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: Given condition.
We are given the condition that: \[ \vec{a} + \vec{b} + \vec{c} = 0 \] This implies: \[ \vec{c} = -(\vec{a} + \vec{b}) \] Step 2: Find the expression for the required dot product.
We need to find: \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \] Substitute \(\vec{c} = -(\vec{a} + \vec{b})\) into this expression: \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \vec{a} \cdot \vec{b} + \vec{b} \cdot (-(\vec{a} + \vec{b})) + (-(\vec{a} + \vec{b})) \cdot \vec{a} \] Simplifying each term: \[ \vec{a} \cdot \vec{b} + \vec{b} \cdot (-(\vec{a} + \vec{b})) + (-(\vec{a} + \vec{b})) \cdot \vec{a} = \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} \] Now, combine like terms: \[ \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{b} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} \] \[ = - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{b} \] Step 3: Substitute the magnitudes of vectors.
We know the magnitudes of the vectors: \[ |\vec{a}| = 3, \quad |\vec{b}| = 4, \quad |\vec{c}| = 2 \] So: \[ \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 3^2 = 9 \] \[ \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 4^2 = 16 \] Now, substitute into the equation: \[ - \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b} - \vec{a} \cdot \vec{b} = -9 - 16 - \vec{a} \cdot \vec{b} \] Step 4: Use the fact that \(\vec{c} = -(\vec{a} + \vec{b})\).
Since \(\vec{c} = -(\vec{a} + \vec{b})\), we have: \[ |\vec{c}|^2 = (-(\vec{a} + \vec{b})) \cdot (-(\vec{a} + \vec{b})) = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) \] Expanding this: \[ |\vec{c}|^2 = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \] Since \( |\vec{c}| = 2 \), we know \( |\vec{c}|^2 = 4 \). Thus: \[ 4 = \vec{a} \cdot \vec{a} + 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \] Substitute the values of \( \vec{a} \cdot \vec{a} = 9 \) and \( \vec{b} \cdot \vec{b} = 16 \): \[ 4 = 9 + 2 \vec{a} \cdot \vec{b} + 16 \] Simplifying: \[ 4 = 25 + 2 \vec{a} \cdot \vec{b} \] \[ 2 \vec{a} \cdot \vec{b} = 4 - 25 = -21 \] \[ \vec{a} \cdot \vec{b} = -\frac{21}{2} \] Step 5: Final calculation.
Now substitute \( \vec{a} \cdot \vec{b} = -\frac{21}{2} \) into the earlier expression: \[ -9 - 16 - \vec{a} \cdot \vec{b} = -9 - 16 + \frac{21}{2} \] \[ = -25 + \frac{21}{2} = \frac{-50 + 21}{2} = \frac{-29}{2} \] Final Answer: The value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) is \( \frac{-29}{2} \).