into the second condition:
\[ |2\vec{a} + k(\vec{b} + \vec{c})| = 3 \implies |2\vec{a} + k(-\vec{a})| = 3 \] \[ |(2-k)\vec{a}| = 3 \] Since \(|\vec{a}| = 1\):
\[ |2-k| = 3 \implies 2-k = 3 \text{ or } 2-k = -3 \] \[ k = -1 \text{ or } k = 5 \] Since \(k\) must be positive, \(k = 5\).
Step 4: Final Answer:
The positive value of \(k\) is 5.