Vector Algebra
Dot Product
Grade None
Question:
<p>\(\vec{a}, \vec{b}, \vec{c}\) are 3 vectors, such that \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), \(|\vec{a}| = 1\), \(|\vec{b}| = 2\), \(|\vec{c}| = 3\), then \(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}\) is equal to</p>
<p>0</p>
<p>\(-7\)</p>
<p>7</p>
<p>1</p>
Step-by-Step Solution
Key Concept: Since $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, square both sides to get $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$, which directly isolates the required dot product sum.
Step 1: Use the given constraint $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ Step 2: Square both sides: $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$ Step 3: Expand the left side: $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ Step 4: Substitute magnitudes: $1^2 + 2^2 + 3^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ Step 5: Simplify: $1 + 4 + 9 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ Step 6: Solve: $14 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$ $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = -7$ ∴ Answer: B
Correct Answer: B