# Consider the lines L_1 : (4, -1, -2) + s<6, -18, 3> L_2: (3, 17, 0) + t<-5, 0, -5> What is the angle between these intersecting lines in radians. I have gotten this so far...

Consider the lines L_1 : (4, -1, -2) + s<6, -18, 3> L_2: (3, 17, 0) + t<-5, 0, -5>

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### 1 Answer

You need to evaluate the dot product of direction vectors of the given lines, such that:

`bar v_1*bar v_2 = |bar v_1|*|bar v_2|*cos alpha`

alpha represents the angle between the direction vectors of the lines

`bar v_1*bar v_2 = <6,-18,3>*<-5,0,-5>`

`bar v_1*bar v_2 = 6*(-5) + (-18)*0 + 3*(-5)`

`bar v_1*bar v_2 = -45`

`|bar v_1| = sqrt(6^2 + (-18)^2 + 3^2)`

`|bar v_1| = sqrt(369)`

`|bar v_2| = sqrt(25 + 0 + 25) => |bar v_2| = sqrt50 = 5sqrt2`

`cos alpha = (bar v_1*bar v_2)/(|bar v_1|*|bar v_2|)`

`cos alpha = -45/(5sqrt738) => cos alpha = -9/27.16 = -0.331 =>` `alpha ~~ (11pi)/18 radians` .

**Hence, evaluating the angle between the given lines, yields alpha ~~ (11pi)/18 radians.**