Ex 11.3, 12 - Find angle between planes r.(2i + 2j - 3k) = 5 - Ex 11.3

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Ex 11.3, 12 Find the angle between the planes whose vector equations are 𝑟﷯ . (2 𝑖﷯ + 2 𝑗﷯ – 3 𝑘﷯) = 5 and 𝑟﷯ . (3 𝑖﷯ – 3 𝑗﷯ + 5 𝑘﷯) = 3 . Angle between two planes 𝑟﷯ . 𝑛﷮1﷯﷯ = d1 and 𝑟﷯. 𝑛2﷯ = d2 is given by cos 𝜃 = 𝒏𝟏﷯. 𝒏𝟐﷯﷮ 𝒏𝟏﷯﷯ 𝒏𝟐﷯﷯﷯﷯ Given, the two planes are So, cos θ = 2 𝑖﷯ + 2 𝑗﷯ − 3 𝑘﷯﷯ . 3 𝑖﷯ − 3 𝑗﷯ + 5 𝑘﷯﷯﷮ ﷮17﷯ × ﷮43﷯﷯﷯ = 2 × 3﷯ + 2 × −3﷯ + (−3 × 5)﷮ ﷮17 × 43﷯﷯﷯ = 6 − 6 − 15﷮ ﷮731﷯﷯﷯ = −15﷮ ﷮731﷯﷯﷯ = 15﷮ ﷮731﷯﷯ So, cos θ = 15﷮ ﷮731﷯﷯ ∴ θ = cos−1 𝟏𝟓﷮ ﷮𝟕𝟑𝟏﷯﷯﷯ Therefore, the angle between the planes is cos−1 15﷮ ﷮731﷯﷯﷯.

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