Question

If three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) satisfying the condition \(\vec{a} + \vec{b} + \vec{c} = 0\). If \(|\vec{a}| = 3\), \[|\vec{b}| = 4 \text{ and } |\vec{c}| = 2, \text{ then find the value of } \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}.\] 5

Hint:

{Key Formula:} \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \] When \(\vec{a} + \vec{b} + \vec{c} = 0\), LHS = 0, so: \[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2S = 0 \] \[ S = -\frac{|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2}{2} \]

Answer 1

Step 1: Use the given condition \(\vec{a} + \vec{b} + \vec{c} = 0\) From \(\vec{a} + \vec{b} + \vec{c} = 0\), we have: \[ \vec{a} + \vec{b} = -\vec{c} \]
Step 2: Square both sides \[ |\vec{a} + \vec{b}|^2 = |-\vec{c}|^2 = |\vec{c}|^2 \]
Step 3: Expand using dot product \[ |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) \] Therefore: \[ |\vec{a}|^2 + |\vec{b}|^2 + 2(\vec{a} \cdot \vec{b}) = |\vec{c}|^2 \]
Step 4: Substitute the given magnitudes \[ 3^2 + 4^2 + 2(\vec{a} \cdot \vec{b}) = 2^2 \] \[ 9 + 16 + 2(\vec{a} \cdot \vec{b}) = 4 \] \[ 25 + 2(\vec{a} \cdot \vec{b}) = 4 \] \[ 2(\vec{a} \cdot \vec{b}) = 4 - 25 = -21 \] \[ \vec{a} \cdot \vec{b} = -\frac{21}{2} \]
Step 5: Similarly, use \(\vec{b} + \vec{c} = -\vec{a}\) \[ |\vec{b} + \vec{c}|^2 = |-\vec{a}|^2 = |\vec{a}|^2 \] \[ |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{b} \cdot \vec{c}) = |\vec{a}|^2 \] \[ 4^2 + 2^2 + 2(\vec{b} \cdot \vec{c}) = 3^2 \] \[ 16 + 4 + 2(\vec{b} \cdot \vec{c}) = 9 \] \[ 20 + 2(\vec{b} \cdot \vec{c}) = 9 \] \[ 2(\vec{b} \cdot \vec{c}) = 9 - 20 = -11 \] \[ \vec{b} \cdot \vec{c} = -\frac{11}{2} \]
Step 6: Similarly, use \(\vec{c} + \vec{a} = -\vec{b}\) \[ |\vec{c} + \vec{a}|^2 = |-\vec{b}|^2 = |\vec{b}|^2 \] \[ |\vec{c}|^2 + |\vec{a}|^2 + 2(\vec{c} \cdot \vec{a}) = |\vec{b}|^2 \] \[ 2^2 + 3^2 + 2(\vec{c} \cdot \vec{a}) = 4^2 \] \[ 4 + 9 + 2(\vec{c} \cdot \vec{a}) = 16 \] \[ 13 + 2(\vec{c} \cdot \vec{a}) = 16 \] \[ 2(\vec{c} \cdot \vec{a}) = 16 - 13 = 3 \] \[ \vec{c} \cdot \vec{a} = \frac{3}{2} \]
Step 7: Find the required sum \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \left(-\frac{21}{2}\right) + \left(-\frac{11}{2}\right) + \left(\frac{3}{2}\right) \] \[ = \frac{-21 - 11 + 3}{2} = \frac{-29}{2} \]
Step 8: Final answer \[ \boxed{-\frac{29}{2}} \] Alternative Method: Square \(\vec{a} + \vec{b} + \vec{c} = 0\): \[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \] \[ |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \] \[ 9 + 16 + 4 + 2S = 0 \] \[ 29 + 2S = 0 \] \[ S = -\frac{29}{2} \] where \( S = \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \). This is a more direct approach!