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Section 12.3 : Equations of Planes

6. Determine if the line given by \(\vec r\left( t \right) = \left\langle { - 2t,2 + 7t, - 1 - 4t} \right\rangle \) intersects the plane given by \(4x + 9y - 2z = - 8\) or show that they do not intersect.

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Start Solution

If the line and the plane do intersect then there must be a value of \(t\) such that if we plug that \(t\) into the equation of the line we’d get a point that lies on the plane. We also know that if a point \(\left( {x,y,z} \right)\) is on the plane then the coordinates will satisfy the equation of the plane.

Show Step 2

If you think about it the coordinates of all the points on the line can be written as,

\[\left( { - 2t,2 + 7t, - 1 - 4t} \right)\]

for all values of \(t\).

Show Step 3

So, let’s plug the “coordinates” of the points on the line into the equation of the plane to get,

\[4\left( { - 2t} \right) + 9\left( {2 + 7t} \right) - 2\left( { - 1 - 4t} \right) = - 8\] Show Step 4

Let’s solve this for \(t\) as follows,

\[63t + 20 = - 8\hspace{0.5in} \to \hspace{0.5in}t = - \frac{4}{9}\] Show Step 5

We were able to find a \(t\) from this equation. What that means is that this is the value of \(t\) that, once we plug into the equation of the line, gives the point of intersection of the line and plane.

So, the line and plane do intersect and they will intersect at the point \(\left( {\frac{8}{9}, - \frac{{10}}{9},\frac{7}{9}} \right)\).

Note that all we did to get the point is plug \(t = - \frac{4}{9}\) into the general form for points on the line we wrote down in Step 2.