1. ## Plane in 3 dimensions (Please check my work)

Find scalar equations of the form $ax + by + cz = 0$ for the following plane in $R^3$:

The plane through the points:

• $P(0;1;1)$
• $Q(1;0;1)$
• $R(1;1;0)$

${PQ}^{\rightarrow} = (1;-1;0)$

${QR}^{\rightarrow} = (0;1;-1)$

The perpendicular vector $= {PQ}^{\rightarrow} \times {QR}^{\rightarrow}$

Then I just picked point $Q$ to substitute into the equation:

$1(x -1) + 1(y - 0) + 1(z - 1) = 0$

Multiplying it out:

$x + y + z = 2$

And that should be the equation of the plane?

2. Originally Posted by janvdl
Find scalar equations of the form $ax + by + cz = 0$ for the following plane in $R^3$:

The plane through the points:

• $P(0;1;1)$
• $Q(1;0;1)$
• $R(1;1;0)$

${PQ}^{\rightarrow} = (1;-1;0)$

${QR}^{\rightarrow} = (0;1;-1)$

The perpendicular vector $= {PQ}^{\rightarrow} \times {QR}^{\rightarrow}$

Then I just picked point $Q$ to substitute into the equation:

$1(x -1) + 1(y - 0) + 1(z - 1) = 0$

Multiplying it out:

$x + y + z = 2$

And that should be the equation of the plane?

Correct.

3. That's fine. Notice that (1) x+y+z = 2 is a linear equation, so it represents a plane; (2) the equation is obviously satisfied when (x,y,z) is P, Q or R. So that has to be the right answer!

4. As Opalq said, it is easy to check it yourself.
Do P: (0, 1, 1), Q: (1, 0, 1), and R: (1, 1, 0) all satisfy x+ y+ z= 2? If yes, that is the equation of the plane containing those three points and if not, it is not.

(In fact, you could have found the equation just by noting that the three components of each of those points sums to 2.)

5. Originally Posted by HallsofIvy
As Opalq said, it is easy to check it yourself.
Do P0, 1, 1), Q1, 0, 1), and R1, 1, 0) all satisfy x+ y+ z= 2? If yes, that is the equation of the plane containing those three points and if not, it is not.

(In fact, you could have found the equation just by noting that the three components of each of those points sums to 2.)
or : ( ...... The space makes a difference if you don't disable the smilies .... *ahem* or did I mean : )