Question

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

• A. 8
• B. $\sqrt{2}$
• C. 2
• D. 5

Hint:

For coplanar $\vec{a}, \vec{b}, \vec{c}$, the magnitude of the triple product is simply $|\vec{a}| |\vec{b} \times \vec{c}|$.

Answer 1

The Correct Option is A

Step 1: Concept
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: Analysis
Alternatively, since $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$, the vector $(\vec{b} \times \vec{c})$ is perpendicular to the plane containing $\vec{a}$.
Thus, the angle between $\vec{a}$ and $(\vec{b} \times \vec{c})$ is $90^{\circ}$.

Step 3: Calculation
$|\vec{a} \times (\vec{b} \times \vec{c})| = |\vec{a}| |\vec{b} \times \vec{c}| \sin 90^{\circ} = 1 \times |\vec{b} \times \vec{c}|$.
From $\vec{b} \cdot \vec{c} = 8$: $|\vec{b}| |\vec{c}| \cos 45^{\circ} = 8 \implies 2 |\vec{c}| (1/\sqrt{2}) = 8 \implies |\vec{c}| = 4\sqrt{2}$.
$|\vec{b} \times \vec{c}| = |\vec{b}| |\vec{c}| \sin 45^{\circ} = 2 \times 4\sqrt{2} \times (1/\sqrt{2}) = 8$.

Step 4: Conclusion
Hence, the magnitude is 8. Final Answer: (A)