Misc 2 - Chapter 11 3D Geometry Class 12 CBSE NCERT - Angle between two lines - Direction ratios or cosines

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  1. Chapter 11 Class 12 Three Dimensional Geometry
  2. Serial order wise
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Misc 2 (Method 1) If 𝑙﷮1﷯ , 𝑚﷮1﷯, 𝑛﷮1﷯ and 𝑙﷮2﷯ , 𝑚﷮2﷯, 𝑛﷮2﷯ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯ , 𝑛﷮1﷯ 𝑙﷮2﷯ – 𝑛﷮2﷯ 𝑙﷮1﷯ , 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯. We know that 𝑎﷯ × 𝑏﷯ is perpendicular to both 𝑎﷯ & 𝑏﷯ So, required line is cross product of lines having direction cosines 𝑙﷮1﷯ , 𝑚﷮1﷯, 𝑛﷮1﷯ and 𝑙﷮2﷯ , 𝑚﷮2﷯, 𝑛﷮2﷯ Required line = 𝑖﷯﷮ 𝑗﷯﷮ 𝑘﷯﷮ 𝑙﷮1﷯﷮ 𝑚﷮1﷯﷮ 𝑛﷮1﷯﷮ 𝑙﷮2﷯﷮ 𝑚﷮2﷯﷮ 𝑛﷮2﷯﷯﷯ = 𝑖﷯ ( 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯) – 𝑗﷯ ( 𝑙﷮1﷯ 𝑛﷮2﷯ – 𝑙﷮2﷯ 𝑛﷮1﷯) + 𝑘﷯( 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯) = ( 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯) 𝑖﷯ – ( 𝑙﷮2﷯ 𝑛﷮1﷯− 𝑙﷮1﷯ 𝑛﷮2﷯) 𝑗﷯ + ( 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯) 𝑘﷯ Hence, direction cosines = 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯ , 𝑛﷮1﷯ 𝑙﷮2﷯ – 𝑛﷮2﷯ 𝑙﷮1﷯ , 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯ ∴ Direction cosines of the line perpendicular to both of these are 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯ , 𝑛﷮1﷯ 𝑙﷮2﷯ – 𝑛﷮2﷯ 𝑙﷮1﷯ , 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯. Hence proved Misc 2 (Method 2) If 𝑙﷮1﷯ , 𝑚﷮1﷯, 𝑛﷮1﷯ and 𝑙﷮2﷯ , 𝑚﷮2﷯, 𝑛﷮2﷯ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are 𝑚﷮1﷯ 𝑛﷮2﷯ – 𝑚﷮2﷯ 𝑛﷮1﷯ , 𝑛﷮1﷯ 𝑙﷮2﷯ – 𝑛﷮2﷯ 𝑙﷮1﷯ , 𝑙﷮1﷯ 𝑚﷮2﷯ – 𝑙﷮2﷯ 𝑚﷮1﷯. Two lines having direction cosines 𝑙﷮1﷯ , 𝑚﷮1﷯, 𝑛﷮1﷯ and 𝑙﷮2﷯ , 𝑚﷮2﷯, 𝑛﷮2﷯ are mutually perpendicular if 𝒍﷮𝟏﷯ 𝒍﷮𝟐﷯+ 𝒎﷮𝟏﷯ 𝒎﷮𝟐﷯+ 𝒏﷮𝟏﷯ 𝒏﷮𝟐﷯=𝟎 Let a line having direction cosines 𝑙﷮3﷯ , 𝑚﷮3﷯, 𝑛﷮3﷯ be perpendicular to both the given lines. So, 𝒍﷮𝟑﷯ 𝒍﷮𝟏﷯+ 𝒎﷮𝟑﷯ 𝒎﷮𝟏﷯+ 𝒏﷮𝟑﷯ 𝒏﷮𝟏﷯=𝟎 & 𝒍﷮𝟑﷯ 𝒍﷮𝟐﷯+ 𝒎﷮𝟑﷯ 𝒎﷮𝟐﷯+ 𝒏﷮𝟑﷯ 𝒏﷮𝟐﷯=𝟎 From (2) and (3), So, l3 = 𝜆 (m1n2 – m2 n1) m3 = 𝜆 (n1l2 – n2 l1) n3 = 𝜆 (l1m2 – l2 m1) We have to find the value of 𝜆 Also, we know that the sum of square of direction cosines of a line is 1. 