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Section 12.3 : Equations of Planes

2. Write down the equation of the plane containing the point \(\left( {3,0, - 4} \right)\) and orthogonal to the line given by \(\vec r\left( t \right) = \left\langle {12 - t,1 + 8t,4 + 6t} \right\rangle \).

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Start Solution

We know that we need a point on the plane and a vector that is normal to the plane. We’ve were given a point that is in the plane so we’re okay there.

Show Step 2

The normal vector for the plane is actually quite simple to get.

We are told that the plane is orthogonal to the line given in the problem statement. This means that the plane will also be orthogonal to any vector that just happens to be parallel to the line.

From the equation of the line we know that the coefficients of the \(t\)’s are the components of a vector that is parallel to the line. So, a vector parallel to the line is then,

\[\vec v = \left\langle { - 1,8,6} \right\rangle \]

Now, as mentioned above because this vector is parallel to the line then it will also need to be orthogonal to the plane and hence be normal to the plane. So, a normal vector for the plane is,

\[\vec n = \left\langle { - 1,8,6} \right\rangle \] Show Step 3

Now all we need to do is write down the equation. The equation of the plane is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{ - \left( {x - 3} \right) + 8\left( {y - 0} \right) + 6\left( {z + 4} \right) = 0\hspace{0.5in} \to \hspace{0.5in}\, - x + 8y + 6z = - 27}}\]