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1.6: Lines and Planes

  • Page ID
    598
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    Lines

    Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line. The figure (shown in 2D for simplicity) shows that if P is a point on the line then

    \[ \langle x,y \rangle = P + tv\nonumber \]

    for some number \(t\).

    linpln20.gif

    The picture is the same for 3D. The formula is given below.

    Parametric Equations of a Line

    The parametric equations for the line through the point \((a,b,c)\) and parallel to the vector v are

    \[\langle x,y,z\rangle = \langle a,b,c\rangle + t\textbf{v}.\nonumber \]

    Example \(\PageIndex{1}\)

    Find the parametric equations of the line that passes through the point \((1, 2, 3)\) and is parallel to the vector \(\langle 4, -2, 1\rangle\).

    Solution

    We write:

    \[\langle x, y, z\rangle = \langle 1, 2, 3\rangle + t \langle 4, -2, 1\rangle = \langle 1 + 4t, 2 - 2t, 3 + t\rangle \nonumber \]

    or

    \[ x(t) = 1 + 4t,\nonumber \]

    \[y(t) = 2 - 2t,\nonumber \]

    \[z(t) = 3 + t.\nonumber \]

    Exercise \(\PageIndex{1}\)

    Find the parametric equations of the line through the two points \((2,1,7)\) and \((1,3,5)\).

    Hint: a vector parallel to the line has tail at \((2,1,7)\) and head at \((1,3,5)\).

    Planes

    If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane. Suppose that n is a normal vector to a plane and \((a,b,c)\) is a point on the plane. Let \((x,y,z)\) be a general point on the plane, then

    \[ \langle x - a, y - b, z - c\rangle \nonumber \]

    is parallel to the plane, hence

    \[\vec{n} \cdot \langle x - a, y - b, z - c\rangle = 0.\nonumber \]

    This defines the equation of the plane.

    linpln21.gif

    Example \(\PageIndex{2}\)

    Find the equation of the plane that contains the point \((2,1,0)\) and has normal vector \(\langle 1,2,3\rangle \).

    Solution

    We have

    \[\langle 1,2,3\rangle \cdot \langle x - 2,y - 1,z - 0\rangle = 0\nonumber \]

    so that

    \[1(x - 2) + 2(y - 1) + 3z = 0\nonumber \]

    or

    \[ x + 2y + 3z = 4.\nonumber \]

    Example \(\PageIndex{3}\)

    Find the equation of the plane through the points

    • \(P = (0,0,1)\)
    • \(Q = (2,1,0)\)
    • \(R = (1,1,1)\)

    Solution

    Let

    \[\textbf{v} = Q-P = \langle 2,1,-1\rangle \nonumber \]

    and

    \[\textbf{w} = R-P = \langle 1,1,0\rangle \nonumber \]

    then to find a vector normal to the plane, we find the cross product of \(v\) and \(w\):

    \[v \times w = \begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{i}} & \hat{\textbf{i}} \\ 2 & 1 &-1 \\ 1 &1 0 \end{vmatrix} = \hat{\textbf{i}} - \hat{\textbf{j}} + \hat{\textbf{k}} \nonumber \]

    or

    \[\langle 1, -1, 1\rangle .\nonumber \]

    We can now use the formula:

    \[\langle 1, -1, 1\rangle \cdot \langle x, y, z - 1\rangle = 0\nonumber \]

    or

    \[x - y + z - 1 = 0\nonumber \]

    or

    \[x - y + z = 1\nonumber \]

    Distance Between a Point and a Plane

    Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by

    Distance Between a Point \(P\) and a Plane With Normal Vector n

    Let \(Q\) be a point on the plane with normal vector \(\vec{n}\). The the distance from the point \(P\) to this plane is given by

    \[ Proj_nPQ = \dfrac{ ||PQ \cdot \vec{n}|| }{ || \vec{n}|| }.\nonumber \]

    Example \(\PageIndex{4}\)

    Find the distance between the point \((1,2,3)\) and the plane

    \[2x - y - 2z = 5.\nonumber \]

    Solution

    The normal vector can be read off from the equation as

    \[\vec{n} = \langle 2, -1, -2\rangle .\nonumber \]

    linpln25.gif

    Now find a convenient point on the plane such as \(Q = (0, -5, 0)\). We have

    \[Q = \langle -1, -7, -3\rangle \nonumber \]

    and

    \[ \vec{n} \cdot PQ = -2 + 7 + 6 = 11.\nonumber \]

    We find the magnitude of n by taking the square root of the sum of the squares. The sum is

    \[4 + 1 + 4 = 9\nonumber \]

    so

    \[|| \vec{n} || = 3.\nonumber \]

    Hence the distance from the point to the plane is \(\frac{11}{3}\).

    The Angle Between Two Planes

    The angle between two planes is given by the angle between the normal vectors.

    Example \(\PageIndex{5}\)

    Find the angle between the two planes

    \[ 3x - 2y + 5z = 1\nonumber \]

    and

    \[4x + 2y - z = 4.\nonumber \]

    We have the two normal vectors are

    \[ \vec{n} = \langle 3,-2,5\rangle \nonumber \]

    and

    \[\vec{m} = \langle 4,2,-1\rangle .\nonumber \]

    We have

    \[\vec{n} \cdot \vec{ m} = 3\nonumber \]

    \[ || \vec{n} || = \sqrt{38}\nonumber \]

    \[ || \vec{m} || = \sqrt{21}\nonumber \]

    hence the angle is

    \[ \cos^{-1} \left(\dfrac{3}{\sqrt{38}\sqrt{21}} \right) = 1.46 \, rad.\nonumber \]

    Contributors and Attributions


    This page titled 1.6: Lines and Planes is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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