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By: Jeffrey Bivin Lake Zurich High School Last Updated: October 9, 2007.

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Presentation on theme: "By: Jeffrey Bivin Lake Zurich High School Last Updated: October 9, 2007."— Presentation transcript:

1 By: Jeffrey Bivin Lake Zurich High School Last Updated: October 9, 2007

2 What is the intersection of 3 planes? We will be looking at the case where they intersect in a point. Jeff Bivin -- LZHS

3 I think I’ll do some substitution. 3z = 12 Jeff Bivin -- LZHS 2x + 3y - 5z = y + z = -1

4 3z = 12 z = 4 - 5y + 4 = -1 2x + 3(1) – 5(4) = 13 2x + 3 – 20 = 13 2x - 17 = 13 x = y + 4 = y = -5 y = 1 2x = 30 Jeff Bivin -- LZHS 2x + 3y - 5z = y + z = -1 3z = 12

5 Decide which variable you want to eliminate. I think I’ll choose to eliminate the y variable. 2x + 5y + 8z = 8 3x - 2y + 4z = 26 2x + 4y + 3z = -3 Jeff Bivin -- LZHS

6 2x + 5y + 8z = 8 3x - 2y + 4z = 26 2x + 4y + 3z = -3 6x - 4y + 8z = 52 2x + 4y + 3z = -3 8x + 11z = 49 4x + 10y + 16z = 16 15x - 10y + 20z = x + 36z = x - 288z = x + 209z = z = -237 z = 3 19x + 36(3) = x = 38 x = 2 2(2) + 5y + 8(3) = y + 24 = 8 5y = -20 y = -4 Jeff Bivin -- LZHS

7 Decide which variable you want to eliminate. I think I’ll choose to eliminate the x variable. 5x + 2y + 7z = 19 3x - 3y + 5z = 34 x + 3y - 5z = -22 Jeff Bivin -- LZHS

8 3x - 3y + 5z = 34 -3x - 9y + 15z = y + 20z = 100 5x + 2y + 7z = 19 -5x - 15y + 25z = y + 32z = y - 384z = y + 260z = z = -248 z = 2 -13y + 32(2) = y = 65 y = -5 5x + 2(-5) + 7(2) = 19 5x = 19 5x = 15 x = 3 5x + 2y + 7z = 19 3x - 3y + 5z = 34 x + 3y - 5z = -22 Jeff Bivin -- LZHS

9 Decide which variable you want to eliminate. I think I’ll choose to eliminate the z variable. 2x + 3y - 5z = -12 4x - 5y + 3z = 12 2x - 3y = -7 Jeff Bivin -- LZHS

10 6x + 9y - 15z = x - 25y + 15z = 60 26x - 16y = x + 39y = 91 23y = 115 y = 5 26x – 80 = 24 26x = 104 x = 4 2(4) + 3(5) – 5z = – 5z = z = -35 z = 7 2x + 3y - 5z = -12 4x - 5y + 3z = 12 2x - 3y = -7 26x – 16(5) = 24 Jeff Bivin -- LZHS


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