Ex 11.2, 12

Transcript

Ex 11.2, 12 Find the values of 𝑝 so that the lines 1 − 𝑥﷮3﷯ = 7𝑦 − 14﷮2𝑝﷯ = 𝑧 − 3﷮2﷯ and 7 − 7𝑥﷮3𝑝﷯ = 𝑦 − 5﷮1﷯ = 6 − 𝑧﷮5﷯ are at right angles. Two lines 𝑥 − 𝑥1﷮𝑎1﷯ = 𝑦 − 𝑦1﷮𝑏1﷯ = 𝑧 − 𝑧1﷮𝑐1﷯ and 𝑥 − 𝑥2﷮𝑎2﷯ = 𝑦 − 𝑦2﷮𝑏2﷯ = 𝑧 − 𝑧2﷮𝑐2﷯ are at right angles to each other if 𝒂𝟏 𝒂𝟐 + 𝒃𝟏 𝒃𝟐 + 𝒄𝟏 𝒄𝟐 = 0 Since the lines are perpendicular 𝑎1𝑎2+𝑏1𝑏2+𝑐1𝑐2 = 0 3× − 3𝑝﷮7﷯﷯ + 2𝑝﷮7﷯×1 ﷯ + (2 × 5) = 0 9𝑝﷮7﷯ + 2𝑝﷮7﷯ − 10 = 0 11𝑝﷮7﷯ = 10 p = 10 × 7﷮11﷯ ∴ p = 𝟕𝟎﷮𝟏𝟏﷯