Compute $t$ such that $\langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = 0$.
["SEO-Optimized Article: Solving $\langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = 0$ — Step-by-Step Explanation", "---", "Understanding When Two Vectors Are Orthogonal: A Step-by-Step Guide", "When studying linear algebra or vector geometry, one common task is determining when two vectors are orthogonal (perpendicular). Orthogonality occurs when their dot product equals zero. In this article, we’ll solve the equation:", "$$\n\langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = 0\n$$", "to find all values of $ t $ that make the vectors orthogonal. This problem is a great example of applying vector dot product properties in real-world mathematical modeling.", "---", "### What Does the Dot Product Represent?", "The dot product of two vectors $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$ is defined as:", "$$\n\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2\n$$", "Geometrically, it measures the projection of one vector onto another and equals zero when the vectors are perpendicular.", "---", "### Set Up the Equation", "We compute the dot product of $\langle 2t, t^2 \rangle$ and $\langle 1, -1 \rangle$:", "$$\n\langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = (2t)(1) + (t^2)(-1) = 2t - t^2\n$$", "Set this equal to zero:", "$$\n2t - t^2 = 0\n$$", "---", "### Solve the Quadratic Equation", "Rewriting the equation:", "$$\n-t^2 + 2t = 0\n$$", "Factor out $ -t $:", "$$\n-t(t - 2) = 0\n$$", "Set each factor equal to zero:", "$$\n-t = 0 \quad \ ext{or} \quad t - 2 = 0\n$$", "$$\nt = 0 \quad \ ext{or} \quad t = 2\n$$", "---", "### Interpret the Solution", "The values $ t = 0 $ and $ t = 2 $ are the solutions to the equation. These are the values for which the vectors $ \langle 2t, t^2 \rangle $ and $ \langle 1, -1 \rangle $ are orthogonal.", "- When $ t = 0 $: the vector becomes $ \langle 0, 0 \rangle $, which is technically orthogonal to every vector, though trivial.\n- When $ t = 2 $: the vector becomes $ \langle 4, 4 \rangle $, and its dot product with $ \langle 1, -1 \rangle $ is:", "$$\n\langle 4, 4 \rangle \cdot \langle 1, -1 \rangle = 4(1) + 4(-1) = 4 - 4 = 0\n$$", "So both values satisfy the orthogonality condition.", "---", "### Practical Use and Summary", "This kind of equation appears in physics (e.g., force and displacement vectors), computer graphics (checking perpendicularity of motion paths), and optimization problems.", "Final Answer:\n$$\n\boxed{t = 0 \quad \ ext{or} \quad t = 2}\n$$", "These are the values for which $ \langle 2t, t^2 \rangle $ is orthogonal to $ \langle 1, -1 \rangle $.", "---", "Keywords:\n- $ \langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = 0 $\n- orthogonal vectors\n- dot product equation\n- solve for $ t $\n- vector geometry tutorial\n- quadratic equation in vectors", "Meta Description:\nLearn how to solve $ \langle 2t, t^2 \rangle \cdot \langle 1, -1 \rangle = 0 $ by computing the dot product, solving a quadratic, and interpreting the geometric meaning — essential for students and professionals in math, physics, and engineering.", "---", "Table of Contents\n1. Introduction to Vector Orthogonality\n2. Dot Product Definition\n3. Step-by-Step Solution\n4. Interpretation of Solutions\n5. Real-World Applications\n6. Summary", "---", "Clear, structured, and SEO-friendly content ensures better readability and increased search visibility—ideal for students seeking help with vector calculus and related topics."]
