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Ex 11.1, 5 - Find direction cosines of sides of triangle

Ex 11.1, 5 - Chapter 11 Class 12 Three Dimensional Geometry - Part 2

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Ex 11.1, 5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, −4), ( − 1, 1, 2) and ( −5, − 5, − 2).Direction ratios of a line passing through two points P(x1, y1, z1,), & Q (x2, y2, z2) = (x2 – x1), (y2 − y1), (z2 − z1) Direction cosines = (𝑥2 − 𝑥1)/𝑃𝑄 , (𝑦2 − 𝑦1)/𝑃𝑄 , (𝑧2 − 𝑧1)/𝑃𝑄 where, PQ = √((𝑥2−𝑥1)^2+(𝑦2−𝑦1)^2+(𝑧2−𝑧1)^2 ) AB A (3, 5, −4) B( −1, 1, 2) Direction ratios = −1 − 3, 1 − 5, 2 − (−4) = − 4, − 4, 6 AB = √68 = √(4 × 17 ) = 2√17 Direction cosines = ( −4)/(2√17) , ( −4)/(2√17) , 6/(2√17) = (−𝟐)/√𝟏𝟕 , (−𝟐)/√𝟏𝟕 , 𝟑/√𝟏𝟕 BC B ( −1, 1, 2) C ( −5, −5, −2) Direction ratios = −5 − (−1), −5 − 1, −2 − 2 = –4, −6, −4 BC = √68 = √(4 × 17 ) = 2√17 Direction cosines = ( −4)/(2√17) , (−6)/(2√17) , ( −4)/(2√17) = (−𝟐)/√𝟏𝟕 , (−𝟑)/√𝟏𝟕 , (−𝟐)/√𝟏𝟕 CA C ( −5, −5, −2) A (3, 5, − 4) Direction ratios = 3 − (−5), 5 − (−5), −4 − (−2) = 8, 10 , –2 CA = √168 = √(4 × 42 )= 2√42 Direction cosines = ( 8)/(2√42) , ( 10)/(2√42) , (−2)/(2√42) = ( 𝟒)/√𝟒𝟐 , ( 𝟓)/√𝟒𝟐 , (−𝟏)/√𝟒𝟐

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.