Vector Algebra
Cross product of vectors
Grade 12
Question:
<p>We have \(|\vec{a}\times\vec{b} - \vec{a}\times\vec{c}|^2 = |\vec{a}\times(\vec{b}-\vec{c})|^2\). If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors and the angle between \(\vec{b}\) and \(\vec{c}\) is \(\dfrac{\pi}{3}\), and \(\vec{a}\cdot(\vec{b}-\vec{c})=0\), find \(|\vec{a}\times\vec{b}-\vec{a}\times\vec{c}|^2\).</p>
Step-by-Step Solution
Key Concept: Use the distributive property of cross product: $\vec{a}\times\vec{b} - \vec{a}\times\vec{c} = \vec{a}\times(\vec{b}-\vec{c})$, then apply the formula $|\vec{a}\times\vec{v}|^2 = |\vec{a}|^2|\vec{v}|^2 - (\vec{a}\cdot\vec{v})^2$ along with the orthogonality condition.
Step 1: Use the distributive property of cross product: $|\vec{a}\times\vec{b}-\vec{a}\times\vec{c}|^2 = |\vec{a}\times(\vec{b}-\vec{c})|^2$ Step 2: Apply the vector identity for magnitude of cross product: $|\vec{a}\times\vec{v}|^2 = |\vec{a}|^2|\vec{v}|^2 - (\vec{a}\cdot\vec{v})^2$ where $\vec{v} = \vec{b}-\vec{c}$ Step 3: Use the given condition $\vec{a}\cdot(\vec{b}-\vec{c})=0$: $|\vec{a}\times(\vec{b}-\vec{c})|^2 = |\vec{a}|^2|\vec{b}-\vec{c}|^2 - 0^2 = |\vec{a}|^2|\vec{b}-\vec{c}|^2$ Step 4: Since $\vec{a}$ is a unit vector, $|\vec{a}|^2 = 1$. Calculate $|\vec{b}-\vec{c}|^2$: $|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2\vec{b}\cdot\vec{c}$ $= 1 + 1 - 2(1)(1)\cos\frac{\pi}{3} = 2 - 2\cdot\frac{1}{2} = 2 - 1 = 1$ Step 5: Therefore: $|\vec{a}\times\vec{b}-\vec{a}\times\vec{c}|^2 = 1 \cdot 1 = 1$ ∴ Answer: 1
Correct Answer: 1