Question

If \( \vec{a}, \vec{b}, \vec{c} \) are three vectors such that \[ \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c}), \] \( |\vec{a}| = 1,\; |\vec{b}| = 4,\; |\vec{c}| = 2 \) and the angle between \( \vec{b} \) and \( \vec{c} \) is \( 60^\circ \), then find \( |\vec{a} \cdot \vec{c}| \):

• A. 4
• B. 1
• C. 2
• D. \( \dfrac{1}{2} \)

Hint:

When cross products of a common vector are proportional, the angles made with that vector are equal.

Answer 1

The Correct Option is B

To solve the given problem, we need to find \( |\vec{a} \cdot \vec{c}| \) given the information:

• \(\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})\) 
• \(|\vec{a}| = 1, |\vec{b}| = 4, |\vec{c}| = 2\)
• The angle between \(\vec{b}\) and \(\vec{c}\) is \(60^\circ\)

Let's start by simplifying the equation \(\vec{a} \times \vec{b} = 2(\vec{a} \times \vec{c})\).

Using the property of the vector cross product, we know:

• \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta\)

where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\). Similarly,

• \(|\vec{a} \times \vec{c}| = |\vec{a}| |\vec{c}| \sin \phi\)

where \(\phi\) is the angle between \(\vec{a}\) and \(\vec{c}\).

Substituting these into the given equation, we have:

\(|\vec{a}| |\vec{b}| \sin \theta = 2 |\vec{a}| |\vec{c}| \sin \phi\)

Given \(|\vec{a}| = 1\), this simplifies to:

\(4 \sin \theta = 4 \sin \phi \implies \sin \theta = \sin \phi\)

Hence, \(\theta = \phi\), indicating that vectors \(\vec{b}\) and \(\vec{c}\) are at the same angle with respect to \(\vec{a}\).

Now, to find \(|\vec{a} \cdot \vec{c}|\), we use the dot product property:

\(|\vec{a} \cdot \vec{c}| = |\vec{a}| |\vec{c}| \cos \phi\)

Using \(|\vec{a}| = 1\) and \(|\vec{c}| = 2\), this becomes:

\(= 2 \cos \phi\)

From the given data, the angle between \(\vec{b}\) and \(\vec{c}\) is \(60^\circ\), and since \(\theta = \phi\), we use \(\phi = 60^\circ\):

\(= 2 \times \cos 60^\circ = 2 \times \frac{1}{2} = 1\)

Therefore, \(|\vec{a} \cdot \vec{c}| = 1\).

Conclusion: The correct answer is 1.