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 Question:
We are given geometric properties of three coplanar vectors. We must evaluate the magnitude of a specific vector triple product.

Step 2: Detailed Explanation:
First, let's determine the magnitude of vector $\vec{c}$ using the dot product formula:
$\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta$
$8 = (2) \times |\vec{c}| \times \cos(45^\circ)$
$8 = 2 \times |\vec{c}| \times \frac{1}{\sqrt{2}}$
$8 = \sqrt{2} |\vec{c}| \implies |\vec{c}| = \frac{8}{\sqrt{2}} = 4\sqrt{2}$
Now, let's analyze the vector expression: $\vec{v} = \vec{a} \times (\vec{b} \times \vec{c})$.
By definition, the cross product $(\vec{b} \times \vec{c})$ generates a vector that is perfectly perpendicular (normal) to the plane containing both $\vec{b}$ and $\vec{c}$.
The problem explicitly states that $\vec{a}, \vec{b}$, and $\vec{c}$ are coplanar.
This means vector $\vec{a}$ lies completely flat inside the exact same plane as $\vec{b}$ and $\vec{c}$.
Therefore, the vector $(\vec{b} \times \vec{c})$, being perpendicular to the entire plane, must naturally be perfectly perpendicular to vector $\vec{a}$ as well.
The angle ($\phi$) between $\vec{a}$ and $(\vec{b} \times \vec{c})$ is exactly $90^\circ$.
The magnitude of a cross product is given by $|\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \phi$.
So:
$|\vec{a} \times (\vec{b} \times \vec{c})| = |\vec{a}| \times |\vec{b} \times \vec{c}| \times \sin(90^\circ)$
Since $\sin(90^\circ) = 1$, we just need to find $|\vec{b} \times \vec{c}|$:
$|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin \theta$
$|\vec{b} \times \vec{c}| = (2) \times (4\sqrt{2}) \times \sin(45^\circ)$
$|\vec{b} \times \vec{c}| = 8\sqrt{2} \times \frac{1}{\sqrt{2}} = 8$
Substitute this back into our magnitude equation:
$|\vec{a} \times (\vec{b} \times \vec{c})| = (1) \times (8) \times (1) = 8$.

Step 3: Final Answer:
The magnitude is 8, matching option (a).