Question

If $\vec{a}$, $\vec{b}$, $\vec{c}$ are three coplanar vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 2$, $\vec{b} \cdot \vec{c} = 8$, the angle between $\vec{b}$ and $\vec{c}$ is $45^\circ$, then $|\vec{a} \times (\vec{b} \times \vec{c})| = $ ______.

• A. 8
• B. $4\sqrt{2}$
• C. $\sqrt{2}$
• D. $8\sqrt{2}$

Hint:

Using geometric intuition is faster than using the Vector Triple Product expansion ($\vec{a}\times(\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$). Realizing that $\vec{a}$ is $90^\circ$ to $\vec{b}\times\vec{c}$ instantly collapses the problem!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We use the Vector Triple Product formula: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.

Step 2: Formula Application:
First find $|\vec{c}|$: $\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos 45^\circ = 8 \implies 2 \cdot |\vec{c}| \cdot \frac{1}{\sqrt{2}} = 8 \implies |\vec{c}| = 4\sqrt{2}$.

Step 3: Explanation:
The cross product $\vec{b} \times \vec{c}$ is a vector perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. Since $\vec{a}$ is coplanar with them, $\vec{a}$ is perpendicular to $(\vec{b} \times \vec{c})$. The magnitude $|\vec{a} \times \vec{v}| = |\vec{a}||\vec{v}|\sin 90^\circ$ where $\vec{v} = \vec{b} \times \vec{c}$. $|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}|\sin 45^\circ = 2 \cdot 4\sqrt{2} \cdot \frac{1}{\sqrt{2}} = 8$. Result $= 1 \times 8 \times 1 = 8$.

Step 4: Final Answer:
The magnitude is 8.