Question

If \( |\vec{a}| = 12 \) and the projection of \( \vec{a} \) on \( \vec{b} \) is \( 6\sqrt{3} \), then the angle between \( \vec{a} \) and \( \vec{b} \) is

• A. \( \frac{\pi}{2} \)
• B. \( \frac{\pi}{6} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{2\pi}{3} \)
• E. \( \frac{3\pi}{4} \)

Hint:

Projection magnitude = \( |\vec{a}| \cos\theta \) (no need for \( \vec{b} \) magnitude).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The projection of a vector \(\vec{a}\) on another vector \(\vec{b}\) (also known as the scalar projection or component of \(\vec{a}\) along \(\vec{b}\)) is a measure of how much of vector \(\vec{a}\) points in the direction of \(\vec{b}\).
Step 2: Key Formula or Approach:
The scalar projection of \(\vec{a}\) on \(\vec{b}\) is given by the formula: \[ \text{Projection} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \] Using the definition of the dot product, \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta\), this simplifies to: \[ \text{Projection} = \frac{|\vec{a}||\vec{b}|\cos\theta}{|\vec{b}|} = |\vec{a}|\cos\theta \] where \(\theta\) is the angle between the vectors.
Step 3: Detailed Explanation:
We are given:
The magnitude of \(\vec{a}\) is \(|\vec{a}| = 12\).
The projection of \(\vec{a}\) on \(\vec{b}\) is \(6\sqrt{3}\).
Using the formula \(\text{Projection} = |\vec{a}|\cos\theta\), we can set up the equation: \[ 6\sqrt{3} = 12 \cos\theta \] Now, we solve for \(\cos\theta\): \[ \cos\theta = \frac{6\sqrt{3}}{12} \] \[ \cos\theta = \frac{\sqrt{3}}{2} \] We need to find the angle \(\theta\) (assuming \(0 \le \theta \le \pi\)) for which \(\cos\theta = \frac{\sqrt{3}}{2}\). The principal value is: \[ \theta = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6} \text{ radians} \] Step 4: Final Answer:
The angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\).