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Section 1-3 : Equations of Planes

3. Write down the equation of the plane containing the point \(\left( { - 8,3,7} \right)\) and parallel to the plane given by \(4x + 8y - 2z = 45\).

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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.

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

We are told that the plane is parallel to the plane given in the problem statement. This means that any vector normal to one plane will be normal to both planes.

From the equation of the plane we were given we know that the coefficients of the \(x\), \(y\) and \(z\) are the components of a vector that is normal to the plane. So, a vector normal to the given plane is then,

\[\vec n = \left\langle {4,8, - 2} \right\rangle \]

Now, as mentioned above because this vector is normal to the given plane then it will also need to be normal to the plane we want to find the equation for.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{4\left( {x + 8} \right) + 8\left( {y - 3} \right) - 2\left( {z - 7} \right) = 0\hspace{0.5in} \to \hspace{0.5in}\,4x + 8y - 2z = - 22}}\]