Vector Algebra
Cross product magnitude
Grade None
Question:
<p>If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are unit vectors such that \(\vec{a}+2\vec{b}+2\vec{c}=\vec{0}\), then \(|\vec{a}\times\vec{c}|\) is equal to</p>
<p>\(\dfrac{\sqrt{15}}{4}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{15}{16}\)</p>
<p>\(\dfrac{\sqrt{15}}{16}\)</p>
Step-by-Step Solution
Key Concept: Use the constraint $\vec{a}+2\vec{b}+2\vec{c}=\vec{0}$ to express one vector in terms of others, then apply the dot product condition $|\vec{a}|=|\vec{b}|=|\vec{c}|=1$ to find $\vec{a}\cdot\vec{c}$, which determines $|\vec{a}\times\vec{c}|$.
Step 1: From $\vec{a}+2\vec{b}+2\vec{c}=\vec{0}$, we get $\vec{a}=-2(\vec{b}+\vec{c})$ Step 2: Take dot product with itself: $|\vec{a}|^2 = 4|\vec{b}+\vec{c}|^2$ Since $|\vec{a}|=1$: $1 = 4(|\vec{b}|^2 + 2\vec{b}\cdot\vec{c} + |\vec{c}|^2) = 4(1 + 2\vec{b}\cdot\vec{c} + 1) = 4(2 + 2\vec{b}\cdot\vec{c})$ Therefore: $\vec{b}\cdot\vec{c} = -\frac{3}{4}$ Step 3: From $\vec{a}=-2(\vec{b}+\vec{c})$, take dot product with $\vec{c}$: $\vec{a}\cdot\vec{c} = -2(\vec{b}\cdot\vec{c} + |\vec{c}|^2) = -2(-\frac{3}{4} + 1) = -2(\frac{1}{4}) = -\frac{1}{2}$ Step 4: Use $|\vec{a}\times\vec{c}|^2 = |\vec{a}|^2|\vec{c}|^2 - (\vec{a}\cdot\vec{c})^2 = 1\cdot1 - (-\frac{1}{2})^2 = 1 - \frac{1}{4} = \frac{3}{4}$ ∴ $|\vec{a}\times\vec{c}| = \frac{\sqrt{3}}{2}$
Correct Answer: A