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Ex 11.3

Ex 11.3, 1 (a)
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Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

Last updated at Aug. 18, 2020 by Teachoo

Ex 11.3, 13 In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. (a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0 7x + 5y + 6z + 30 = 0 7x + 5y + 6z = −30 −(7x + 5y + 6z) = 30 −7x – 5y – 6z = 30 Comparing with A1x + B1y + C1z = d1 Direction ratios of normal = –7, –5, –6 A1 = –7 , B1 = –5 , C1 = –6 3x − y − 10z + 4 = 0 3x − y − 10z = −4 −(3x − y − 10z) = 4 −3x + y + 10z = 4 Comparing with A2x + B2y + C2z = d2 Direction ratios of normal = –3, 1, 10 A2 = –3 , B2 = 1 , C2 = 10 Check parallel Two lines with direction ratios 𝐴_1, 𝐵_1, 𝐶_1 and 𝐴_2, 𝐵_2, 𝐶_2 are parallel if 𝑨_𝟏/𝑨_𝟐 = 𝑩_𝟏/𝑩_𝟐 = 𝑪_𝟏/𝑪_𝟐 So, 𝐴_1/𝐴_2 = (−7)/(−3) = 7/3, 𝐵_1/𝐵_2 = (−5)/1 = −5 , 𝐶_1/𝐶_2 = (−6)/10 = (−3)/5 Since 𝐴_1/𝐴_2 ≠ 𝐵_1/𝐵_2 ≠ 𝐶_1/𝐶_2 So, the two normal are not parallel. ∴ Given two planes are not parallel. Check perpendicular Two lines with direction ratios 𝐴_1, 𝐵_1, 𝐶_1 and 𝐴_2, 𝐵_2, 𝐶_2 are perpendicular if 𝑨_𝟏 𝑨_𝟐 + 𝑩_𝟏 𝑩_𝟐 + 𝑪_𝟏 𝑪_𝟐 = 0 𝐴_1 𝐴_2 + 𝐵_1 𝐵_2 + 𝐶_1 𝐶_2 = (−7 × −3) + (−5 × 1) + (−6 × 10) = 21 + (−5) + (−60) = −44 Since, 𝐴_1 𝐴_2 + 𝐵_1 𝐵_2 + 𝐶_1 𝐶_2 ≠ 0 Therefore, the two normal are not perpendicular. Hence, the given two planes are not perpendicular. Finding angle Now, the angle between two planes 𝐴_1x + 𝐵_1 𝑦 + 𝐶_1 𝑧 = d1 and 𝐴_2x + 𝐵_2 𝑦 + 𝐶_2 𝑧 = d2 is given by cos θ = |(𝑨_𝟏 𝑨_𝟐 + 𝑩_𝟏 𝑩_𝟐 + 𝑪_𝟏 𝑪_𝟐)/(√(〖𝑨_𝟏〗^𝟐 + 〖𝑩_𝟏〗^𝟐 + 〖𝑪_𝟏〗^𝟐 ) √(〖𝑨_𝟐〗^𝟐 + 〖𝑩_𝟐〗^𝟐 + 〖𝑪_𝟐〗^𝟐 ))| = |((−7 × −3) + (−5 × 1) + (−6 × 10) )/(√((−7)^2+(−5)^2+(−6)^2 ) √((−3)^2 + 1^2 + 〖10〗^2 ))| = |(21 − 5 − 60)/(√(49 + 25 + 36) √(9 + 1 + 100))| = |(−44)/(√110 √110)| = |(−44)/110| = |(−2)/5| = 2/5 Hence, cos θ = 2/5 ∴ θ = 〖𝐜𝐨𝐬〗^(−𝟏)(𝟐/𝟓) Hence, angle between two planes is cos^(−1)(2/5)