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Question

Find the acute angle between the planes r·i^-2j^-2k^ =1 and r·3i^-6j^+2k^ = 0.

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Solution

For given plane
r.i^-2j^-2k^=1 and r.3i^-6j^+2k^=0Using formula for plane r.n1=d1 and r.n2=d2cosθ=n1.n2n1n2 where θ is angle between two given planesi.e. cosθ=i^-2j^-2k^.3i^-6j^+2k^i^-2j^-2k^3i^-6j^+2k^i.e. cosθ=3+12-41+4+4 9+36+4i.e. cosθ=119 49=113×7i.e. cosθ=1121i.e. θ=cos-11121

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