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July 7th, 2015, 01:44 PM   #1
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Scalar Equation of a Plane

Determine a scalar equation for the plane through the points M(1, 2, 3) and N(3 ,2, -1) that is perpendicular to the N plane with equation 3x + 2y + 6z + 1 = 0.

I get that Ax + By + Cz + D = 0 is needed but im doing e learning and am really not sure how to apply it to solve this question
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July 7th, 2015, 03:29 PM   #2
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The plane 3x+ 2y+ 6z+ 1= 0 has normal vector of the form 3i+ 2j+ 6k. The normal vector to the plane Ax+ By+ Cz+ D= 0 is of the form Ai+ Bj+ Ck and must be perpendicular to the first normal vector.

That is, we must have 3A+ 2B+ 6C= 0. The fact that M(1, 2, 3) is in the plane means that we must have A(1)+ B(2)+ C(3)+ D= A+ 2B+ 3C+ D= 0. The fact that N(3, 2, -1) is in the plane means that we must have A(3)+ B(2)+ C(-1)+ D= 3A+ 2B- C+ D= 0.

So we have 3A+ 2B+ 6C= 0, A+ 2B+ 3C+ D= 0, and 3A+ 2B- C+ D= 0.

That is only three equation but if we were to multiply or divide each number in Ax+ By+ Cz+ D= 0 we would have an equivalent equation so an equation for that same plane. That means that we could, for example, take D= 1 and have three equations to solve for A, B, and C.
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July 7th, 2015, 03:30 PM   #3
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Quote:
Originally Posted by holdyuh View Post
Determine a scalar equation for the plane through the points M(1, 2, 3) and N(3 ,2, -1) that is perpendicular to the N plane with equation 3x + 2y + 6z + 1 = 0.

I get that Ax + By + Cz + D = 0 is needed but i'm doing e learning and am really not sure how to apply it to solve this question
Dividing by $\displaystyle D$ you get

$\displaystyle \dfrac{Ax}{D}+\dfrac{By}{D}+\dfrac{Cz}{D}+1=0$

Now let $\displaystyle P=\dfrac{A}{D}$, $\displaystyle Q=\dfrac{B}{D}$, $\displaystyle R=\dfrac{C}{D}$ so your plane becomes

$\displaystyle Px+Qy+Rz+1=0$

You can now get three equations (for your three unknowns $\displaystyle P,Q,R$ by using the two given points and taking the dot product of the normals to the planes. Your final answer should be $\displaystyle 2x-6y+z+7=0$.
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