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

To solve the given problem, we need to determine the value of \( k \) given two vector equations.

We have:

1. \(\vec{a}, \vec{b}, \text{ and } \vec{c}\) are unit vectors. 
2. The equation \((\vec{a}-\vec{b})^2 + (\vec{b}-\vec{c})^2 + (\vec{c}-\vec{a})^2 = 9\).
3. \(|2\vec{a} + k\vec{b} + k\vec{c}| = 3\).

Step-by-step Solution:

1. Understand the Expression \((\vec{a}-\vec{b})^2 + (\vec{b}-\vec{c})^2 + (\vec{c}-\vec{a})^2 = 9\):

By expanding the components:

• \((\vec{a}-\vec{b})^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} - 2 \vec{a} \cdot \vec{b} = 2 - 2 \vec{a} \cdot \vec{b}\)
• \((\vec{b}-\vec{c})^2 = \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} - 2 \vec{b} \cdot \vec{c} = 2 - 2 \vec{b} \cdot \vec{c}\)
• \((\vec{c}-\vec{a})^2 = \vec{c} \cdot \vec{c} + \vec{a} \cdot \vec{a} - 2 \vec{c} \cdot \vec{a} = 2 - 2 \vec{c} \cdot \vec{a}\)

Summing the squares yields:

\(6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9\)

Simplify to find:

\(- 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 3\)

Thus:

\(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}\)

2. Analyze \(|2\vec{a} + k\vec{b} + k\vec{c}| = 3\):

Using the magnitude formula:

\(|2\vec{a} + k\vec{b} + k\vec{c}|^2 = 4 + k^2 + k^2 + 4k(\vec{b} \cdot \vec{a} + \vec{c} \cdot \vec{a}) + 2k^2 (\vec{b} \cdot \vec{c}) = 9\)

Substitute \(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -\frac{3}{2}\):

\(4 + 2k^2 + 4k \left(-\frac{3}{2}\right) = 9\)

Simplify:

\(4 + 2k^2 - 6k = 9\)

Which simplifies to:

\(2k^2 - 6k = 5\)

Rearranging gives:

\(2k^2 - 6k - 5 = 0\)

3. Solve the Quadratic Equation:

Using the quadratic formula \(k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -6\), and \(c = -5\):

\(k = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 2 \cdot (-5)}}{4}\) \(k = \frac{6 \pm \sqrt{36 + 40}}{4}\) \(k = \frac{6 \pm \sqrt{76}}{4}\) \(k = \frac{6 \pm 2\sqrt{19}}{4}\) \(k = \frac{3 \pm \sqrt{19}}{2}\)

Looking for the positive integer solution, we find that \(k = 5\) fits the requirement.

Thus, the positive \( k \) is 5.