Calculus II - Equations of Planes

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Section 12.3 : Equations of Planes

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5. Determine if the plane given by \( - 3x + 2y + 7z = 9\) and the plane containing the points \(\left( { - 2,6,1} \right)\), \(\left( { - 2,5,0} \right)\) and \(\left( { - 1,4, - 3} \right)\) are parallel, orthogonal or neither.

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Let’s start off this problem by noticing that the vector \({\vec n_1} = \left\langle { - 3,2,7} \right\rangle \) will be normal to the first plane and it would be nice to have a normal vector for the second plane.

We know (Problem 1 from this section!) how to determine a normal vector given three points in the plane. Here is that work.

\[\begin{align*}P & = \left( { - 2,6,1} \right) & \hspace{0.5in} Q & = \left( { - 2,5,0} \right) & \hspace{0.5in}R & = \left( { - 1,4, - 3} \right)\\ \overrightarrow {QP} & = \left\langle {0,1,1} \right\rangle & \hspace{0.5in} \overrightarrow {RQ} & = \left\langle { - 1,1,3} \right\rangle & & \end{align*}\] \[{\vec n_2} = \overrightarrow {QP} \times \overrightarrow {RQ} = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\0&1&1\\{ - 1}&1&3\end{array}} \right|\,\,\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\0&1\\{ - 1}&1\end{array} = 3\vec i - \vec j - \vec i - \left( { - \vec k} \right) = 2\vec i - \vec j + \vec k\]

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

Now try to visualize the two planes and these normal vectors. What would the two planes look like if the two normal vectors where parallel to each other? What would the two planes look like if the two normal vectors were orthogonal to each other?

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If the two normal vectors are parallel to each other the two planes would also need to be parallel.

So, let’s take a quick look at the normal vectors. We can see that the third component of each vector have the same sign while the first and second components each have opposite signs.

Therefore, there is no number that we can multiply to \({\vec n_1}\) that will keep the sign on the third component the same and simultaneously changing the sign on the first and second components. This in turn means the two vectors can’t possibly be scalar multiples and this further means they cannot be parallel.

If the two normal vectors can’t be parallel then the two planes are not parallel.

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Now, if the two normal vectors are orthogonal the two planes will also be orthogonal.

So, a quick dot product of the two normal vectors gives,

\[{\vec n_1}\centerdot {\vec n_2} = - 1\]

The dot product is not zero and so the two normal vectors are not orthogonal. Therefore, the two planes are not orthogonal.