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### Section 1-3 : Equations of Planes

8. Find the line of intersection of the plane given by \(3x + 6y - 5z = - 3\) and the plane given by \( - 2x + 7y - z = 24\).

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Start SolutionOkay, we know that we need a point and vector parallel to the line in order to write down the equation of the line. In this case neither has been given to us.

First let’s note that any point on the line of intersection must also therefore be in both planes and it’s actually pretty simple to find such a point. Whatever our line of intersection is it must intersect at least one of the coordinate planes. It doesn’t have to intersect all three of the coordinate planes but it will have to intersect at least one.

So, let’s see if it intersects the \(xy\)-plane. Because the point on the intersection line must also be in both planes let’s set \(z = 0\) (so we’ll be in the \(xy\)-plane!) in both of the equations of our planes.

Doing this gives,

\[\begin{align*}3x + 6y & = - 3\\ - 2x + 7y & = 24\end{align*}\] Show Step 2This is a simple system to solve so we’ll leave it to you to verify that the solution is,

\[x = - 5\hspace{0.5in}\,\,\,\,y = 2\]The fact that we were able to find a solution to the system from Step 1 means that the line of intersection does in fact intersect the \(xy\)-plane and it does so at the point \(\left( { - 5,2,0} \right)\). This is also then a point on the line of intersection.

Note that if the system from Step 1 didn’t have a solution then the line of intersection would not have intersected the \(xy\)-plane and we’d need to try one of the remaining coordinate planes.

Show Step 3Okay, now we need a vector that is parallel to the line of intersection. This might be a little hard to visualize, but if you think about it the line of intersection would have to be orthogonal to both of the normal vectors from the two planes. This in turn means that any vector orthogonal to the two normal vectors must then be parallel to the line of intersection.

Nicely enough we know that the cross product of any two vectors will be orthogonal to each of the two vectors. So, here are the two normal vectors for our planes and their cross product.

\[{\vec n_1} = \left\langle {3,6, - 5} \right\rangle \hspace{0.75in}{\vec n_2} = \left\langle { - 2,7, - 1} \right\rangle \] \[\begin{align*}{{\vec n}_1} \times {{\vec n}_2} & = \left| {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\3&6&{ - 5}\\{ - 2}&7&{ - 1}\end{array}} \right|\,\,\,\,\,\,\begin{array}{*{20}{c}}{\vec i}&{\vec j}\\3&6\\{ - 2}&7\end{array}\\ & = - 6\vec i + 10\vec j + 21\vec k - \left( { - 3\vec j} \right) - \left( { - 35\vec i} \right) - \left( { - 12\vec k} \right) = 29\vec i + 13\vec j + 33\vec k\end{align*}\]Note that we used the “trick” discussed in the notes to compute the cross product here.

Show Step 4So, we now have enough information to write down the equation of the line of intersection of the two planes. The equation is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle { - 5,2,0} \right\rangle + t\left\langle {29,13,33} \right\rangle = \left\langle { - 5 + 29t,2 + 13t,33t} \right\rangle }}\]