XII GM ANSWER KEY FINAL

CHENNAI DISTRICT XII STANDARD – MATHEMATICS QUARTERLY EXAMINATION 2026 – 27 ANSWER KEY 1 (d) [ 𝟒 𝟐 − 𝟏 𝟏 ] (1) 2 (a) 11 (1) 3 (b) 𝟏 (1) 4 (b) 𝟏 𝟐 | 𝒛 | 𝟐 (1) 5 (c) 1 (1) 6 (b) 𝟏 (1) 7 (b) − 𝒒 𝒓 (1) 8 (c) 2 (1) 9 (d) N (1) 10 (d) 𝝅 𝟐 − 𝒙 (1) 11 (d) 𝝅 𝟐 (1) 12 (c) Unique solution (1) 13 (d) 𝟐 √ 𝟑 (1) 14 (b) 𝒙 + 𝟐𝒚 = 𝟑 (1) 15 (d) ( 𝟗 𝟐 √ 𝟐 , 𝟏 √ 𝟐 ) (1) 16 (c) 𝟎 ∘ (1) 17 (d) 𝟎 (1) 18 (a) √ 𝟕 𝟐 √ 𝟐 (1) 19 (a) 𝟏𝟖 𝟓 (1) 20 (b) − 𝟏 (1) PART – II 21 𝑨 = [ 𝟐 − 𝟏 𝟑 𝟐 ] , 𝑿 = [ 𝒙 𝒚 ] , 𝑩 = [ 𝟖 − 𝟐 ] ; hence 𝑨𝑿 = 𝑩 | 𝑨 | = 𝟕 , 𝑨 − 𝟏 = 𝟏 𝟕 [ 𝟐 𝟏 − 𝟑 𝟐 ] 𝑿 = 𝑨 − 𝟏 𝑩 = 𝟏 𝟕 [ 𝟐 𝟏 − 𝟑 𝟐 ] [ 𝟖 − 𝟐 ] = [ 𝟐 − 𝟒 ] . Therefore 𝒙 = 𝟐 , 𝒚 = − 𝟒 (2*) 22 𝒊 𝟏𝟕𝟐𝟗 = 𝒊 𝟒 ( 𝟒𝟑𝟐 ) + 𝟏 𝒊 𝟒 = 𝟏 ⇒ 𝒊 𝟏𝟕𝟐𝟗 = ( 𝒊 𝟒 ) 𝟒𝟑𝟐 𝒊 ∴ 𝒊 𝟏𝟕𝟐𝟗 = 𝒊 (1) (1) 23 𝒛 = 𝒙 + 𝒊𝒚 ⇒ 𝒛 ̅ = 𝒙 − 𝒊𝒚 𝒊 𝒛 ̅ = 𝒊 ( 𝒙 − 𝒊𝒚 ) = 𝒚 + 𝒊𝒙 ∴ 𝐑𝐞 ( 𝒊 𝒛 ̅ ) = 𝒚 (1) (1) 24 Required equation: ( 𝒙 − 𝟐 ) ( 𝒙 − 𝟏 𝟐 ) ( 𝒙 − 𝟏 ) = 𝟎 ( 𝒙 − 𝟐 ) ( 𝟐 𝒙 − 𝟏 ) ( 𝒙 − 𝟏 ) = 𝟎 ∴ 𝟐 𝒙 𝟑 − 𝟕 𝒙 𝟐 + 𝟕 𝒙 − 𝟐 = 𝟎 (1) (1) 25 𝐬𝐢𝐧 𝟓 𝝅 𝟔 = 𝐬𝐢𝐧 ( 𝝅 − 𝝅 𝟔 ) Principal range of 𝐬𝐢𝐧 − 𝟏 