30-43in13.2

30-43in13.2 - S E C T I O N 13.2 Vectors in Three...

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Unformatted text preview: S E C T I O N 13.2 Vectors in Three Dimensions (ET Section 12.2) 319 SOLUTION Since v = 7 i + 4 k = h 7 , , 4 i we obtain the following parametrization: r ( t ) = −→ O P + t v = h 4 , , 8 i + t h 7 , , 4 i = h 4 + 7 t , , 8 + 4 t i 30. Passes through O , direction vector v = h 4 , 2 , − 1 i SOLUTION Using the vector parametrization of the line we get r ( t ) = h , , i + t h 4 , 2 , − 1 i = h 4 t , 2 t , − t i 31. Passes through ( 1 , 1 , 1 ) and ( 3 , − 5 , 2 ) SOLUTION We use the equation of the line through two points P and Q : r ( t ) = ( 1 − t ) −→ O P + t −→ OQ Since −→ O P = h 1 , 1 , 1 i and −→ OQ = h 3 , − 5 , 2 i we obtain r ( t ) = ( 1 − t ) h 1 , 1 , 1 i + t h 3 , − 5 , 2 i = h 1 − t , 1 − t , 1 − t i + h 3 t , − 5 t , 2 t i = h 1 + 2 t , 1 − 6 t , 1 + t i 32. Passes through ( − 2 , , − 2 ) and ( 4 , 3 , 7 ) SOLUTION Using the equation of the line through two points P and Q , with −→ O P = h− 2 , , − 2 i and −→ OQ = h 4 , 3 , 7 i we obtain r ( t ) = ( 1 − t ) −→ O P + t −→ OQ = ( 1 − t ) h− 2 , , − 2 i + t h 4 , 3 , 7 i = h− 2 ( 1 − t ), , − 2 ( 1 − t ) i + h 4 t , 3 t , 7 t i = h− 2 + 6 t , 3 t , − 2 + 9 t i 33.33....
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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30-43in13.2 - S E C T I O N 13.2 Vectors in Three...

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