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Section 12.3 : Equations of Planes

1. Write down the equation of the plane containing the points \(\left( {4, - 3,1} \right)\), \(\left( { - 3, - 1,1} \right)\) and \(\left( {4, - 2,8} \right)\).

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Start Solution

To make the work on this problem a little easier let’s “name” the points as,

\[P = \left( {4, - 3,1} \right)\hspace{0.5in}Q = \left( { - 3, - 1,1} \right)\hspace{0.5in}R = \left( {4, - 2,8} \right)\]

Now, we know that in order to write down the equation of a plane we’ll need a point (we have three so that’s not a problem!) and a vector that is normal to the plane.

Show Step 2

We’ll need to do a little work to get a normal vector.

First, we’ll need two vectors that lie in the plane and we can get those from the three points we’re given. Note that there are lots of possible vectors that we could use here. Here are the two that we’ll use for this problem.

\[\overrightarrow {PQ} = \left\langle { - 7,2,0} \right\rangle \hspace{0.75in}\overrightarrow {PR} = \left\langle {0,1,7} \right\rangle \] Show Step 3

Now, these two vectors lie in the plane and we know that the cross product of any two vectors will be orthogonal to both of the vectors. Therefore, the cross product of these two vectors will also be orthogonal (and so normal!) to the plane.

So, let’s get the cross product.

\[\vec n = \overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{ - 7}&2&0\\0&1&7\end{array}} \right|\,\,\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\{ - 7}&2\\0&1\end{array} = 14\vec i - 7\vec k - \left( { - 49\vec j} \right) = 14\vec i + 49\vec j - 7\vec k\]

Note that we used the “trick” discussed in the notes to compute the cross product here.

Show Step 4

Now all we need to do is write down the equation.

We have three points to choose form here. We’ll use the first point simply because it is the first point listed. Any of the others could also be used.

The equation of the plane is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{14\left( {x - 4} \right) + 49\left( {y + 3} \right) - 7\left( {z - 1} \right) = 0\hspace{0.5in} \to \hspace{0.5in}\,14x + 49y - 7z = - 98}}\]

Note that depending on your choice of vectors in Step 2, the order you chose to use them in the cross product computation in Step 3 and the point chosen here will all affect your answer. However, regardless of your choices the equation you get will be an acceptable answer provided you did all the work correctly.