Question

If \( \mathbf{a} \) and \( \mathbf{b} \) are two non-zero vectors such that the angle between them is \( 60^\circ \), what is the probability that the dot product \( \mathbf{a} \cdot \mathbf{b} \) is positive?

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{3} \)
• C. 1
• D. \( \frac{1}{4} \)

Hint:

Remember that the dot product of two vectors is positive when the angle between them is less than \( 90^\circ \). The probability can be calculated by comparing the favorable angle range with the total possible angle range.

Answer 1

The Correct Option is C

Step 1: Recall the formula for the dot product of two vectors:

The dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is given by:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]

Where:

• \( \mathbf{a} \) and \( \mathbf{b} \) are the two vectors,
• \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors,
• \( \theta \) is the angle between the vectors.

 

Step 2: Substitute the known values into the formula:

The given angle between the vectors is \( \theta = 60^\circ \), so we substitute this into the formula:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos 60^\circ \]

Since \( \cos 60^\circ = \frac{1}{2} \), the dot product becomes:

\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \times \frac{1}{2} \]

Step 3: Determine when the dot product is positive:

The dot product \( \mathbf{a} \cdot \mathbf{b} \) is positive when the cosine of the angle is positive. For \( \cos \theta = \frac{1}{2} \) at \( \theta = 60^\circ \), the dot product is positive because \( |\mathbf{a}| |\mathbf{b}| \) is always positive (as long as both vectors are non-zero). Therefore, the dot product \( \mathbf{a} \cdot \mathbf{b} \) is positive when \( \theta = 60^\circ \).

Conclusion:

The probability that the dot product \( \mathbf{a} \cdot \mathbf{b} \) is positive is \( \mathbf{1} \), since for the angle \( 60^\circ \), the dot product is always positive.