Then \u00b9 3 1 3 1 5 3 1 23 5 2 Put into 2 4 1 2 5 3 1 2 23 5 2 7 or 63 4

Then ¹ 3 1 3 1 5 3 1 23 5 2 put into 2 4 1 2 5 3 1 2

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Then ¹ = 3 − 1 − ¸ = 3 − 1 − 5 − 3 + 1 = 23 − 5 − 2. Put into (2): 4 = 1 + 2 5 − 3 + 1 − 2 23 − 5 − 2 , or 4 = −63 + 45 + 7, or: 63 + 4 − 45 = 7.
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9/2/2015 6 HOW CAN WE GET THE PERPENDICULAR VECTOR OF A PLANE IF WE ONLY HAVE 2 VECTORS ON THE PLANE?? ³ = (1, 2, 2) ´ = (1. −2, 1) Vectors ³ and ´ is ON the plane. How can we find vector 0 perpendicular to the plane? SOLUTION: find the cross product of a and b: ³ 3 ´: 1 2 2 1 2 1 − 2 1 1 − 2 = ( 6, 1, −4) TEST IF (A X B) IS PERPENDICULAR Perpendicular vectors gives °. 0 = 0 Test : °. 0 = 1, 2, 2 . 6, 1, −4 = 6 + 2 − 8 = 0 YES it is true for ° Test : ±. 0 = 1, −2, 1 . 6, 1, −4 = 6 − 2 − 4 = 0 YES it is true for ± SO: ° 3 ± is perpendicular to the plane!!
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9/2/2015 7 EXAMPLE A point L = (4, 5, −7) lies on a plane. Find the scalar equation of the plane that is perpendicular to the directional vector 0 = (1, 2, −3) ² − ² : . 0 = 0 SOLUTION: ² = 3, 4, 5 °08 ² : = 4,5, −7 °08 0 = (1,2, −3) Then: (3, 4, 5 − (4,5, −7)). (1,2, −3) = 0 1 3 − 4 + 2 4 − 5 − 3 5 + 7 = 0 3 − 4 + 24 − 10 − 35 − 21 = 0 3 + 24 − 35 = 35 NOTE: The values in front of x, y and z is the same as the vector n. ANGLE BETWEEN PLANES . Can we now find the angle between the planes 3 + 4 + 5 = 2 and 23 − 5 = 0 as well?
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9/2/2015 8 ANGLE BETWEEN PLANES The angle between the planes is the same as the angle between the perpendicular vectors. The planes are 3 + 4 + 5 = 2 and 23 − 5 = 0 . Vectors perpendicular to each are 3 = 1 1 1 and 4 = 2 0 −1 . With 3 = 3, 4 = 5, 3 . 4 = 1 2 + 1 0 + 1 −1 = 1 the angle ] between them is given by cos ^ = 3 . 4 34 = 1 3 5 = 0,2582, ^ = 75,04°.
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9/2/2015 9 NOW YOU: Find the vector equation, the parametric equations and the scalar equation of the plane which contains vectors a and b and a point P 0 ° = 1,2,2 ; ± = 1, −2,1 ; L : = (1,1,0)
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