Are the Points in the Plane?

p {margin: 0; padding: 0;} .ft00{font-size:20px;font-family:Arial;color:#000000;} .ft01{font-size:18px;font-family:ArialMT;color:#000000;} .ft02{font-size:18px;font-family:CambriaMath;color:#000000;} .ft03{font-size:12px;font-family:CambriaMath;color:#000000;} .ft04{font-size:18px;line-height:23px;font-family:ArialMT;color:#000000;} .ft05{font-size:18px;line-height:22px;font-family:ArialMT;color:#000000;} Are the Points in the Plane? So the general idea is that either they are on opposite sides, or one of them might beon the plane. Let's try to see if one of them is in the plane.So if we take the point Q0, which is at (minus 1, 2, 2), and plug that into the planeequation, we get x plus 2y plus 4z equals minus 1 plus 2 times 2 plus 4 times 2, or 4+ 8 = 12 – 1 = 11. 𝑄 0 : − 1, 2, 2 ( ) 𝑧 + 2𝑦 + 4𝑧 = − 1 ( ) + 2. 2 + 4. 2 = 11 > 7 Let us try the point (1, 3, negative 1). If we plug this into x plus 2y plus 4z andsimplify, we get 1 plus 2 times 3, which equals 7. However, 4 times negative 1 equalsnegative 4. Subtracting this from 7 leaves us with 3, less than 7. 𝑄 1 : 1, 3, − 1 ( ) 𝑧 + 2𝑦 + 4𝑧 = 1 + 2. 3 + 4. − ( ) = 3 < 7 So what about answer number two? Let's think about it. These points are not in theplane, but they're not in a plane in different ways and one of them somehowovershoots. We get 11. The other one, we only get 3. That's less than 7. If you thinkabout how a plane splits space into two half spaces on either side (one of them isgoing to be the point where x plus 2y plus 4z is less than 7), so that's somehow thisside and that's where Q1 is. To go from one to the other, well, x plus 2y plus 4z needs through the value 7. Ifyou're moving along any path from Q0 to Q1, this thing will change continuously from11 to 3. At some time, it has to go through 7. Does that make sense?