## Vector Equations of a Plane [9951407c5570] ### Parametric Form: r = a + λb + μc [33de4cc5081f] The **vector equation of a plane** in its **parametric **form is given by: $r = a + λb + μc$ Where: * $r$ is the position vector of any point on the plane * $a$ is the position vector of a fixed point on the plane * $b$ and $c$ are non-parallel vectors lying within the plane * $λ$ and $μ$ are scalar parameters > This equation represents all points on the plane as a **linear combination** of two non-parallel vectors ($b$ and $c$) added to a fixed point ($a$). [{"_key":"SKegYEUhosFtAcmMhjG6MX","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG6Ph","_type":"span","marks":[],"text":"The vectors $b$ and $c$ must be non-parallel to ensure they span the entire plane."}],"markDefs":[],"style":"normal"}] #### Interpretation [5fc334134fd6] * The vector $a$ **anchors **the plane at a specific point in space. * Vectors $b$ and $c$ define the "**directions**" in which the plane extends. * By varying $λ$ and $μ$, we can reach any point on the plane. [{"_key":"12bd4a569349","_type":"block","asset":null,"children":[{"_key":"c30101571a570","_type":"span","marks":[],"text":"Let \$\\mathbf{a} = (1,2,3) \$, \$\\mathbf{b} = (1,0,1) \$, and \$\\mathbf{c} = (0,1,-1) \$. The equation of the plane is: \\[\\mathbf{r} = (1,2,3) + \\lambda (1,0,1) + \\mu (0,1,-1) \\] This can be written in component form as: \\[\\begin{cases} x = 1 + \\lambda\\\\ y = 2 + \\mu\\\\ z = 3 + \\lambda - \\mu\\end{cases} \\]"}],"markDefs":[],"style":"normal"}] [{"_key":"40b9a114336f","_type":"block","asset":null,"children":[{"_key":"6480f9e559420","_type":"span","marks":["strong"],"text":"Understanding Vector Equations: Lines vs. Planes"}],"markDefs":[],"style":"h4"},{"_key":"df9b0afb2dea","_type":"block","asset":null,"children":[{"_key":"ba14b66607780","_type":"span","marks":[],"text":"A vector line equation extends a point "},{"_key":"ba14b66607781","_type":"span","marks":["strong"],"text":"one-dimensionally"},{"_key":"ba14b66607782","_type":"span","marks":[],"text":" using a "},{"_key":"ba14b66607783","_type":"span","marks":["strong"],"text":"single direction vector"},{"_key":"ba14b66607784","_type":"span","marks":[],"text":", allowing movement along that direction. "}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"},{"_key":"9e0740b026ca","_type":"block","asset":null,"children":[{"_key":"4b5707b10081","_type":"span","marks":[],"text":"In contrast, a vector plane equation extends a point "},{"_key":"ba14b66607785","_type":"span","marks":["strong"],"text":"two-dimensionally"},{"_key":"ba14b66607786","_type":"span","marks":[],"text":" using "},{"_key":"ba14b66607787","_type":"span","marks":["strong"],"text":"two direction vectors"},{"_key":"ba14b66607788","_type":"span","marks":[],"text":", defining a surface rather than a single path."}],"level":1,"listItem":"bullet","markDefs":[],"style":"normal"}] ### Normal Form: (r-a) • n = 0 [929aabbef83f] Another way to represent a plane is using its **normal vector**: $(r-a) • n = 0$ This can be restructured to be: $r • n = a • n$ Where: * $r$ is the position vector of any point on the plane * $n$ is a vector normal (perpendicular) to the plane * $a$ is the position vector of any point on the plane > This equation states that the dot product of the normal vector with any vector from $a$ to a point on the plane ($r - a$) is zero, which is the definition of perpendicularity. [{"_key":"SKegYEUhosFtAcmMhjG6if","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG6lp","_type":"span","marks":[],"text":"To find a normal vector to a plane given in the form $r = a + λb + μc$, you can use the cross product: $n = b × c$"}],"markDefs":[],"style":"normal"}] #### Interpretation [a17306ff55c8] * The normal vector $n$ is **perpendicular **to every vector lying in the plane. * The equation ensures that the **projection **of $(r - a)$ onto $n$ is always **zero**. [{"_key":"SKegYEUhosFtAcmMhjG6oz","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG6s9","_type":"span","marks":[],"text":"Given a plane with normal vector $n = (2, -1, 3)$ passing through the point $(1, 4, -2)$:"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG6yT","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG71d","_type":"span","marks":[],"text":"$r • (2, -1, 3) = (1, 4, -2) • (2, -1, 3)$"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG77x","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7B7","_type":"span","marks":[],"text":"Simplifying the right side: $r • (2, -1, 3) = 2(1) + (-1)(4) + 3(-2) = -8$"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG7HR","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7Kb","_type":"span","marks":[],"text":"So the equation of the plane is: $2x - y + 3z = -8$"}],"markDefs":[],"style":"normal"}] ### Cartesian Equation: ax + by + cz = d [be083bf26cce] The **Cartesian equation of a plane** is: $ax + by + cz = d$ Where: * $(a, b, c)$ are components of the **normal vector** to the plane * $d$ is a constant > This form is directly related to the normal form, as $(a, b, c)$ corresponds to the normal vector $n$. #### Interpretation [dbf85a20433a] [{"_key":"SKegYEUhosFtAcmMhjG7Nl","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7Qv","_type":"span","marks":[],"text":"Students often confuse the coefficients $(a, b, c)$ with a point on the plane. Remember, these form the normal vector, not a point!"}],"markDefs":[],"style":"normal"}] #### Converting Between Forms [d6f204824bce] To **convert **from the normal vector form to the cartesian form: 1. Expand $(r - a)• n = 0$ to $r • n = a • n$ 2. Write $r$ in its component form of $(x, y, z)$ 3. Simplify [{"_key":"SKegYEUhosFtAcmMhjG7U5","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7XF","_type":"span","marks":[],"text":"Converting $r • (2, -1, 3) = -4$ to Cartesian form:"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG7dZ","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7gj","_type":"span","marks":[],"text":"$(x, y, z) • (2, -1, 3) = -4$ $2x - y + 3z = -4$"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG7n3","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7qD","_type":"span","marks":[],"text":"This is already in the Cartesian form $ax + by + cz = d$"}],"markDefs":[],"style":"normal"}] ### Applications and Problem Solving [aaa3985bdedc] Understanding these different forms allows for solving various geometric problems involving planes: 1. Finding the **intersection **of planes 2. Determining if a point **lies on a plane** 3. Calculating the **distance **from a point to a plane 4. Finding the **angle **between two planes [{"_key":"SKegYEUhosFtAcmMhjG7tN","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG7wX","_type":"span","marks":[],"text":"To find if a point $P(2, 1, 3)$ lies on the plane $2x - y + 3z = 5$:"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG82r","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG861","_type":"span","marks":[],"text":"Substitute the coordinates into the equation: $2(2) - 1 + 3(3) = 5$ $4 - 1 + 9 = 12$ $12 ≠ 5$"}],"markDefs":[],"style":"normal"},{"_key":"SKegYEUhosFtAcmMhjG8CL","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG8FV","_type":"span","marks":[],"text":"Therefore, the point does not lie on the plane."}],"markDefs":[],"style":"normal"}] [{"_key":"SKegYEUhosFtAcmMhjG8If","_type":"block","asset":null,"children":[{"_key":"SKegYEUhosFtAcmMhjG8Lp","_type":"span","marks":[],"text":"The ability to switch between different representations of a plane is crucial for solving complex problems efficiently."}],"markDefs":[],"style":"normal"}]