Question

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a} \times \vec{b} = \vec{c}$, $\vec{a} \cdot \vec{c} = 2$ and $\vec{b} \cdot \vec{c} = 1$. If $|\vec{b}| = 1$, then the value of $|\vec{a}|$ is:
 

• A. 2
• B. 1
• C. 4
• D. 3

Hint:

When dealing with cross and dot products, remember the relationships between the magnitudes and angles of vectors.

Answer 1

The Correct Option is A

We are given that:
\[\begin{array}{rl} \bullet & \text{$\vec{a} \times \vec{b} = \vec{c}$, which means the vector $\vec{c}$ is perpendicular to both $\vec{a}$ and $\vec{b}$.} \\ \bullet & \text{$\vec{a} \cdot \vec{c} = 2$, indicating the dot product between $\vec{a}$ and $\vec{c}$ is 2.} \\ \bullet & \text{$\vec{b} \cdot \vec{c} = 1$, indicating the dot product between $\vec{b}$ and $\vec{c}$ is 1.} \\ \bullet & \text{$|\vec{b}| = 1$, so the magnitude of $\vec{b}$ is 1.} \\ \end{array}\] From the cross product $\vec{a} \times \vec{b} = \vec{c}$, we know that $|\vec{c}| = |\vec{a}| |\vec{b}| \sin \theta$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. Since $|\vec{b}| = 1$, this simplifies to: \[ |\vec{c}| = |\vec{a}| \sin \theta \] Additionally, from the dot product $\vec{a} \cdot \vec{c} = 2$, we have: \[ |\vec{a}| |\vec{c}| \cos \theta = 2 \] Now, substituting $|\vec{c}| = |\vec{a}| \sin \theta$ into this equation: \[ |\vec{a}|^2 \sin \theta \cos \theta = 2 \] Using the identity $\sin 2\theta = 2 \sin \theta \cos \theta$, we get: \[ \frac{1}{2} |\vec{a}|^2 \sin 2\theta = 2 \] \[ |\vec{a}|^2 \sin 2\theta = 4 \] Thus, the value of $|\vec{a}|$ is 2.