𝒙 is [ − 𝝅 𝟐 , 𝝅 𝟐 ] ∴ 𝐬𝐢𝐧 − 𝟏 ( 𝐬𝐢𝐧 𝟓 𝝅 𝟔 ) = 𝝅 𝟔 (1) (1) 26 For 𝐜𝐨𝐬 − 𝟏 ( 𝟐 + 𝐬𝐢𝐧 𝒙 𝟑 ) , require − 𝟏 ≤ 𝟐 + 𝐬𝐢𝐧 𝒙 𝟑 ≤ 𝟏 − 𝟏 ≤ 𝐬𝐢𝐧 𝒙 ≤ 𝟏 ⇒ 𝟏 𝟑 ≤ 𝟐 + 𝐬𝐢𝐧 𝒙 𝟑 ≤ 𝟏 ∴ Domain = ℝ (1) (1) 27 𝒚 = 𝟐 √ 𝟐 𝒙 + 𝒄 ⇒ 𝟐 √ 𝟐 𝒙 − 𝒚 + 𝒄 = 𝟎 Condition substitution Distance from ( 𝟎 , 𝟎 ) to tangent = radius: | 𝒄 | √ 𝟖 + 𝟏 = 𝟒 | 𝒄 | = 𝟏𝟐 . Therefore 𝒄 = ± 𝟏𝟐 (2*) 28 Vertex ( 𝒉 , 𝒌 ) = ( 𝟏 , − 𝟐 ) and focus ( 𝟒 , − 𝟐 ) give 𝒂 = 𝟑 Standard form: ( 𝒚 − 𝒌 ) 𝟐 = 𝟒 𝒂 ( 𝒙 − 𝒉 ) ∴ ( 𝒚 + 𝟐 ) 𝟐 = 𝟏𝟐 ( 𝒙 − 𝟏 ) (1) (1) 29 𝑽 = | ∣ ∣ ∣ ∣ 𝟐 − 𝟑 𝟒 𝟏 𝟐 − 𝟏 𝟑 − 𝟏 𝟐 ∣ ∣ ∣ ∣ | = | 𝟐 ( 𝟒 − 𝟏 ) + 𝟑 ( 𝟐 + 𝟑 ) + 𝟒 ( − 𝟏 − 𝟔 ) | = | − 𝟕 | . Therefore 𝑽 = 𝟕 cubic units. (1) (1) 30 Plane: 𝟔 𝒙 + 𝟒 𝒚 − 𝟑 𝒛 = 𝟏𝟐 𝒚 = 𝒛 = 𝟎 ⇒ 𝒙 = 𝟐 ; 𝒙 = 𝒛 = 𝟎 ⇒ 𝒚 = 𝟑 𝒙 = 𝒚 = 𝟎 ⇒ 𝒛 = − 𝟒 Intercepts: ( 𝟐 , 𝟎 , 𝟎 ) , ( 𝟎 , 𝟑 , 𝟎 ) , ( 𝟎 , 𝟎 , − 𝟒 ) (2*) PART – III 31 [ 𝟎 𝟓 𝟏 𝟎 − 𝟏 𝟔 𝟎 𝟏 ] 𝑹 𝟏 ↔ 𝑹 𝟐 , 𝑹 𝟏 → − 𝑹 𝟏 ⇒ [ 𝟏 − 𝟔 𝟎 − 𝟏 𝟎 𝟓 𝟏 𝟎 ] 𝑹 𝟐 → 𝟏 𝟓 𝑹 𝟐 ⇒ [ 𝟏 − 𝟔 𝟎 − 𝟏 𝟎 𝟏 𝟏 𝟓 𝟎 ] 𝑹 𝟏 → 𝑹 𝟏 + 𝟔 𝑹 𝟐 ∴ 𝑨 − 𝟏 = [ 𝟔 𝟓 − 𝟏 𝟏 𝟓 𝟎 ] (1) (2*) 32 Let 𝑩 = 𝐚𝐝𝐣 𝑨 = [ 𝟏 𝟎 𝟏 𝟎 𝟐 𝟎 − 𝟏 𝟎 