# Consider the lines L_1 : (4, -1, -2) + s<6, -18, 3> L_2: (3, 17, 0) + t<-5, 0, -5>The angle between these intersecting lines is ______ radians.

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### 1 Answer

You need to evaluate the angle between the lines `L_1` and `L_2` , hence, you need to find the dot product of direction vectors of the lines, such that:

`bar u = <6, -18, 3>` (direction vector of `L_1` )

`bar v = <-5, 0, -5>` (direction vector of `L_2` )

`bar v*bar u = |bar v|*|bar u|*cos alpha`

`alpha = hat(bar v, bar u)`

`|bar v| = sqrt((-5)^2 + 0^2 + (-5)^2) => |bar v| = 5sqrt2`

`|bar u| = sqrt((6)^2 + (-18)^2 + (3)^2) =>|bar u| = sqrt369 = 3sqrt41`

`bar v*bar u = 6*(-5) + 0*(-18) + 3*(-5)`

`bar v*bar u = -45`

`cos alpha = (bar v*bar u)/(|bar v|*|bar u|) => cos alpha = -45/(15sqrt82) = -3/sqrt82 = -0.331 = (2pi)/5`

**Hence, evaluating the angle between the given lines, in radians, yields `alpha = (2pi)/5.` **