Questions: What is the angle between vectors a and b if vector a has a magnitude of 4 and a direction of -30° and b has components <-2,-5>? Give your answer rounded to the nearest degree.

Solution Steps

To find the angle between two vectors, we can use the dot product formula. First, we need to express vector \( a \) in component form using its magnitude and direction. Then, we calculate the dot product of vectors \( a \) and \( b \). Finally, we use the dot product formula to find the cosine of the angle between the vectors and then use the inverse cosine function to find the angle itself.

The direction of vector \( a \) is given as \( -30^\circ \). To convert this to radians, we use the conversion factor \( \frac{\pi}{180} \): \[ \text{direction\_a\_radians} = -30 \times \frac{\pi}{180} = -\frac{\pi}{6} \approx -0.524 \]

Using the magnitude \( |a| = 4 \) and the direction in radians, we find the components of vector \( a \): \[ a_x = |a| \cdot \cos(\text{direction\_a\_radians}) = 4 \cdot \cos\left(-\frac{\pi}{6}\right) = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} \approx 3.464 \] \[ a_y = |a| \cdot \sin(\text{direction\_a\_radians}) = 4 \cdot \sin\left(-\frac{\pi}{6}\right) = 4 \cdot \left(-\frac{1}{2}\right) = -2 \] Thus, vector \( a \) can be expressed as: \[ a = \left[2\sqrt{3}, -2\right] \]

The components of vector \( b \) are given as \( b = \langle -2, -5 \rangle \). We calculate the dot product \( a \cdot b \): \[ \text{dot\_product} = a_x \cdot b_x + a_y \cdot b_y = (2\sqrt{3})(-2) + (-2)(-5) = -4\sqrt{3} + 10 \approx 3.072 \] Next, we calculate the magnitude of vector \( b \): \[ |b| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29} \approx 5.385 \]

Using the dot product and magnitudes, we find the cosine of the angle \( \theta \) between vectors \( a \) and \( b \): \[ \cos(\theta) = \frac{\text{dot\_product}}{|a| \cdot |b|} = \frac{3.072}{4 \cdot \sqrt{29}} \approx 0.143 \]

To find the angle \( \theta \), we take the inverse cosine: \[ \theta = \cos^{-1}(0.143) \approx 81.801 \] Rounding this to the nearest degree gives: \[ \theta \approx 82^\circ \]

Final Answer