# need help fast

• Mar 17th 2009, 02:48 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, 04:11 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

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

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

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

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

$\displaystyle A + 2B + 4C = D$

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

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

Solve for $\displaystyle A, B, C$ (your answer will be in terms of $\displaystyle D$).
• Mar 17th 2009, 04:29 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, 04:47 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 $\displaystyle R_2 + 2R_1$ and $\displaystyle R_3 - 2R_1$.

$\displaystyle A + 2B + 4C = D$

$\displaystyle 4B + 2C = 2D$

$\displaystyle -3B - 5C = -D$

Do $\displaystyle \frac{R_2}{4}$

$\displaystyle A + 2B + 4C = D$

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

$\displaystyle -3B - 5C = -D$

Do $\displaystyle R_3 + 3R_2$

$\displaystyle A + 2B + 4C = D$

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

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

I take it now you can solve for $\displaystyle A, B, C$?
• Mar 17th 2009, 04:57 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 $\displaystyle R_2 + 2R_1$ and $\displaystyle R_3 - 2R_1$.

$\displaystyle A + 2B + 4C = D$

$\displaystyle 4B + 2C = 2D$

$\displaystyle -3B - 5C = -D$

Do $\displaystyle \frac{R_2}{4}$

$\displaystyle A + 2B + 4C = D$

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

$\displaystyle -3B - 5C = -D$

Do $\displaystyle R_3 + 3R_2$

$\displaystyle A + 2B + 4C = D$

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

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

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

nvm, finished, thanks a lot