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Discrete Math Section 12.5 Apply vectors in three dimensions Given points A(x 1,y 1,z 1 ) and B(x 2,y 2,z 2 ) Vector = Absolute value of = √((x 2 – x 1.

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Presentation on theme: "Discrete Math Section 12.5 Apply vectors in three dimensions Given points A(x 1,y 1,z 1 ) and B(x 2,y 2,z 2 ) Vector = Absolute value of = √((x 2 – x 1."— Presentation transcript:

1 Discrete Math Section 12.5 Apply vectors in three dimensions Given points A(x 1,y 1,z 1 ) and B(x 2,y 2,z 2 ) Vector = Absolute value of = √((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 + (z 2 – z 1 ) 2 ) Midpoint of = (x 2 + x 1, y 2 + y 1, z 2 + z 1 ) 2 2 2 Equation of sphere (x – x 1 ) 2 + (y – y 1 ) 2 + (z – z 1 ) 2 =r 2 center at (x 1, y 1, z 1 ) radius = r Vector Equation (x,y,z) = (x 0,y 0,z 0 ) + t Dot product x 1 x 2 + y 1 y 2 + z 1 z 2

2 Examples Add the vectors and Find the absolute value of V =

3 If v = and u = find v∙u Find a vector equation through (7,4,2) and parallel to (x,y,z) = (1,0,-2) + t

4 Find the parametric equations of (x,y,z) = (4,5,-1) + t Find the angle between the vectors and

5 A sphere has points A(8,-2,3) and B(4,0,7) as endpoints of a diameter. a. Find the center C and the radius r of the sphere. b. Find the equation of the sphere. c. Where does a particle with the vector equation (x,y,z) = (5,2,7) + t intersect the sphere?

6 Assignment Page 450 Problems 2,4,8,9,12,16,22,24,26,28,42,43


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