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November 26, 2014

November 26, 2014

Posted by **j** on Thursday, May 17, 2012 at 9:18am.

a) contains the point A(6,2,3), B(5,1,-3), and C(5,7,-2)

b) parallel to 2x + 6y + 4z = 1 and contains the point P(3,2,1)

- Calculus -
**Reiny**, Thursday, May 17, 2012 at 9:37amvector AB = (-1,-1,-6)

vector AC = (-1,5,-5)

a normal to these two vectors is (35, 1, -6) , I assume you know how to take a cross-product

the plane has equation

35x+ y - 6z = c

plug in a point, e.g. A(6,2,3)

210 + 2 - 18 = c

c =194

plane equation: 35x + y - 6z = 194

check by subbing in points B and C, (they work)

b) Since they are parallel, they must differ only in the constant

so new equation: 2x + 6y + 4z = c

plug in your point to find c

- Calculus -
**MathMate**, Thursday, May 17, 2012 at 9:38am(a)

1. Find vectors AB and AC.

2. Calculate the cross product of ABxAC, which gives the normal vector to the required plane.

3. Form the equation of the plane. The components of the normal vector correspond to the coefficients of x, y and z of the equation of the plane.

4. Find the constant term of the plane by passing the plane through one of the three points, i.e. substitute (xa,ya,za) into the equation and find the constant term.

5. Optionally, check that the other points pass through the equation of the plane.

(b)

Parallel to 2x + 6y + 4z = 1 means that the normal is <2,6,4>.

Take the left-hand side of the equation and evaluate with the point P(3,2,1), i.e. pass a plane parallel to 2x + 6y + 4z = 1 through the given point:

2(3)+6(2)+4(1)=6+12+4=16

So the required equation of the plane is

2x + 6y + 4z = 16

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