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Section 1-3 : Equations of Planes

4. Determine if the plane given by \(4x - 9y - z = 2\) and the plane given by \(x + 2y - 14z = - 6\) are parallel, orthogonal or neither.

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Let’s start off this problem by noticing that the vector \({\vec n_1} = \left\langle {4, - 9, - 1} \right\rangle \) will be normal to the first plane and the vector \({\vec n_2} = \left\langle {1,2, - 14} \right\rangle \) will be normal to the second plane.

Now try to visualize the two planes and these normal vectors. What would the two planes look like if the two normal vectors where parallel to each other? What would the two planes look like if the two normal vectors were orthogonal to each other?

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If the two normal vectors are parallel to each other the two planes would also need to be parallel.

So, let’s take a quick look at the normal vectors. We can see that the first component of each vector have the same sign and the same can be said for the third component. However, the second component of each vector has opposite signs.

Therefore, there is no number that we can multiply to \({\vec n_1}\) that will keep the sign on the first and third component the same and simultaneously changing the sign on the second component. This in turn means the two vectors can’t possibly be scalar multiples and this further means they cannot be parallel.

If the two normal vectors can’t be parallel then the two planes are not parallel.

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Now, if the two normal vectors are orthogonal the two planes will also be orthogonal.

So, a quick dot product of the two normal vectors gives,

\[{\vec n_1}\centerdot {\vec n_2} = 0\]

The dot product is zero and so the two normal vectors are orthogonal. Therefore, the two planes are orthogonal.