Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at April 16, 2024 by Teachoo

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