𝑙﷮1﷯﷮2﷯ + 𝑚﷮1﷯﷮2﷯ + 𝑛﷮1﷯﷮2﷯ = 1 𝑙﷮2﷯﷮2﷯ + 𝑚﷮2﷯﷮2﷯ + 𝑛﷮2﷯﷮2﷯ = 1 & 𝑙﷮3﷯﷮2﷯ + 𝑚﷮3﷯﷮2﷯ + 𝑛﷮3﷯﷮2﷯ = 1 Substituting values of l3 , m3 and n3 in (6) 𝜆2(m1n2 – m2 n1)2 + 𝜆2 (n1l2 – n2 l1)2 + 𝜆2 (l1m2 – l2 m1)2 = 1 𝜆2 (m1n2 – m2 n1)2 + (n1l2 – n2 l1)2 + (l1m2 – l2 m1)2﷯ = 1 ∴ 𝜆2 = 𝟏﷮ 𝒎﷮𝟏﷯ 𝒏﷮𝟐﷯ − 𝒎﷮𝟐﷯ 𝒏﷮𝟏﷯﷯﷮𝟐﷯ + 𝒍﷮𝟐﷯ 𝒏﷮𝟏﷯ − 𝒍﷮𝟏﷯ 𝒏﷮𝟐﷯﷯﷮𝟐﷯ + 𝒍﷮𝟏﷯ 𝒎﷮𝟐﷯ − 𝒍﷮𝟐﷯ 𝒎﷮𝟏﷯﷯﷮𝟐﷯﷯ Solving denominator, 𝑚﷮1﷯ 𝑛﷮2﷯ − 𝑚﷮2﷯ 𝑛﷮1﷯﷯﷮2﷯ + 𝑙﷮2﷯ 𝑛﷮1﷯ − 𝑙﷮1﷯ 𝑛﷮2﷯﷯﷮2﷯ + 𝑙﷮1﷯ 𝑚﷮2﷯ − 𝑙﷮2﷯ 𝑚﷮1﷯﷯﷮2﷯ = 𝑚﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯+ 𝑚﷮2﷯﷮2﷯ 𝑛﷮1﷯﷮2﷯+ 𝑙﷮2﷯﷮2﷯ 𝑛﷮1﷯﷮2﷯+ 𝑙﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ + 𝑙﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑙﷮2﷯﷮2﷯ 𝑚﷮1﷯﷮2﷯ −2𝑚1𝑛2𝑚2𝑛1−2𝑙2𝑛1𝑙1𝑛2 − 2𝑙2𝑚2𝑙2𝑚1 Method 1: Now, using equations (1), (4) and (5), i.e. 𝑙﷮1﷯ 𝑙﷮2﷯+ 𝑚﷮1﷯ 𝑚﷮2﷯+ 𝑛﷮1﷯ 𝑛﷮2﷯=0 …(1) 𝑙﷮2﷯﷮2﷯ + 𝑚﷮2﷯﷮2﷯ + 𝑛﷮2﷯﷮2﷯ = 1 …(2) & 𝑙﷮3﷯﷮2﷯ + 𝑚﷮3﷯﷮2﷯ + 𝑛﷮3﷯﷮2﷯ = 1 …(3) Solving (l12 + m12 + n12) (l22 + m22 + n22) − (l1l2 + m1m2 + n1n2)2 = (1)(1) − (0)2 = 1 − 0 = 1 Solving again (l12 + m12 + n12) (l22 + m22 + n22) − (l1l2 + m1m2 + n1n2)2 = l12 (l22 + m22 + n22) + m12 (l22 + m22 + n22) + n12 (l22 + m22 + n22) − (l12l22 + m12m22 + n12n22 + 2l1l2m1m2 + 2m1m2n1n2 + 2l1l2n1n2) = l12l22 + l12m22 + l12n22 + m12l22 + m12m22 +m12n22 + n12l22 + n12m22 + n12 n22 − l12l22 – m12m22 – n12n22 – 2l1l2m1m2 – 2 m1m2n1n2 – 2 l1l2n1n2 = l12m22 + l12n22 + m12l22 +m12n22 + n12l22 + n12m22 – 2l1l2m1m2 – 2 m1m2n1n2 – 2 l1l2n1n2 Equation (9) is same as (7) And from equation (8) 𝑚﷮1﷯ 𝑛﷮2﷯ − 𝑚﷮2﷯ 𝑛﷮1﷯﷯﷮2﷯ + 𝑙﷮2﷯ 𝑛﷮1﷯ − 𝑙﷮1﷯ 𝑛﷮2﷯﷯﷮2﷯ + 𝑙﷮1﷯ 𝑚﷮2﷯ − 𝑙﷮2﷯ 𝑚﷮1﷯﷯﷮2﷯ = 1 1﷮ 𝑚﷮1﷯ 𝑛﷮2﷯ − 𝑚﷮2﷯ 𝑛﷮1﷯﷯﷮2﷯ + 𝑙﷮2﷯ 𝑛﷮1﷯ − 𝑙﷮1﷯ 𝑛﷮2﷯﷯﷮2﷯ + 𝑙﷮1﷯ 𝑚﷮2﷯ − 𝑙﷮2﷯ 𝑚﷮1﷯﷯﷮2﷯﷯ = 1 𝜆2 = 1 ∴ 𝜆 = 1 Therefore, the direction cosines are So, l3 = 