# Find the value of k so that the lines (x-3)/(3k+1)= (y+6)/2 = (z+3)/2k and (x+7)/3 = (y+8)/-2k = (z+9)/-3 are perpendicular

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`A (x=3, y=-6, z=-3)` belongs to the first line because its coordinates satisfy `(x-3)/(3k+1)= (y+6)/2 = (z+3)/2k=0`

`B (x=3+3k+1, y=-6+2, z=-3+2k)`belongs to the first line too because its coordinates satisfy `(x-3)/(3k+1)= (y+6)/2 = (z+3)/2k=1`

`vec(AB) (3k+1,2,2k)` is parallel to the line.

Now lets find 2 point on the second line.

`C(-7,-8,-9) ` belongs to the second line because `(x+7)/3 = (y+8)/-2k = (z+9)/-3=0`

and `D(-7+3, -8-2k,-9-3)` belongs to the line because `(x+7)/3 = (y+8)/(-2k) = (z+9)/(-3)=1`

the vector `vec(CD) (3,-2k,-3)` is parallel to the second line.

The 2 lines are perpendicular iff `vec(AB)_|_vec(CD)`

iff their dot product is 0.

In general the dot product of u(x,y,z) and v(X,Y,Z) is given by the formula `xX+yY+zZ`

In our case, `vec(AB).vec(CD)=3(3+1)+2(-2k)+2k(-3)=0`

iff `9k+3-4k-6k=0`

iff `-k+3=0`

iff `k=3`

**Answer: The lines are perpendicular iff k=3**

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