# Find an equation for the plane containing the (non-skew) lines L_1 : (x+11)/3 = (y+14)/3 = (z+23)/5 and L_2 : (x, y, z) = (8,-3,0) + t<-4,4,2>

sciencesolve | Certified Educator

You need to write the given equations of lines in vectorial form such that:

`L_1 = <-11,-14,-23> + <3,3,5>s`

`L_2 = <8,-3,0> + <-4,4,2>t`

You need to solve for s and t the following system of equations, to evaluate if the lines `L_1` and `L_2` intersect each other, such that:

`x = -11 + 3s = 8 - 4t`

`y = -14 + 3s = -3 + 4t`

`z = -23 + 5s = 2t`

Using the equation `-23 + 5s = 2t` you may write t in terms of s, such that:

`t = -23/2 + (5/2)s`

Substituting `-23/2 + (5/2)s` for t in one of the equations `-11 + 3s = 8 - 4t` or `-14 + 3s = -3 + 4` t, yields:

`-11 + 3s = 8 - 4(-23/2 + (5/2)s)`

Opening the brackets and isolating the terms that contain s to the left side, yields:

`3s + 10s = 11 + 8 + 46 => 13s = 65 => s = 5`

`t = -23/2 + (5/2)s => t = (-23 + 25)/2 => t = 1`

You may find the coordinates of the point of intersection of the lines L_1 and L_2, such that:

`x = 8 - 4*1 = 4`

`y = -3 + 4*1 = 1`

`z = 2*1 = 2`

Hence, the line` L_1` intercepts the line `L_2` at the point `(4,1,2)` .

You need to find the normal vector to the directional vectors of lines contained by the plane, such that:

`bar(L_1) = <3,3,5>`

`bar(L_2) = <-4,4,2>`

You need to evaluate the cross product `bar(L_1) x bar(L_2)` to find the normal vector `bar n` , such that:

`bar n = [(bar i, bar j, bar k),(3,3,5),(-4,4,2)]`

`bar n = 6bar i + 12 bar k - 20 bar j + 12 bar k - 20 bar i - 6 bar j`

`bar n = -14 bar i - 26 bar j + 24 bar k`

You may write the equation of the plane containing the lines `L_1` and `L_2` , using the normal vector and the point of intersection of lines, such that:

`P: -14(x - 4) - 26(y - 1) + 24(z - 2)`

Hence, evaluating the equation of the plane that contains the lines` L_1` and `L_2` , yields `P: -14(x - 4) - 26(y - 1) + 24(z - 2).`