𝜆 (m1n2 – m2 n1) = 1 × (m1n2 – m2 n1) = m1n2 – m2 n1 m3 = 𝜆 (n1l2 – n2 l1) = 1 × (n1l2 – n2 l1) = n1l2 – n2 l1 n3 = 𝜆 (l1m2 – l2 m1) = 1 × (l1m2 – l2 m1) = l1m2 – l2 m1 Hence proved Method 2: = 𝑚﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯+ 𝑚﷮2﷯﷮2﷯ 𝑛﷮1﷯﷮2﷯+ 𝑙﷮2﷯﷮2﷯ 𝑛﷮1﷯﷮2﷯+ 𝑙﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ + 𝑙﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑙﷮2﷯﷮2﷯ 𝑚﷮1﷯﷮2﷯ −2𝑚1𝑛2𝑚2𝑛1−2𝑙2𝑛1𝑙1𝑛2 − 2𝑙2𝑚2𝑙2𝑚1 Adding and subtracting 𝑙﷮1﷯﷮2﷯ 𝑙﷮2﷯﷮2﷯, 𝑚﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ and 𝑛﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯, = 𝑙﷮1﷯﷮2﷯( 𝑙﷮2﷯﷮2﷯+ 𝑚﷮2﷯﷮2﷯+ 𝑛﷮2﷯﷮2﷯) − 𝑙﷮1﷯﷮2﷯ 𝑙﷮2﷯﷮2﷯ + 𝑚﷮1﷮2﷯ ( 𝑙﷮2﷯﷮2﷯+ 𝑚﷮2﷯﷮2﷯+ 𝑛﷮2﷯﷮2﷯) − 𝑚﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑛﷮1﷮2﷯ ( 𝑙﷮2﷯﷮2﷯+ 𝑚﷮2﷯﷮2﷯+ 𝑛﷮2﷯﷮2﷯) − 𝑛﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ −2𝑚1𝑛2𝑚2𝑛1−2𝑙2𝑛1𝑙1𝑛2 − 2𝑙2𝑚2𝑙2𝑚1 = 𝑙﷮1﷯﷮2﷯(1) − 𝑙﷮1﷯﷮2﷯ 𝑙﷮2﷯﷮2﷯ + 𝑚﷮1﷮2﷯ (1) − 𝑚﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑛﷮1﷮2﷯ (1) − 𝑛﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ −2𝑚1𝑛2𝑚2𝑛1−2𝑙2𝑛1𝑙1𝑛2 − 2𝑙2𝑚2𝑙2𝑚1 = ( 𝑙﷮1﷯﷮2﷯ + 𝑚﷮1﷯﷮2﷯ + 𝑛﷮1﷯﷮2﷯) − 𝑙﷮1﷯﷮2﷯ 𝑙﷮2﷯﷮2﷯ + 𝑚﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑛﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ +2𝑚1𝑛2𝑚2𝑛1﷮+2𝑙2𝑛1𝑙1𝑛2 + 2𝑙2𝑚2𝑙2𝑚1 ﷯﷯ = 1 − 𝑙﷮1﷯﷮2﷯ 𝑙﷮2﷯﷮2﷯ + 𝑚﷮1﷯﷮2﷯ 𝑚﷮2﷯﷮2﷯ + 𝑛﷮1﷯﷮2﷯ 𝑛﷮2﷯﷮2﷯ +2𝑚1𝑛2𝑚2𝑛1﷮+2𝑙2𝑛1𝑙1𝑛2 + 2𝑙2𝑚2𝑙2𝑚1 ﷯﷯ = 1 − (l1l2 + m1m2 + n1n2)2 = 1 − 02 = 1 − 0 = 1 𝑚﷮1﷯ 𝑛﷮2﷯ − 𝑚﷮2﷯ 𝑛﷮1﷯﷯﷮2﷯ + 𝑙﷮2﷯ 𝑛﷮1﷯ − 𝑙﷮1﷯ 𝑛﷮2﷯﷯﷮2﷯ + 𝑙﷮1﷯ 𝑚﷮2﷯ − 𝑙﷮2﷯ 𝑚﷮1﷯﷯﷮2﷯ = 1 1﷮ 𝑚﷮1﷯ 𝑛﷮2﷯ − 𝑚﷮2﷯ 𝑛﷮1﷯﷯﷮2﷯ + 𝑙﷮2﷯ 𝑛﷮1﷯ − 𝑙﷮1﷯ 𝑛﷮2﷯﷯﷮2﷯ + 𝑙﷮1﷯ 𝑚﷮2﷯ − 𝑙﷮2﷯ 𝑚﷮1﷯﷯﷮2﷯﷯ = 1 𝜆2 = 1 ∴ 𝜆 = 1 Therefore, the direction cosines are So, l3 = 𝜆 (m1n2 – m2 n1) = 1 × (m1n2 – m2 n1) = m1n2 – m2 n1 m3 = 𝜆 (n1l2 – n2 l1) = 1 × (n1l2 – n2 l1) = n1l2 – n2 l1 n3 = 𝜆 (l1m2 – l2 m1) = 1 × (l1m2 – l2 m1) = l1m2 – l2 m1 Hence proved

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