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October 2, 2014

October 2, 2014

Posted by **Derek** on Wednesday, July 9, 2008 at 2:23pm.

a.) x + 2y + 3z = −4

2x + 4y + 6z = 10

b.)2x − y + 2z = −8

4x − 2y + 4z = −16

c.) x + 3y − 5z = −12

2x + 3y − 4z = −6

- Calc. -
**Reiny**, Wednesday, July 9, 2008 at 3:18pmfor the first one, did you notice that the normals are

(1,2,3) and (2,4,6) or 2(1,2,3) ?

so the normals are parallel, which means that the planes are parallel.

Does that make sense? OK!

But the constants are -4 and 10 which are not in the same ratio as 1:2:3 to 2:4:6

So the first question represents two parallel and distinct planes.

For the second one, did you notice that multiplying the first equation by 2 gives you the second equation?

So the two equations actually represent the same plane

- Calc. -
**Reiny**, Wednesday, July 9, 2008 at 3:33pmFor the third one, the two normals are not scalar multiples of each other, so the two planes are not parallel.

They must therefore intersect in a straight line.

(Think of two pages of an open book intersecting in the spine of the book)

subtract the first equation from the second to get

x+z=6

now let's pick any value for z

z=0, then x = 6 and from the first equation y=-6

z=6, then x=0 and from the first equation y = 6

So we now have two points (0,6,6) and (6,-6,0) which lie on the line of intersection.

notice the two points satisfy both plane equations.

That means we can find the direction vector of our line which is (6,-12,-6) or reduced to (1,-2,-1)

So now we have the direction of our line and a point on the line.

Then a possible equation of the line in vector form is

vector r = (0,6,6) + t(1,-2,-1)

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