Question

The modulus of two vectors \(\vec{a}\) and \(\vec{b}\) are \(\sqrt{3}\) and 4 respectively, and \(\vec{a} \cdot \vec{b} = 6\). Then find the angle between the vectors \(\vec{a}\) and \(\vec{b}\).

Hint:

The dot product formula is a crucial tool for problems involving angles between vectors. Remember \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\). If the dot product is positive, the angle is acute; if negative, obtuse; if zero, the vectors are orthogonal.

Answer 1

Step 1: Understanding the Concept:
The dot product (or scalar product) of two vectors is a scalar quantity that relates the magnitudes of the vectors and the cosine of the angle between them. This relationship can be used to find the angle if the magnitudes and the dot product are known.
Step 2: Key Formula or Approach:
The formula for the dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \] where \(|\vec{a}|\) and \(|\vec{b}|\) are the magnitudes (modulus) of the vectors and \(\theta\) is the angle between them. We can rearrange this formula to solve for \(\cos\theta\).
Step 3: Detailed Explanation:
We are given the following information:
Modulus of \(\vec{a}\): \(|\vec{a}| = \sqrt{3}\)
Modulus of \(\vec{b}\): \(|\vec{b}| = 4\)
Dot product: \(\vec{a} \cdot \vec{b} = 6\)
We substitute these values into the dot product formula: \[ 6 = (\sqrt{3})(4) \cos\theta \] Now, we solve for \(\cos\theta\): \[ \cos\theta = \frac{6}{4\sqrt{3}} \] Simplify the fraction: \[ \cos\theta = \frac{3}{2\sqrt{3}} \] To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{3}\), or recognize that \(3 = \sqrt{3} \times \sqrt{3}\): \[ \cos\theta = \frac{\sqrt{3} \times \sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \] Now, we find the angle \(\theta\) whose cosine is \(\frac{\sqrt{3}}{2}\). The principal value for the angle is: \[ \theta = \arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \text{ radians} \] This is equivalent to 30°.
Step 4: Final Answer:
The angle between the vectors \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\) or 30°.