Question

If \(\vec{a}, \vec{b}, \vec{c}\) are three unit vectors such that \[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 15, \] then \[ |\vec{a} - \vec{b} - \vec{c}|^2 - 4(\vec{b} \cdot \vec{c}) = ? \]

• A. 6
• B. 15
• C. 12
• D. 10

Hint:

Use vector identities and dot product properties to simplify expressions.

Answer 1

The Correct Option is C

Step 1: Expand sum of squares
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2 \vec{a} \cdot \vec{b} \] Similarly for others. Since \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors: \[ |\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1 \] Step 2: Write out given expression
\[ |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 3 \times 2 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 2S = 15 \] where \[ S = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \] Step 3: Solve for \(S\)
\[ 6 - 2S = 15 \implies -2S = 9 \implies S = -\frac{9}{2} \] Step 4: Calculate required expression
\[ |\vec{a} - \vec{b} - \vec{c}|^2 = (\vec{a} - \vec{b} - \vec{c}) \cdot (\vec{a} - \vec{b} - \vec{c}) = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) + 2 \vec{b} \cdot \vec{c} \] \[ = 3 - 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) + 2 \vec{b} \cdot \vec{c} \] Step 5: Using \(S\)
\[ S = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \] So, \[ \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = S - \vec{b} \cdot \vec{c} \] Therefore, \[ |\vec{a} - \vec{b} - \vec{c}|^2 = 3 - 2(S - \vec{b} \cdot \vec{c}) + 2 \vec{b} \cdot \vec{c} = 3 - 2S + 4 \vec{b} \cdot \vec{c} \] Step 6: Compute required value
\[ |\vec{a} - \vec{b} - \vec{c}|^2 - 4(\vec{b} \cdot \vec{c}) = 3 - 2S = 3 - 2 \times \left(-\frac{9}{2}\right) = 3 + 9 = 12 \]