𝟏 ] | 𝑩 | = 𝟒 = ( | 𝑨 | ) 𝟐 For a 𝟑 × 𝟑 matrix, 𝐚𝐝𝐣 ( 𝐚𝐝𝐣 𝑨 ) = | 𝑨 | 𝑨 = | 𝑨 | 𝟐 𝑩 − 𝟏 ∴ 𝐚𝐝𝐣 ( 𝐚𝐝𝐣 𝑨 ) = 𝐚𝐝𝐣 𝑩 = [ 𝟐 𝟎 − 𝟐 𝟎 𝟐 𝟎 𝟐 𝟎 𝟐 ] (Any alternate method) (1) (1) (1) 33 | 𝒛 + 𝒊 | 𝟐 = 𝒙 𝟐 + ( 𝒚 + 𝟏 ) 𝟐 | 𝒛 − 𝟏 | 𝟐 = ( 𝒙 − 𝟏 ) 𝟐 + 𝒚 𝟐 𝒙 𝟐 + ( 𝒚 + 𝟏 ) 𝟐 = ( 𝒙 − 𝟏 ) 𝟐 + 𝒚 𝟐 ∴ 𝒙 + 𝒚 = 𝟎 (1) (1) (1) 34 Put 𝒚 = 𝒙 𝟐 : 𝒚 𝟐 − 𝟏𝟒 𝒚 + 𝟒𝟓 = 𝟎 ( 𝒚 − 𝟓 ) ( 𝒚 − 𝟗 ) = 𝟎 𝒙 𝟐 = 𝟓 or 𝒙 𝟐 = 𝟗 ∴ 𝒙 = ± √ 𝟓 , ± 𝟑 (1) (1) (1) 35 𝐭𝐚𝐧 − 𝟏 ( − 𝟏 ) = − 𝝅 𝟒 𝐜𝐨𝐬 − 𝟏 ( 𝟏 𝟐 ) = 𝝅 𝟑 𝐬𝐢𝐧 − 𝟏 ( − 𝟏 𝟐 ) = − 𝝅 𝟔 Sum = − 𝝅 𝟒 + 𝝅 𝟑 − 𝝅 𝟔 = − 𝝅 𝟏𝟐 (2*) (1) 36 Direction of line 𝒅 ⃗ = ( 𝟏 , 𝟐 , − 𝟐 ) ; normal to plane 𝒏 ⃗ = ( 𝟔 , 𝟑 , 𝟐 ) 𝐬𝐢𝐧 𝜽 = | 𝒅 ⃗ ⃗ ⃗ ⋅ 𝒏 ⃗ ⃗ ⃗ | | 𝒅 ⃗ ⃗ ⃗ | | 𝒏 ⃗ ⃗ ⃗ | 𝒅 ⃗ ⋅ 𝒏 ⃗ = 𝟖 , | 𝒅 ⃗ | = 𝟑 , | 𝒏 ⃗ | = 𝟕 ∴ 𝜽 = 𝐬𝐢𝐧 − 𝟏 ( 𝟖 𝟐𝟏 ) (1) (2*) 37 𝒙 𝟐 𝟑𝟐 + 𝒚 𝟐 𝟖 = 𝟏 , 𝒂 = 𝟒 √ 𝟐 , 𝒃 = 𝟐 √ 𝟐 At 𝜽 = 𝝅 𝟒 , point is ( 𝒂 𝐜𝐨𝐬 𝜽 , 𝒃 𝐬𝐢𝐧 𝜽 ) = ( 𝟒 , 𝟐 ) Tangent: 𝟒 𝒙 𝟑𝟐 + 𝟐 𝒚 𝟖 = 𝟏 ⇒ 𝒙 + 𝟐 𝒚 = 𝟖 Normal slope = 𝟐 through ( 𝟒 , 𝟐 ) : 𝟐 𝒙 − 𝒚 − 𝟔 = 𝟎 (1) (1) (1) 38 Take vertex ( 𝟎 , 𝟎 ) and focus ( 𝟏 𝟐 , 𝟎 ) ; hence 𝒂 = 𝟏 𝟐 𝒚 𝟐 = 