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I am really confused about this question. It doesnt make sense to me. Any help would be greatly appreciated

What is the relationship of the given line L : (x, y, z) = (1,-2,-3) + t<-4,-1,5> to each of the planes below? Is the relationship that the Plane contain L, that the plane is parallel to L or that the plane interests L?

1. -10x+25y-3z = -51

2. -2x+5y+2z = 4

3. 9x-16y+4z = 26

4. -20x-20y-20z = -4

Billy, Timmy, Sam,

Wish you had more specific questions or could at least list some of the definitions and results presented to you in your course work. The subject of lines and planes in 3-D is covered pretty succinctly in various references; one example that you should look at is

science.kennesaw.edu/~plaval/math2203/linesplanes.pdf.

The key item in determining whether the line is parallel to the planes is to see if it is perpindicular to the normal (= perpindicular) vector to the planes, in other words, being perpindicular to the perpindicular means its parallel. To determine this, take the dot product of the line with the normal vector. If it is zero, the the line is parallel to the plane.

For plane #1, the normal vector is given by the coefficients in the equation, n = <-10,25,-3>. Taking the dot product gives (I assume you know how to take dot products)

(-10,25,-3)･(-4,-1,5) = 40-25-15 = 0,

so it is parallel. Does the plane contain the line? To find out, note that only one point in the line needs to be shown to be in the plane if the line is parallel. A point is in the plane if it satifies the scalar equation for the plane. A convenient point on the line is the point given explicitly in #3 as (1,-2,-3). Plugging this in gives

-10･1+25･(-2)-3･(-3) = -51, which equal to the constant on the right side of the scalar equation. So the line is in fact contained in plane #3.

The other planes can be analyzed in the same way. Let me know how it goes.

Randy

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