help on equation of a plane!

[earless]

A plane has intercepts (4,0,0),(0,6,0), and (0,0,12).Find the equations of another plane through (6,-2,4) that is parallel to this plane.


This is a H.W. question.I remember solving something like this in calculus i think i calculated the cross product using the intercepts and found the normal vector which would be normal to both planes since they are parallel.But we havent covered cross products in this section yet so i think i have to use some other method.This is from my vector analysis class.

 

[daon]

Equations of planes need a normal vector n=<a,b,c> and a point p(x0,y0,z0).

Goal: a(x-x0)+b(y-y0)+c(z-z0)=0.

If two planes are parallel to one another, any vector that is normal to one is normal to the other. You need to use the first plane (hereafter referred to as Plane 1) to find a normal vector for the second (Plane 2).

Steps:

Use the three points of the first plane to construct two vectors. Label the points p1, p2, p3 in any order. Calculate v1=p1-p2, v2=p2-p3.

Calculate u = v1 x v2. Note that v1 and v2 are in Plane #1, so the vector u will be normal to Plane 1 (and hence Plane 2).

Use this normal vector in constructing the equation for the second plane. Note u=<a,b,c>. Use the given point, (x0,y0,z0) = (6,-2,4).

 

[galactus]

Use the given points to find the equation of your plane.


A=(4,0,0), B=(0,6,0), C=(0,0,12)

AB=(0-4)i+(6-0)j+(0-0)k=-4i+6j

AC=(0-4)i+(0-0)j+(12-0)k=-4i+12k

Use the cross product:

\(\displaystyle \L\\\begin{vmatrix}-4&6&0\\-4&0&12\end{vmatrix}=72i+48j+24k=3i+2j+k\)

You have the normal vector. Now use:

\(\displaystyle \L\\a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0\)

Can you finish now?

 

[earless]

Is there a way to solve it without using cross products?Since the book hasnt covered cross products yet i think it wants me to solve it using other methods.

 

[daon]

I thought about this for awhile and have come up with a method that seems to work, although I have not verfied it. A normal vector to a plane is one that does not project any components onto the plane.

So, n=<a,b,c> and \(\displaystyle \L |proj_{v_i}\(n\)| = 0\) for any vector \(\displaystyle v_i\) in the plane.

To solve for n, do the following. Calculate three different vectors using the given points. Call them u,v,w

\(\displaystyle \L |proj_{u}\(n\)| = 0 \,\, \,\, \Rightarrow \,\, \,\, n \cdot u = 0\)

\(\displaystyle \L |proj_{v}\(n\)| = 0 \,\, \,\, \Rightarrow \,\, \,\, n \cdot v = 0\)

\(\displaystyle \L |proj_{w}\(n\)| = 0 \,\, \,\, \Rightarrow \,\, \,\, n \cdot w = 0\)

So...

\(\displaystyle \L au_1 + bu_2+ cu_3 = 0\)

\(\displaystyle \L av_1 + bv_2 + cv_3 = 0\)

\(\displaystyle \L aw_1 + bw_2 + cw_3 = 0\).

Solve this system of equations and you should have your a,b,c. It seems c will be a free variable and a and b will depend upon it...


edit: Selecting C=1 should give the same normal vector as galactus'.