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Question
If x cos θ = `y cos (theta + (2pi)/3) = z cos (theta + (4pi)/3)`, then find the value of xy + yz + zx.
Solution
Note that xy + yz + zx = `xyz (1/x + 1/y + 1/z)`.
If we put x cos θ = `y cos (theta + (2pi)/3)`
= `z cos (theta + (4pi)/3)` = k ...(say)
Then x = `k/costheta`, y = `k/(cos(theta + (2pi)/3)` and z = `k/(cos(theta + (4pi)/3)`
So that `1/x + 1/y + 1/z = 1/"k"[cos theta + cos(theta + (2pi)/3) + cos(theta + (4pi)/3)]`
= `1/k [costheta + costheta cos (2pi)/3 - sin theta sin (2pi)/3 + cos theta cos (4pi)/3 - sin theta sin (4pi)/3]`
= `1/k[cos theta + cos theta ((-1)/2) - sqrt(3)/2 sin theta - 1/2 cos theta + sqrt(3)/2 sin theta]`
= `1/k xx 0`
= 0
Hence, xy + yz + zx = 0
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