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ps2sol - 18.02 Fall 2008 Problem Set 2 Part B Solutions 1 a...

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18.02 Fall 2008 – Problem Set 2, Part B Solutions 1. a) The two planes have normal vector −−→ AB × −−→ CD = ( 8 , 2 , 6 ) × ( 4 , 3 , 1 ) = ( 16 , 32 , 32 ) . Scaling by 1 / 16, a simpler normal vector is vector n = ( 1 , 2 , 2 ) . The plane con- taining A and B has equation x + 2 y 2 z = 7, while the plane containing C and D has equation x + 2 y 2 z = 11. To find the distance between these planes, we can compute the component along vector n of a vector joining the two planes, for example −−→ AC : −−→ AC · vector n | vector n | = ( 2 , 6 , 2 ) · ( 1 , 2 , 2 ) radicalbig 1 2 + 2 2 + ( 2) 2 = 18 3 = 6 . Hence the distance between the planes (which is the same as the distance between the lines ( AB ) and ( CD )) is 6. b) A parametric equation for ( AB ) is: x = 3+8 t, y = 1+2 t, z = 1+6 t . A parametric equation for ( CD ) is: x = 1+4 t , y = 5 3 t , z = 1 t . If P 1 = ( 3+8 t 1 , 1+2 t 1 , 1+6 t 1 ) and P 2 = ( 1 + 4 t 2 , 5 3 t 2 , 1 t 2 ) are points on the two lines, then −−−→ P 1 P 2 · −−→ AB = ( 2 8 t 1 + 4 t 2 , 6 2 t 1 3
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