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Lines, Planes, and Vectors

$ \newcommand{\vecb}[1]{{\bf #1}} $ In this tutorial, we will use vector methods to represent lines and planes in 3-space.

Displacement Vector

The displacement vector $\vecb{v}$ with initial point $(x_{1},y_{1},z_{1})$ and terminal point $(x_{2},y_{2},z_{2})$ is $$ \vecb{v}=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1}). $$

Why?

That is, if vector $\vecb{v}$ were positioned with its initial point at the origin, then its terminal point would be at $(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})$.

Example

The vector $\vecb{v}$ with initial point $(-1,4,5)$ and final point $(4,-3,2)$ is $$ \vecb{v} = \left( 4-(-1),-3-4,2-5 \right) = (5,-7,-3). $$

Parametric Equations for a Line in 3-space

The line throught the point $(x_{0},y_{0},z_{0})$ and parallel to the non-zero vector $\vecb{v} = (a,b,c)$ has parametric equations \begin{eqnarray*} x & = & x_{0} + at \\ y & = & y_{0} + bt \\ z & = & z_{0} + ct. \end{eqnarray*}

Why?

Example

The line through $(2,-1,3)$ and parallel to the vector $\vecb{v}=(3,-7,4)$ has parametric equations \begin{eqnarray*} x &= & 2+3t \\ y & =& -1-7t \\ z & =& 3+4t. \end{eqnarray*} Notice that when $t=0$, we are at the point $(2,-1,3)$. As $t$ increases or decreases from 0, we move away from this point parallel to the direction indicated by $(3,-7,4)$.

If you know two points $p_{1} = (x_{1},y_{1},z_{1})$ and $p_{2}=(x_{2},y_{2},z_{2})$ that a line passes through, you can find a parametrization for the lilne. First, find the displacement vector $\vecb{v}=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})$. then write down parametric equations for the line through either $p_{1}$ or $p_{2}$ and parallel to $\vecb{v}$.

Equation of a Plane in 3-space

The equation of the plane containing the point $(x_{0},y_{0},z_{0})$ with normal vector $\vecb{n} = (a,b,c)$ is $$ a(x-x_{0})+ b(y-y_{0})+c(z-z_{0})=0. $$

Why?

Thus, the graph of the equation $$ ax+by+cz=d $$ is a plane with normal vector $ (a,b,c)$.

Example

The equation of the plane containing $(2,4,-1)$ and normal to the vector $\vecb{n} = (3,5,-2)$ is $$ 3(x-2)+5(y-4)-2(z-(-1))=0. $$ Simplifying, $$ 3x+5y-2z=28. $$ With a little extra work, we can use this procedure to find the equation of the plane defined by any thee points. First, compute displacement vectors $\vecb{u}$ and $\vecb{v}$ between two pairs of these points. Then $\vecb{n} = \vecb{u} \times \vecb{v}$ in normal to the plane. Now, use one of the points and the vector $\vecb{n} = \vecb{u} \times \vecb{v}$ to obtain the equation of the plane.


Key Concepts

  • Displacement Vector

    The displacement vector $\vecb{v}$ with initial point $(x_{1},y_{1},z_{1})$ and terminal point $(x_{2},y_{2},z_{2})$ is $\vecb{v}=(x_{2}-x_{1},y_{2}-y_{1},z_{2}-z_{1})$.

  • Parametric Equations for a line in 3-space

    The line throught he point $(x_{0},y_{0},z_{0})$ and parallel to the non-zero vector $\vecb{v} = (a,b,c)$ has parametric equations \begin{eqnarray*} x & = & x_{0} + at \\ y & = & y_{0} + bt \\ z & = & z_{0} + ct. \end{eqnarray*}

  • Equation of a plane in 3-space

    The equation of the plane containing the point $(x_{0},y_{0},z_{0})$ with normal vector $\vecb{n} = (a,b,c)$ is $$ a(x-x_{0})+ b(y-y_{0})+c(z-z_{0})=0. $$