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### Section 1-3 : Equations of Planes

9. Determine if the line given by \(x = 8 - 15t\), \(y = 9t\), \(z = 5 + 18t\) and the plane given by \(10x - 6y - 12z = 7\) are parallel, orthogonal or neither.

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Start SolutionLet’s start off this problem by noticing that the vector \(\vec v = \left\langle { - 15,9,18} \right\rangle \) will be parallel to the line and the vector \(\vec n = \left\langle {10, - 6, - 12} \right\rangle \) will be normal to the plane.

Now try to visualize the line and plane and their corresponding vectors. What would the line and plane look like if the two vectors were orthogonal to each other? What would the line and plane look like if the two vectors were parallel to each other?

Show Step 2If the two vectors are orthogonal to each other the line would be parallel to the plane. If you think about this it does make sense. If \(\vec v\) is orthogonal to \(\vec n\) then it must be parallel to the plane because \(\vec n\) is orthogonal to the plane. Then because the line is parallel to \(\vec v\) it must also be parallel to the plane.

So, let’s do a quick dot product here.

\[\vec v\centerdot \vec n = - 420\]The dot product is not zero and so the two vectors aren’t orthogonal to each other. Therefore, the **line and plane are not parallel**.

If the two vectors are parallel to each other the line would be orthogonal to the plane. If you think about this it does make sense. The line is parallel to \(\vec v\) which we’ve just assumed is parallel to \(\vec n\). We also know that \(\vec n\) is orthogonal to the plane and so anything that is parallel to \(\vec n\) (the line for instance) must also be orthogonal to the plane.

In this case it looks like we have the following relationship between the two vectors.

\[\vec v = - \frac{3}{2}\vec n\]The two vectors are parallel and so the **line and plane are orthogonal**.