Vector Algebra
Scalar triple product
Grade 12
Question:
<p>We have <span>\((2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b})]\)</span>. If <span>\(|\vec{a}| = 1\)</span>, <span>\(|\vec{b}| = 1\)</span>, <span>\(\vec{a} \cdot \vec{b} = 1\)</span>, find the value of the expression.</p>
<p>\(-5\)</p>
<p>\(5\)</p>
<p>\(0\)</p>
<p>\(-5(\vec{a})^2(\vec{b})^2 + 5(\vec{a}\cdot\vec{b})^2\)</p>
Step-by-Step Solution
Key Concept: Use the scalar triple product property and the constraint that $\vec{a} \cdot \vec{b} = 1$ with $|\vec{a}| = |\vec{b}| = 1$, which implies $\vec{a} = \vec{b}$. This makes $(\vec{a} \times \vec{b}) = \vec{0}$, immediately collapsing the entire expression to zero.
Step 1: Analyze the constraint. Given $|\vec{a}| = 1$, $|\vec{b}| = 1$, and $\vec{a} \cdot \vec{b} = 1$. By the definition of dot product: $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 1 \cdot 1 \cdot \cos\theta = 1$ This gives $\cos\theta = 1$, so $\theta = 0°$, meaning $\vec{a} \parallel \vec{b}$ and $\vec{a} = \vec{b}$. Step 2: Evaluate the cross product. Since $\vec{a} = \vec{b}$: $(\vec{a} \times \vec{b}) = \vec{a} \times \vec{a} = \vec{0}$ Step 3: Evaluate the entire expression: $(2\vec{a} - \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} + 2\vec{b})] = (2\vec{a} - \vec{b}) \cdot [\vec{0} \times (\vec{a} + 2\vec{b})]$ $= (2\vec{a} - \vec{b}) \cdot \vec{0} = 0$ ∴ Answer: 0
Correct Answer: A