𝟒 𝒂𝒙 ⇒ 𝒚 𝟐 = 𝟒 𝟖 𝒙 equation 𝒚 𝟐 = 𝟒 𝟖 𝒙 At opening, 𝒚 = ± 𝟐 𝟓 : 𝟔 𝟐𝟓 = 𝟒 𝟖 𝒙 Depth 𝒙 = 𝟔 𝟐𝟓 𝟒 𝟖 ≈ 𝟏 𝟑𝟎 m;. (1) (1) (1) 39 𝑨 = ( 𝟐 , 𝟑 , 𝟒 ) , 𝑩 = ( − 𝟏 , 𝟒 , 𝟓 ) , 𝑪 = ( 𝟖 , 𝟏 , 𝟐 ) 𝑨𝑩 ⃗ ⃗ ⃗ ⃗ = ( − 𝟑 , 𝟏 , 𝟏 ) 𝑨𝑪 ⃗ ⃗ ⃗ = ( 𝟔 , − 𝟐 , − 𝟐 ) ALTERNATE METHOD 𝑨𝑪 ⃗ ⃗ ⃗ = − 𝟐 𝑨𝑩 ⃗ ⃗ ⃗ ⃗ . Hence the points are collinear. (2) (1) 40 Let 𝜽 = 𝐬𝐞𝐜 − 𝟏 𝒙 , so 𝐬𝐞𝐜 𝜽 = 𝒙 𝐭𝐚𝐧 𝟐 𝜽 = 𝐬𝐞𝐜 𝟐 𝜽 − 𝟏 = 𝒙 𝟐 − 𝟏 Hence 𝐜𝐨𝐭 𝜽 = 𝟏 √ 𝒙 𝟐 − 𝟏 on the relevant principal branch. ∴ 𝐜𝐨𝐭 − 𝟏 ( 𝟏 √ 𝒙 𝟐 − 𝟏 ) = 𝐬𝐞𝐜 − 𝟏 𝒙 (1) (2*) PART – IV 41(a) 𝑨 = [ 𝟐 𝟑 − 𝟏 𝟏 𝟏 𝟏 𝟑 − 𝟏 − 𝟏 ] , 𝑿 = [ 𝒙 𝒚 𝒛 ] , 𝑩 = [ 𝟗 𝟗 𝟏 ] | 𝑨 | = 𝟏𝟔 ≠ 𝟎 𝑪 = [ 𝟎 𝟒 − 𝟒 𝟒 𝟏 𝟏𝟏 𝟒 − 𝟑 − 𝟏 ] 𝐚𝐝𝐣 𝑨 = 𝑪 𝑻 = [ 𝟎 𝟒 𝟒 𝟒 𝟏 − 𝟑 − 𝟒 𝟏𝟏 − 𝟏 ] 𝑨 − 𝟏 = 𝟏 𝟏𝟔 𝐚𝐝𝐣 𝑨 , 𝑿 = 𝑨 − 𝟏 𝑩 𝑿 = [ 𝟒𝟎 / 𝟏𝟔 𝟒𝟐 / 𝟏𝟔 𝟔𝟐 / 𝟏𝟔 ] . Hence 𝒙 = 𝟓 𝟐 , 𝒚 = 𝟐𝟏 𝟖 , 𝒛 = 𝟑𝟏 𝟖 (1) (1) (1) (2*) 41(b) [ 𝟏 − 𝟏 𝟐 𝟐 𝟐 𝟏 𝟒 𝟕 𝟒 − 𝟏 𝟏 𝟒 ] 𝑹 𝟐 → 𝑹 𝟐 − 𝟐 𝑹 𝟏 ⇒ [ 𝟎 , 𝟑 , 𝟎 | 𝟑 ] 𝑹 𝟑 → 𝑹 𝟑 − 𝟒 𝑹 𝟏 ⇒ [ 𝟎 , 𝟑 , − 𝟕 | − 𝟒 ] 𝑹 𝟑 → 𝑹 𝟑 − 𝑹 𝟐 ⇒ [ 𝟎 , 𝟎 , − 𝟕 | − 𝟕 ] 𝒛 = 𝟏 , 𝒚 = 𝟏 , 𝒙 = 𝟏 . Therefore ( 𝒙 , 𝒚 , 𝒛 ) = ( 𝟏 , 𝟏 , 𝟏 ) Each