# Find the distance between the line L: (x,y,z) = (0,4,-3) + t<4,28,-24> and each of the following lines below. c) L_3 : (x,y,z) = (1,3,-5) + t<-2,4,-5> The distance between L and...

Find the distance between the line L: (x,y,z) = (0,4,-3) + t<4,28,-24> and each of the following lines below.

c) L_3 : (x,y,z) = (1,3,-5) + t<-2,4,-5>

The distance between L and L_3 is ?

*print*Print*list*Cite

### 1 Answer

You need to find the smallest distance between the line L and the line `L_3` , hence, you need to use the following formula, such that:

`D = |bar(AB)*bar n|/||bar n||`

You need to consider `A = (0,4,-3)` and `B = (1,3,-5)` , hence, you may evaluate `bar(AB)` such that:

`bar(AB) = <x_B - x_A,y_B - y_A,z_B - z_A>`

`bar(AB) = <1 - 0,3 - 4,-5 - (-3)> => bar(AB) = <1,-1,-2>`

You need to evaluate bar n using the cross product of the direction vectors of the lines` L` and `L_2` , such that:

`bar u = <4,28,-24>` and `bar v = <-2,4,-5>`

`bar u x bar v = [(bar i, bar j, bar k),(4,28,-24),(-2,4,-5)]`

`bar u x bar v = -140 bar i + 48 bar j + 16 bar k + 56 bar k + 96 bar i + 20 bar j`

`bar u x bar v = - 44 bar i + 68 bar j + 72 bar k`

`bar u x bar v = 4(- 11 bar i + 17 bar j + 18 bar k)`

Hence, evaluating the normal vector `bar n` yields:

`bar n = <-11,17,18>`

You need to evaluate the dot product `bar(AB)*bar n` such that:

`bar(AB)*bar n = 1*(-11) + (-1)*17 + (-2)*18`

`bar(AB)*bar n = -11 - 17 - 36 = -64 => |bar(AB)*bar n| = 64`

`|bar n| = sqrt((-11)^2 + 17^2 + 18^2) = sqrt734`

Using the found values, you may evaluate the distance, such that:

`D = 64/sqrt734`

**Hence, evaluating the smallest distance between the given lines, yields `D = 64/sqrt734` .**