# Determine if the line `(x-1)/2`=`(y-2)/3`=`(z+3)/-2` and `(x-1)/-1`=`(y-2)/2`=`(z+3)/2` are perpendicular.

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You need to convert the symmetric forms of equation of first line in parametric form such that:

`(x-1)/2 = t =gt x = 1 + 2t`

`(y-2)/3 = t =gt y = 2 + 3t`

`(z+3)/(-2) = t =gt z = -3 - 2t`

Hence, the parametric equations of the first line are `x = 1 + 2t ; y = 2 + 3t ;z = -3 - 2t` and the direction vector of the line is `bar v_1 = lt2,3,-2gt.`

You need to convert the symmetric forms of equation of second line in parametric form such that:

`(x-1)/(-1) =s =gt x = 1- s`

`(y-2)/2 =s =gt y = 2 + 2s`

`(z+3)/2 = s =gt z = -3 - 2s`

The direction vector of the second line is `barv_2 = lt-1,2,-2gt`

You need to remember that these two lines are orthogonal if the cross product of direction vectors is zero, hence you should check if `lt2,3,-2gt*lt-1,2,-2gt = 0` .

`barv_1*barv_1 = 0`

`lt2,3,-2gt*lt-1,2,-2gt = 2*(-1) + 3*2 + (-2)*(-2)`

`lt2,3,-2gt*lt-1,2,-2gt = -2 + 6 - 4`

`lt2,3,-2gt*lt-1,2,-2gt = 0`

**Hence, since `barv_1*barv_1 = 0` , then the lines are orthogonal.**