Step 1 Mark 42(a) 𝒛 − 𝒊 = 𝒙 + 𝒊 ( 𝒚 − 𝟏 ) , 𝒛 + 𝟐 = ( 𝒙 + 𝟐 ) + 𝒊𝒚 𝒛 − 𝒊 𝒛 + 𝟐 = ( 𝒙 + 𝒊 ( 𝒚 − 𝟏 ) ) ( ( 𝒙 + 𝟐 ) − 𝒊𝒚 ) ( 𝒙 + 𝟐 ) 𝟐 + 𝒚 𝟐 Real numerator = 𝒙 𝟐 + 𝟐 𝒙 + 𝒚 𝟐 − 𝒚 Imaginary numerator = 𝟐 𝒚 − 𝒙 − 𝟐 Argument = 𝝅 𝟒 ⇒ real part = imaginary part. 𝟐 𝒚 − 𝒙 − 𝟐 = 𝒙 𝟐 + 𝟐 𝒙 + 𝒚 𝟐 − 𝒚 ∴ 𝒙 𝟐 + 𝒚 𝟐 + 𝟑 𝒙 − 𝟑 𝒚 + 𝟐 = 𝟎 (1) (1) (1) (1) (1) 42(b) 𝟏𝟗 + 𝟗 𝒊 𝟓 − 𝟑 𝒊 = 𝟐 + 𝟑 𝒊 𝟖 + 𝒊 𝟏 + 𝟐 𝒊 = 𝟐 − 𝟑 𝒊 Let 𝒘 = 𝟐 + 𝟑 𝒊 ; then 𝒘 ̅ = 𝟐 − 𝟑 𝒊 Expression = 𝒘 𝟏𝟓 − ( 𝒘 ̅ ) 𝟏𝟓 If 𝒘 𝟏𝟓 = 𝒂 + 𝒊𝒃 , then ( 𝒘 ̅ ) 𝟏𝟓 = 𝒂 − 𝒊𝒃 Difference = 𝟐 𝒊𝒃 . Hence it is purely imaginary. (2) (1) (1) (1) 43(a) Divide by 𝒙 𝟐 : 𝟔 ( 𝒙 𝟐 + 𝟏 𝒙 𝟐 ) − 𝟑𝟓 ( 𝒙 + 𝟏 𝒙 ) + 𝟔𝟐 = 𝟎 Put 𝒚 = 𝒙 + 𝟏 𝒙 ; then 𝒙 𝟐 + 𝟏 𝒙 𝟐 = 𝒚 𝟐 − 𝟐 𝟔 ( 𝒚 𝟐 − 𝟐 ) − 𝟑𝟓 𝒚 + 𝟔𝟐 = 𝟎 ⇒ 𝟔 𝒚 𝟐 − 𝟑𝟓 𝒚 + 𝟓𝟎 = 𝟎 ( 𝟑 𝒚 − 𝟏𝟎 ) ( 𝟐 𝒚 − 𝟓 ) = 𝟎 ⇒ 𝒚 = 𝟏𝟎 𝟑 , 𝟓 𝟐 𝒙 + 𝟏 𝒙 = 𝟏𝟎 𝟑 ⇒ 𝒙 = 𝟑 , 𝟏 𝟑 𝒙 + 𝟏 𝒙 = 𝟓 𝟐 ⇒ 𝒙 = 𝟐 , 𝟏 𝟐 Solutions: 𝟑 , 𝟏 𝟑 , 𝟐 , 𝟏 𝟐 (1) (1) (1) (1+1) 43(b) 𝒂 𝟐 − 𝒂 𝟏 = 𝒅 ⇒ 𝐭𝐚𝐧 − 𝟏 𝒅 𝟏 + 𝒂 𝟏 𝒂 𝟐 = 𝐭𝐚𝐧 − 𝟏 𝒂 𝟐 − 𝐭𝐚𝐧 − 𝟏 𝒂 𝟏 𝐭𝐚𝐧 − 𝟏 𝒅 𝟏 + 𝒂 𝟐 𝒂 𝟑 = 𝐭𝐚𝐧 − 𝟏 𝒂 𝟑 − 𝐭𝐚𝐧 − 𝟏 𝒂 𝟐 Proceed