Question

For three unit vectors \( \vec a, \vec b, \vec c \) satisfying \[ |\vec a-\vec b|^2 + |\vec b-\vec c|^2 + |\vec c-\vec a|^2 = 9 \] and \[ |2\vec a + k\vec b + k\vec c| = 3, \] the positive value of \( k \) is:

• A. \(3\)
• B. \(6\)
• C. \(4\)
• D. \(5\)

Hint:

In symmetric vector problems, assuming equal pairwise dot products often simplifies calculations and leads quickly to the correct option.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For unit vectors, \(|\vec{a}|=1\). The expansion of \(|\vec{a}-\vec{b}|^2\) involves dot products.
Step 2: Detailed Explanation:
Expansion: \((|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}) + (|\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b}\cdot\vec{c}) + (|\vec{c}|^2 + |\vec{a}|^2 - 2\vec{c}\cdot\vec{a}) = 9\).
\(6 - 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 9 \Rightarrow \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -\frac{3}{2}\).
We know \(|\vec{a}+\vec{b}+\vec{c}|^2 = 1+1+1 + 2(\sum \vec{a}\cdot\vec{b}) = 3 + 2(-\frac{3}{2}) = 0\).
Thus, \(\vec{a} + \vec{b} + \vec{c} = 0 \Rightarrow \vec{b} + \vec{c} = -\vec{a}\).
Substitute this into the second equation:
\(|2\vec{a} + k(\vec{b} + \vec{c})| = |2\vec{a} + k(-\vec{a})| = 3\).
\(|(2 - k)\vec{a}| = 3 \Rightarrow |2 - k| \cdot 1 = 3\).
\(2 - k = \pm 3 \Rightarrow k = -1\) or \(k = 5\).
Since we need the positive value, \(k = 5\).
Step 3: Final Answer:
The positive value of \(k\) is 5.