       Re: Vectors

• To: mathgroup at smc.vnet.net
• Subject: [mg124259] Re: Vectors
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Sat, 14 Jan 2012 17:14:13 -0500 (EST)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <201201140757.CAA01372@smc.vnet.net>

```r = {Reduce[{x - 1 == (y + 1)/2 == (z - 3)/h, h != 0}, {h, y},
Backsubstitution -> True] // ToRules}

{{z -> 3, x -> 1, y -> -1}, {h -> (-3 + z)/(-1 + x), y -> -3 + 2*x}}

h /. r[]

(-3 + z)/(-1 + x)

eqns = {6 x - 2 y + z == 11, x - 1 == (y + 1)/2};

parametricLine[x_] = {x, y, z} /.
(Reduce[eqns, {y, z}] // ToRules)

{x, -3 + 2 x, 5 - 2 x}

Show[
ContourPlot3D[Evaluate[eqns],
{x, -5, 5}, {y, -10, 10}, {z, -10, 10}],
ParametricPlot3D[parametricLine[x], {x, -5, 5},
PlotStyle ->
{Darker[Blue], AbsoluteThickness}],
AxesLabel -> {x, y, z}]

Bob Hanlon

On Sat, Jan 14, 2012 at 2:57 AM, Harry Har <harryhar800 at gmail.com> wrote:
> Hi All,
>
> I'm a math senior high school teacher, and I wish to understand how
> using mathematica 7 to find "h" and the intersection point between the
> plane 6x-2y+z=11 which contains the line x-1 = (y+1)/2 = (z-3)/h. Also
> how to plot/show it visually in 3D graph.
>
> Many thank's.
>
> Harry.
>

```

• References:
• Vectors
• From: Harry Har <harryhar800@gmail.com>
• Prev by Date: Re: how can one use mathematica get the approximate derivative of {x,y} data points?
• Next by Date: Re: Cross reference