Find the distance between the line L: (x,y,z) = (0,4,-3) + t<4,28,-24> and each of the following lines below. b) L_2: (x,y,z) = (5,5,3) + t<2,1,5> The distance between L and L_2 is ?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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First, you need to evaluate the normal vector to both lines, hence, you need to evaluate the following vector product, such that:

`bar n = <4,28,-24>*<2,1,5> = [(bar i,bar j, bar k),(4,28,-24),(2,1,5)]`

`bar n = 140bar i + 4bar k - 48bar j - 56bar k + 24bar i - 20bar j`

`bar n = 164 bar i - 68bar j + 28 bar k`

`bar n = 4*<41,17,7>`

You need to write the equation of the plane `P` that contains the line `L` , such that:

`P: (0,4,-3)*bar n => (0,4,-3)*<41,17,7> = 4*17 + 7*(-3) = 47`

You need to write the equation of the plane `P_2` that contains the line `L_2` , such that:

`P_2: (5,5,3)*bar n => (5,5,3)*<41,17,7> = 5*41 + 5*17 + 3*7`

`(5,5,3)*<41,17,7> = 205 + 85 + 21 = 311`

You need to evaluate the distance between the lines, such that:

`d = |(0,4,-3)*bar n - (5,5,3)*bar n|/(|bar n|)`

`d = |47 - 311|/(sqrt(41^2 + 17^2 + 7^2))`

`d = 264/(sqrt(1681 + 289 + 49)) => d = 264/(sqrt2019)`

Hence, evaluating the distance between the given lines `L` and `L_2` , yields `d = 264/(sqrt2019).`

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