Let's solve each part step by step, providing both the explanation and the final answer, including sample equations and sketches (where possible using text). --- ## **In \(\mathbb{R}^2\):** ### **(a) Sketch two random lines, label equations, and intersection** **Step 1:** Choose two random linear equations in \(x_1, x_2\). Let’s pick: - \(L_1: x_2 = 2x_1 + 1\) - \(L_2: x_2 = -x_1 + 4\) **Step 2:** Find the intersection: Set \(2x_1 + 1 = -x_1 + 4\): \[ 2x_1 + 1 = -x_1 + 4 \implies 3x_1 = 3 \implies x_1 = 1 \] Plug back into \(L_1\): \(x_2 = 2(1) + 1 = 3\) **Intersection point:** \((1, 3)\) **Final answer:** - Sketch two lines, label as \(x_2 = 2x_1 + 1\) and \(x_2 = -x_1 + 4\), mark intersection at \((1, 3)\). --- ### **(b) Sketch one random line and a parallel line, label equations** **Step 1:** Pick a line and a parallel line (same slope, different intercept). Let’s take: - \(L_1: x_2 = x_1 + 2\) - \(L_2: x_2 = x_1 - 1\) **Final answer:** - Sketch both lines, label as \(x_2 = x_1 + 2\) and \(x_2 = x_1 - 1\). They are parallel (slope = 1). --- ## **In \(\mathbb{R}^3\):** ### **(a) Use equations from \(\mathbb{R}^2\) part (a), now in 3D (planes)** Let’s use: - \(P_1: x_2 = 2x_1 + 1\) \(\implies x_2 - 2x_1 - 1 = 0\) - \(P_2: x_2 = -x_1 + 4\) \(\implies x_2 + x_1 - 4 = 0\) Now, **in 3D**, these are planes: - \(x_2 - 2x_1 - 1 = 0\) - \(x_2 + x_1 - 4 = 0\) **Step 2:** Find line of intersection: Solve the system: \[ \begin{cases} x_2 - 2x_1 - 1 = 0 \\ x_2 + x_1 - 4 = 0 \end{cases} \] Subtract the first from the second: \[ [x_2 + x_1 - 4] - [x_2 - 2x_1 - 1] = (x_1 + 2x_1) + (-4 + 1) = 3x_1 - 3 = 0 \implies x_1 = 1 \] Plug \(x_1 = 1\) into either equation: \(x_2 = 2x_1 + 1 = 2(1) + 1 = 3\) So, for any \(x_3\): \[ (x_1, x_2, x_3) = (1, 3, t), \quad t \in \mathbb{R} \] **Final answer:** - Sketch two planes \(x_2 - 2x_1 - 1 = 0\) and \(x_2 + x_1 - 4 = 0\). - Label the line of intersection: \((1, 3, t)\). --- ### **(b) Plane \(x_1 = 0\) (back wall)** This is the plane where all points have \(x_1 = 0\). **Final answer:** - Sketch the \(x_2 x_3\)-plane (vertical plane at \(x_1 = 0\)), label as \(x_1 = 0\). --- ### **(c) Planes \(x_3 = 0\) and \(x_1 = 0\), and their intersection** - \(x_3 = 0\): the \(x_1 x_2\)-plane (the "floor"). - \(x_1 = 0\): the \(x_2 x_3\)-plane (the "back wall"). **Their intersection:** Both \(x_1 = 0\) and \(x_3 = 0\) \(\implies\) line where \(x_1 = 0, x_3 = 0\), i.e., the \(x_2\)-axis. **Final answer:** - Sketch both planes, label as \(x_1 = 0\) and \(x_3 = 0\). - Their intersection is the \(x_2\)-axis (\((0, t, 0)\), \(t \in \mathbb{R}\)), label it. --- ### **(d) Line through origin not in any coordinate plane (include 3 points)** Choose a direction not parallel to any axis. Pick parametric: \((1, 2, 3)t\). Three points: - At \(t = 0\): \((0, 0, 0)\) - At \(t = 1\): \((1, 2, 3)\) - At \(t = -1\): \((-1, -2, -3)\) **Final answer:** - Sketch line through \((0,0,0)\), \((1,2,3)\), and \((-1,-2,-3)\). Label these points. --- ### **(e) Plane \(2x_1 + 3x_2 + x_3 = 6\), at least 4 points** Find 4 points, none collinear, not all on coordinate axes/planes. Set two variables to zero for intercepts: - \(x_1\)-intercept: \(x_2 = 0, x_3 = 0 \implies 2x_1 = 6 \implies x_1 = 3\) → \((3,0,0)\) - \(x_2\)-intercept: \(x_1 = 0, x_3 = 0 \implies 3x_2 = 6 \implies x_2 = 2\) → \((0,2,0)\) - \(x_3\)-intercept: \(x_1 = 0, x_2 = 0 \implies x_3 = 6\) → \((0,0,6)\) Choose one more point (not on axes): Let \(x_1 = 1, x_2 = 1\): \[ 2(1) + 3(1) + x_3 = 6 \implies 5 + x_3 = 6 \implies x_3 = 1 \] So, \((1, 1, 1)\). **Final answer:** - Sketch plane \(2x_1 + 3x_2 + x_3 = 6\). - Plot and label points: \((3,0,0)\), \((0,2,0)\), \((0,0,6)\), \((1,1,1)\). --- ## **Summary Table** | Part | Equation(s) | Objects | Intersection/Points | |------|-------------|---------|---------------------| | 2D(a) | \(x_2 = 2x_1 + 1\), \(x_2 = -x_1 + 4\) | Lines | \((1,3)\) | | 2D(b) | \(x_2 = x_1 + 2\), \(x_2 = x_1 - 1\) | Parallel Lines | - | | 3D(a) | \(x_2 - 2x_1 - 1 = 0\), \(x_2 + x_1 - 4 = 0\) | Planes | Line: \((1,3,t)\) | | 3D(b) | \(x_1 = 0\) | Plane | - | | 3D(c) | \(x_1=0\), \(x_3=0\) | Planes | Line: \((0, t, 0)\) | | 3D(d) | \((t, 2t, 3t)\) | Line | \((0,0,0), (1,2,3), (-1,-2,-3)\) | | 3D(e) | \(2x_1 + 3x_2 + x_3 = 6\) | Plane | \((3,0,0), (0,2,0), (0,0,6), (1,1,1)\) | --- If you would like to see the **sketches** or have further clarification on any part, let me know!

