Quick linear algebra question

Hey guys,

I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.

What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).

The answer is 5x - 3y + z = 8

How do I go about solving this?

Thanks for yout time!

Go to math forum then...

Moved. Please don't post technical questions in the General Discussion / Chit Chat forum.

Quote:

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Originally Posted by INF3RN0666

Hey guys,

I didn't know who to ask and I need an answer quickly :S. It should be simple, I just cant seem to remember how I did this problem.

What is the equation of the plane parallel to the vector (1, 1, -2) which passes through the points (1, 5, 18) and (4, 2, -6).

The answer is 5x - 3y + z = 8

How do I go about solving this?

Thanks for yout time!

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The equation in vector form is:

R = r + av₁ + bv₂

where bold means vector and a,b are scalars (numbers).

r is a vector from the coordinate origin to a point in the plane, say to (1,5,18)
v₁ and v₂ are any two (linear independent) vectors contained on the plane. For example, v₁ can be the given vector (1,1,-2) and v₂ can be the vector joining the 2 given points:

v₂ = (4,2,-6) - (1,5,18) = (3,-3,-24)

So the equation becomes:

R = (x,y,z) = (1,5,18) + a(1,1,-2) + b(3,-3,-24)

Expressing each component separately:

x = 1 + a +3b
y = 5 + a -3b
z = 18 -2a -24b

Now you make substitutions to find z in terms of x and y:

a = x - 1 - 3b
y = 5 + x -1 -3b -3b = 4 + x -6b
b = (4 + x - y) / 6

z = 18 -2(x - 1 -3b) - 24b = 18 - 2x + 2 + 6b - 24b = 20 - 2x - 16(4 + x - y) / 6 = 20 - 2x -3(4 + x - y) = 20 - 2x - 12 - 3x + 3y = 8 - 5x + 3y

Or, rearranging,

5x - 3y + z = 8