# Determine the value of k for which the planes `x-2y+kz+7=0` and `x-2y-z=0` have an angle of intersection of 60 degrees? The planes x - 2y + kz = -7 and x - 2y - z = 0 intersect at an angle 60 degrees.

The normal to the plane x - 2y + kz = -7 is <1, -2, k> and the normal to x - 2y - z = 0...

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The planes x - 2y + kz = -7 and x - 2y - z = 0 intersect at an angle 60 degrees.

The normal to the plane x - 2y + kz = -7 is <1, -2, k> and the normal to x - 2y - z = 0 is <1, -2, -1>

The dot product to the vectors is `sqrt(1 + 4 + k^2)` *`sqrt(1 + 4)` *cos x = 1 + 4 - k

As x = 60 degrees, `sqrt(5 + k^2)` *`sqrt 5` *0.5 = 5 - k

=> `sqrt(5 + k^2)` *`sqrt 5` = 10 - 2k

=> `(5 + k^2)*5 = 100 + 4k^2 - 40k`

=> 25 + 5k^2 = 100 + 4k^2 - 40k

=> k^2 + 40k - 75 = 0

k1 =` -20 + sqrt 475` and k2 = `-20 - sqrt 475`

The value of k is `-20 + sqrt 475` and `-20 - sqrt 475`

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