# Find the distance between the line L: (x,y,z) = (0,4,-3) + t<4,28,-24> and each of the following lines below.a) L_1 : (x,y,z) = (-5,5,3) + t<-3, -21, 18> The distance between L and L_1...

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

a) L_1 : (x,y,z) = (-5,5,3) + t<-3, -21, 18>

The distance between L and L_1 is ?

### 1 Answer | Add Yours

The line `L: (x,y,z) = (0,4,-3) +t*<4,28,-24>` has the directional vector `V_1=<4,28,-24> =4*<1,7,-6>`

The line `L¨_1: (x,y,z) =(-5,5,3) +t*<-3,-21,18>` has the directional vector `V_2=<-3,-21,18> =3*<-1,-7,6> =-3*<1,7,-6>` ` `

Since the two directional vectors are the same `V =<1,7,-6>` the lines `L` and `L_1` are parallel so the distance between them is equal at any point in space. If we chose points `A=(0,4,3)` on first line and `B =(-5,5,3)` on the second line the distance between lines is (see the figure)

`d = |<AB> xx V|/|V|`

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

`<AB> xx V =|[i,j,k],[5,-1,0],[1,7,-6]| =-6i +30j+36k =<-6,30,36>`

`|<-6,30,36>| =sqrt(2232)`

`|V| =|<1,7,-6>| =sqrt(86)`

Therefore the distance between the two given lines is

`d =sqrt(2232/86) =sqrt(1116/43) ~~5.094`

**Answer: the disatnce between the two lines is `d=sqrt(1116/43)` **

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