On Wed, 13 Feb 2008 17:04:06 -0800 (PST), Citizen
<FlammesSombres@[EMAIL PROTECTED]
> wrote in
<news:65ca7113-85a1-49df-a879-cb02c7402f04@[EMAIL PROTECTED]
>
in alt.algebra.help:
> Hi all,
> Can you please help me with this math problem?
> The plane x - 3y - z = 12 is parallel to the plane x - 3y - z = 0.
> One particular point on this plane is (12, 0, 0). All points all the
> plane have the form (x, y, z) = (a, 0, 0,) + y(b, 1, 0) + z(c, 0, 1).
> what are a, b, and c?
Begin by computing (a, 0, 0) + y(b, 1, 0) + z(c, 0, 1) to
get a better idea of what points in the plane look like
according to the final assertion:
(a, 0, 0) + y(b, 1, 0) + z(c, 0, 1) =
(a + by + cz, y, z).
Now you know that (12, 0, 0) is one of these points, so it
has the form (a + by + cz, y, z) for some values of y and z.
Clearly the only way to get
(12, 0, 0) = (a + by + cz, y, z)
is to set y = z = 0, at which point you have
(12, 0, 0) = (a, 0, 0)
and can infer that a = 12.
Now every point in the plane x - 3y - z = 12 must have the
form (12 + by + cz, y, z); in other words, a point (x, y, z)
is in the plane if and only if x = 12 + by + cz. On the
other hand, you also know that (x, y, z) is in the plane if
and only if x - 3y - z = 12; use this to get a second
equation for x in terms of y and z, and use the pair of
equations to figure out what b and c must be.
[...]
Brian
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