Solution: The angle $\theta$ between $\mathbf{a}$ and $\mathbf{b}$ is given by $\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|}$. Compute the dot product: $2(1) + (-3)(4) = 2 - 12 = -10$. Compute magnitudes: $\|\mathbf{a}\| = \sqrt{2^2 + (-3)^2} = \sqrt{13}$, $\|\mathbf{b}\| = \sqrt{1^2 + 4^2} = \sqrt{17}$. Thus, $\cos\theta = \frac{-10}{\sqrt{13}\sqrt{17}}$. Rationalizing, $\theta = \arccos\left(-\frac{10}{\sqrt{221}}\right)$. $\boxed{\arccos\left(-\dfrac{10}{\sqr
["Understanding the Angle Between Two Vectors Using the Dot Product", "When analyzing geometric relationships in vector algebra, one essential concept is determining the angle $\ heta$ between two vectors $\mathbf{a}$ and $\mathbf{b}$. A fundamental formula in this context is:", "[\n\cos\ heta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}\n]", "This equation expresses the cosine of the angle between two vectors in terms of their dot product and magnitudes. Understanding this relationship not only clarifies geometric interpretations but also supports applications across physics, engineering, and data science.", "---", "### Step 1: Compute the Dot Product", "Given vectors $\mathbf{a} = \langle 2, -3 \rangle$ and $\mathbf{b} = \langle 1, 4 \rangle$, the dot product is calculated component-wise:", "[\n\mathbf{a} \cdot \mathbf{b} = (2)(1) + (-3)(4) = 2 - 12 = -10\n]", "---", "### Step 2: Calculate Vector Magnitudes", "The magnitude of a vector $\mathbf{v} = \langle x, y \rangle$ is given by $|\mathbf{v}| = \sqrt{x^2 + y^2}$.", "- For $\mathbf{a} = \langle 2, -3 \rangle$:", "[\n|\mathbf{a}| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\n]", "- For $\mathbf{b} = \langle 1, 4 \rangle$:", "[\n|\mathbf{b}| = \sqrt{1^2 + 4^2} = \sqrt{1 + 16} = \sqrt{17}\n]", "---", "### Step 3: Compute $\cos\ heta$", "Substitute the dot product and magnitudes into the formula:", "[\n\cos\ heta = \frac{-10}{\sqrt{13} \cdot \sqrt{17}} = \frac{-10}{\sqrt{221}}\n]", "---", "### Step 4: Determine the Angle", "Since $\cos\ heta = -\frac{10}{\sqrt{221}} < 0$, the angle $\ heta$ lies in the second quadrant, between $90^\circ$ and $180^\circ$.", "To express the angle precisely, we write:", "[\n\ heta = \arccos\left(-\dfrac{10}{\sqrt{221}}\right)\n]", "This exact form is preferred over decimal approximations when precision and symbolic clarity are required.", "---", "### Conclusion", "The angle $\ heta$ between vectors $\mathbf{a}$ and $\mathbf{b}$ is fully determined by:", "[\n\boxed{\arccos\left(-\dfrac{10}{\sqrt{221}}\right)}\n]", "This solution highlights how the dot product and magnitudes interrelate to reveal geometric properties, forming a cornerstone in vector geometry."]