Certainly! Below are detailed step-by-step instructions along with sketches for each part. Since I can't produce actual images directly here, I'll describe how you can draw each sketch accurately. You can follow these instructions to sketch on paper or using graphing software like R or GeoGebra. --- ## **In \(\mathbb{R}^2\):** ### **(a) Two random lines and their intersection** **Equations:** - \(L_1: x_2 = 2x_1 + 1\) - \(L_2: x_2 = -x_1 + 4\) **Steps to sketch:** 1. Draw axes \(x_1\) (horizontal) and \(x_2\) (vertical). 2. Plot \(L_1\): - When \(x_1=0\), \(x_2=1\). - When \(x_1=1\), \(x_2=3\). - Draw a straight line through these points. 3. Plot \(L_2\): - When \(x_1=0\), \(x_2=4\). - When \(x_1=1\), \(x_2=3\). - Draw this line. 4. Find intersection: - Set \(2x_1 + 1 = -x_1 + 4 \Rightarrow 3x_1=3 \Rightarrow x_1=1\). - \(x_2=2(1)+1=3\). - Mark the intersection point \((1,3)\). 5. Label each line with their equations. --- ### **(b) One line and a parallel line** **Equations:** - \(L_1: x_2 = x_1 + 2\) - \(L_2: x_2 = x_1 - 1\) **Steps:** 1. Draw axes. 2. Plot \(L_1\): - \(x_1=0\), \(x_2=2\). - \(x_1=1\), \(x_2=3\). 3. Plot \(L_2\): - \(x_1=0\), \(x_2=-1\). - \(x_1=1\), \(x_2=0\). 4. Draw both lines parallel, label with equations. 5. No intersection (parallel). --- ## **In \(\mathbb{R}^3\):** ### **(a) Two planes and their line of intersection** **Equations:** - \(P_1: x_2 = 2x_1 + 1\) \(\Rightarrow\) \(x_2 - 2x_1 - 1=0\) - \(P_2: x_2 = -x_1 + 4\) \(\Rightarrow\) \(x_2 + x_1 - 4=0\) **Steps to sketch:** 1. Draw the three axes: \(x_1\), \(x_2\), \(x_3\). 2. Sketch plane \(P_1\): - For various \(x_1\), find \(x_2\): e.g., at \(x_1=0\), \(x_2=1\). - Draw a plane passing through points \((0,1,0)\) and \((1,3,0)\). - Extend it in the \(x_3\) direction. 3. Sketch plane \(P_2\): - For \(x_1=0\), \(x_2=4\). - For \(x_1=1\), \(x_2=3\). - Draw this second plane. 4. Find the line of intersection: - From the algebra above, the intersection line is \((x_1, x_2) = (1,3)\) with any \(x_3\). - Draw a line through points \((1,3, t)\) for \(t \in \mathbb{R}\). --- ### **(b) Plane \(x_1=0\) (Back wall)** - Draw the plane parallel to the \(x_2 x_3\) axes at \(x_1=0\). - Label it. --- ### **(c) Two planes \(x_3=0\) and \(x_1=0\)** - \(x_3=0\): the \(x_1 x_2\) plane. - \(x_1=0\): the \(x_2 x_3\) plane. - Their intersection: the \(x_2\)-axis, a line along \((0, t, 0)\). - Sketch both planes intersecting along the line. --- ### **(d) Line passing through the origin not in coordinate planes** - Choose parametric form: \[ (x_1, x_2, x_3) = (t, 2t, 3t) \] - Plot points: - \(t=0\): (0,0,0) - \(t=1\): (1,2,3) - \(t=-1\): (-1,-2,-3) - Draw the line passing through these points. --- ### **(e) Plane \(2x_1 + 3x_2 + x_3=6\)** **Points to plot:** - \(x_1=3, x_2=0, x_3=0\) (intercept on \(x_1\)) - \(x_1=0, x_2=2, x_3=0\) (intercept on \(x_2\)) - \(x_1=0, x_2=0, x_3=6\) (intercept on \(x_3\)) - Additional point, e.g., \(x_1=1, x_2=1\): \[ 2(1)+3(1)+x_3=6 \Rightarrow 2+3+x_3=6 \Rightarrow x_3=1 \] So point: \((1,1,1)\). - Sketch the plane through these points: - Draw coordinate axes. - Mark these points clearly. - Connect them to visualize the plane. --- ## **Summary** - Use the axes to position your lines/planes. - Mark key points (intercepts, intersection points). - Label each object with its equation. - For 3D sketches, indicate the orientation of planes and lines clearly, and show intersections explicitly. --- **Tip:** Using graphing software like GeoGebra 3D can help produce accurate visualizations based on these instructions. --- If you'd like, I can also help generate R code to plot these objects for precise visualization!