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How do I find an expression for an equation of a plane that passes through the point (0,0,a) and contains the line x=t, y=1-t, z=0 with the expression being in terms of a?

2007-07-25 13:54:55 · 2 answers · asked by Mr. Smith 1 in Science & Mathematics Mathematics

2 answers

Rewrite the equation of the line L, in vector form.

L:
x = t
y = 1 - t
z = 0

Becomes

L = P + tu
L = <0, 1, 0> + t<1, -1, 0>
where t is a scalar ranging over the real numbers

Now create a second directional vector of the plane with the two points.

v = <0, 0, a> - <0, 1, 0> = <0, -1, a>

The normal vector to the plane n, is orthogonal to the directional vectors of the plane. Take the cross product.

n = u X v = <1, -1, 0> X <0, -1, a> = <-a, -a, -1>

Any non-zero multiple of n is also a normal vector to the plane. Multiply by -1.

n =

With the normal vector and a point in the plane we can write the equation to the plane. Let's use the point (0, 0, a).

a(x - 0) + a(y - 0) + 1(z - a) = 0
ax + ay + z - a = 0

2007-07-27 21:44:19 · answer #1 · answered by Northstar 7 · 0 0

Solve first for x, then substitute the values.

2007-07-25 21:04:58 · answer #2 · answered by Marie 4 · 0 1

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