Let $\vec{a} = \vec{i} + 2\vec{j} + 3\vec{k}, \vec{b} = 2\vec{i} + 3\vec{j} + \vec{k}$ and $\vec{l} = \vec{i} + \vec{m}d$. If $(\vec{a} \times \vec{b}) = (\vec{a} \times \vec{c}) \times \vec{b}$. If $\vec{a} = \vec{0}$, then $|\vec{l}|$ is equal to
Step-by-Step Solution
Key Concept: The magnitude of a cross product relates to the scalar triple product, and vector triple product properties govern the relationships between vectors.
Using the condition $\vec{a} \cdot \vec{b} = 2, |\vec{a}|^2 = 5, |\vec{b}| = 1$, we compute $|\vec{a} \times \vec{b}| = \sqrt{|\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2} = \sqrt{5 \cdot 1 - 4} = 1$. From the given equation $(\vec{a} \times \vec{b}) \cdot \vec{a} = 11(\vec{c} \cdot \vec{a})$, we extract relationships using the vector triple product formula. Computing the cross product components yields $|\vec{x}| = 2\sqrt{3} \approx 3.46$.
Correct Answer: 3.46