Check sibling questions

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Ex 11.2, 1 Show that the three lines with direction cosines 12/13, (−3)/13, ( −4)/13 ; 4/13, 12/13, 3/13 ; 3/13,( − 4)/13, 12/13 are mutually perpendicular. Two lines with direction cosines 𝑙_1, 𝑚_1 , 𝑛_1 & 𝑙_2, 𝑚_2 , 𝑛_2 are perpendicular to each other if 𝒍𝟏 𝒍𝟐 + 𝒎𝟏 𝒎𝟐 + 𝒏𝟏 𝒏𝟐 = 0 𝒍𝟏 𝒍𝟐 + 𝒎𝟏 𝒎𝟐 + 𝒏𝟏 𝒏𝟐 = (12/13×4/13) + ((−3)/13×12/13) + ((−4)/13 ×3/13) = 48/169 + ((−36)/169) + ((−12)/169) Line 1 𝑙1 = 12/13, m1 = (−3)/13 , n1 = (−4)/13 Line 2 𝑙2 = 4/13, m2 = 12/13 , n2 = 3/13 = (48 − 36 − 12)/169 = (48 − 48)/169 = 0 ∴ 𝑙1 𝑙2 + 𝑚1 𝑚2 + 𝑛1 𝑛2 = 0 Hence, Line 1 and Line 2 are perpendicular. Checking Line 2 and Line 3 Line 2 𝑙2 = 4/13, m2 = 12/13 , n2 = 3/13 Line 3 𝑙3 = 3/13, m3 = (−4)/13 , n3 = 12/13 Now, 𝒍𝟐 𝒍𝟑 + 𝒎𝟐 𝒎𝟑 + 𝒏𝟐 𝒏𝟑 = (4/13×3/13) + (12/13×( − 4)/13) + (3/13×12/13) = 12/169 + (( − 48)/169) + 36/169 = (12 − 48 + 36)/169 = (48 − 48)/169 = 0 ∴ 𝑙2 𝑙3 + 𝑚2 𝑚3 + 𝑛2 𝑛3 = 0 Hence, Line 2 and Line 3 are perpendicular. Checking Line 1 and Line 3 Line 1 𝑙1 = 12/13, m1 = (−3)/13 , n1 = (−4)/13 Line 3 𝑙3 = 3/13, m3 = (−4)/13 , n3 = 12/13 Now, 𝒍𝟏 𝒍𝟑 + 𝒎𝟏 𝒎𝟑 + 𝒏𝟏 𝒏𝟑 = (3/13×12/13) + (( −4)/13×(−3)/13) + (12/13×(−4)/13) = 36/169 + 12/169 + ((−48)/169) = (36 + 12 − 48)/169 = (48 − 48)/169 = 0 ∴ 𝑙1 𝑙3 + 𝑚1 𝑚3 + 𝑛1 𝑛3 = 0 Hence, the Line 1 and 3 are perpendicular. Thus, all 3 lines are mutually perpendicular.

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Davneet Singh

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.