Question 13 (a) - Angle between two planes - Chapter 11 Class 12 Three Dimensional Geometry

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Question 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)