# need help fast

• Mar 17th 2009, 03:48 PM
ruthvik
need help fast
This was given in class to us

Equation of a Plane

From geometry, you learned that three distinct non-collinear points will determine a unique plane. Algebraically, the equation of the plane is given by Ax + By + Cz = D; thus, if the three distinct points were given, then we should be able to find the values of A, B, C, and D which will describe the equation of the plane passing through the points.

Suppose the points P(1, 2, 4), Q(-2, 2, -2), and S(2, 1, 3) are the three distinct points determining the plane PQS. Write the algebraic representation for the plane containing these points.

Your solution is due Wednesday, 3/18 by the beginning of class!

i don't exactly understand how to approach this problem
• Mar 17th 2009, 05:11 PM
Prove It
Quote:

Originally Posted by ruthvik
This was given in class to us

Equation of a Plane

From geometry, you learned that three distinct non-collinear points will determine a unique plane. Algebraically, the equation of the plane is given by Ax + By + Cz = D; thus, if the three distinct points were given, then we should be able to find the values of A, B, C, and D which will describe the equation of the plane passing through the points.

Suppose the points P(1, 2, 4), Q(-2, 2, -2), and S(2, 1, 3) are the three distinct points determining the plane PQS. Write the algebraic representation for the plane containing these points.

Your solution is due Wednesday, 3/18 by the beginning of class!

i don't exactly understand how to approach this problem

$(x_1, y_1, z_1) = (1, 2, 4)$

$(x_2, y_2, z_2) = (-2, 2, -2)$

$(x_3, y_3, z_3) = (2, 1, 3)$.

Substitute these values into the equation $Ax + By + Cz = D$ and you'll get 3 equations in 3 unknowns which you can solve.

$A + 2B + 4C = D$

$-2A + 2B - 2C = D$

$2A + B + 3C = D$.

Solve for $A, B, C$ (your answer will be in terms of $D$).
• Mar 17th 2009, 05:29 PM
ruthvik
how many solutions should i be getting ?

Thanks btw

EDIT : im getting 5 = 1, and all these weird answers, not getting anything equal to d :(
• Mar 17th 2009, 05:47 PM
Prove It
Quote:

Originally Posted by ruthvik
how many solutions should i be getting ?

Thanks btw

EDIT : im getting 5 = 1, and all these weird answers, not getting anything equal to d :(

http://www.mathhelpforum.com/math-he...a850cb86-1.gif

http://www.mathhelpforum.com/math-he...b95e02f8-1.gif

http://www.mathhelpforum.com/math-he...ba92d82d-1.gif.

Do $R_2 + 2R_1$ and $R_3 - 2R_1$.

$A + 2B + 4C = D$

$4B + 2C = 2D$

$-3B - 5C = -D$

Do $\frac{R_2}{4}$

$A + 2B + 4C = D$

$B + \frac{1}{2}C = \frac{1}{2}D$

$-3B - 5C = -D$

Do $R_3 + 3R_2$

$A + 2B + 4C = D$

$B + \frac{1}{2}C = \frac{1}{2}D$

$-\frac{7}{3}C = \frac{1}{2}D$

I take it now you can solve for $A, B, C$?
• Mar 17th 2009, 05:57 PM
ruthvik
Quote:

Originally Posted by Prove It
http://www.mathhelpforum.com/math-he...a850cb86-1.gif

http://www.mathhelpforum.com/math-he...b95e02f8-1.gif

http://www.mathhelpforum.com/math-he...ba92d82d-1.gif.

Do $R_2 + 2R_1$ and $R_3 - 2R_1$.

$A + 2B + 4C = D$

$4B + 2C = 2D$

$-3B - 5C = -D$

Do $\frac{R_2}{4}$

$A + 2B + 4C = D$

$B + \frac{1}{2}C = \frac{1}{2}D$

$-3B - 5C = -D$

Do $R_3 + 3R_2$

$A + 2B + 4C = D$

$B + \frac{1}{2}C = \frac{1}{2}D$

$-\frac{7}{3}C = \frac{1}{2}D$

I take it now you can solve for $A, B, C$?

nvm, finished, thanks a lot