# Find the values of "k" so that the lines (which are in symmetric form below) are perpendicular: `(x-1)/(k) = (y+2)/(3) =(z)/(2)` `(x+2)/(k) = (y+7)/(k-2) = z-5`

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The lines given are `(x - 1)/k = (y+2)/3 = z/2` and `(x+2)/k = (y+7)/(k-2) = z-5`

`(x - 1)/k = (y+2)/3 = z/2 = t `

=> x = kt + 1, y = 3t - 2 and z = 2t

The direction vector of this line is <k, 3, 2>

`(x+2)/k = (y+7)/(k-2) = z-5 = t`

=> x = kt - 2, y = (k-2)t - 7 and z = 5 + t

The direction vector if this line is <k, (k-2), 1>

If the lines are perpendicular the dot product is 0.

k*k + 3*(k-2) + 2*1 = 0

=> k^2 + 3k - 6 + 2 = 0

=> k^2 + 3k - 4 = 0

=> k^2 + 4k - k - 4 = 0

=> k(k + 4) - 1(k + 4) = 0

=> (k - 1)(k + 4) = 0

k = 1 and k = -4

The two lines are perpendicular when k is equal to 1 or when k is equal to -4