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Line L goes through B(3,4,-2) and C(7,-1,5).

Calculate the directional vector v and its magnitude.

v = BC = = <7-3, -1-4, 5+2> = <4, -5, 7>

| v | = √(4² + (-5)² + 7²) = √(16 + 25 + 49) = √90

Write the equation of the line.

L = B + tv
L = <3, 4, -2> + t<4, -5, 7>

Calculate the vector AB.

Let
u = AB = = <3-2, 4+6, -2-1> = <1, 10, -3>

Take the cross product.

AB X BC = u X v = <1, 10, -3> X <4, -5, 7> = <55, -19, -45>

| u X v | = √[55² + (-19)² + (-45)²] = √5411

Calculate the distance h, of A from the line L.

h = | u X v | / | v | = √5411 / √90 = √(5411/90) ≈ 7.7538521

2007-07-20 15:10:58 · answer #1 · answered by Northstar 7 · 0 0

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