similarly for all consecutive terms. On addition, all intermediate inverse - tangent terms cancel. Sum = 𝐭𝐚𝐧 − 𝟏 𝒂 𝒏 − 𝐭𝐚𝐧 − 𝟏 𝒂 𝟏 Taking tangent, 𝐭𝐚𝐧 ( sum ) = 𝒂 𝒏 − 𝒂 𝟏 𝟏 + 𝒂 𝟏 𝒂 𝒏 (2*) (1) (2*) 44(a) Given line 𝒙 − 𝒚 + 𝟒 = 𝟎 ⇒ 𝒚 = 𝒙 + 𝟒 m = 1 c = 4 Substitute: 𝒙 𝟐 + 𝟑 𝒚 𝟐 = 𝟏𝟐 𝒂 𝟐 = 𝟒 𝒃 𝟐 = 𝟑 Condition substitution conclusion 𝒙 = − 𝟑 , 𝒚 = 𝟏 . Point of contact = ( − 𝟑 , 𝟏 ) (1) (1) (2*) (1) 44(b) 𝟒 ( 𝒙 𝟐 + 𝟔 𝒙 ) + ( 𝒚 𝟐 − 𝟐 𝒚 ) + 𝟐𝟏 = 𝟎 𝟒 ( 𝒙 + 𝟑 ) 𝟐 + ( 𝒚 − 𝟏 ) 𝟐 = 𝟏𝟔 ⇒ ( 𝒙 + 𝟑 ) 𝟐 𝟒 + ( 𝒚 − 𝟏 ) 𝟐 𝟏𝟔 = 𝟏 Centre = ( − 𝟑 , 𝟏 ) ; 𝒂 = 𝟒 , 𝒃 = 𝟐 with vertical major axis. Vertices: ( − 𝟑 , 𝟓 ) , ( − 𝟑 , − 𝟑 ) 𝒄 = √ 𝒂 𝟐 − 𝒃 𝟐 = 𝟐 √ 𝟑 ; foci: ( − 𝟑 , 𝟏 ± 𝟐 √ 𝟑 ) Length of latus rectum = 𝟐 𝒃 𝟐 𝒂 = 𝟐 . Hence proved. (1) (1) (1) (1) (1) 45(a) Let top - to - centre distance = 𝒕 and base - to - centre distance = 𝟐 𝒕 ; 𝟑 𝒕 = 𝟏𝟓𝟎 ⇒ 𝒕 = 𝟓𝟎 m. Thus 𝒚 = 𝟓𝟎 m at top and 𝒚 = − 𝟏𝟎𝟎 m at base. From 𝒙 𝟐 𝟑𝟎 𝟐 − 𝒚 𝟐 𝟒𝟒 𝟐 = 𝟏 , 𝒙 𝟐 = 𝟗𝟎𝟎 ( 𝟏 + 𝒚 𝟐 𝟏𝟗𝟑𝟔 ) At top: 𝒙 = 𝟑𝟎 √ 𝟏 + 𝟐𝟓𝟎𝟎 𝟏𝟗𝟑𝟔 ≈ 𝟒𝟓 𝟑𝟖 m. Top diameter ≈ 𝟗𝟎 𝟕𝟔 m At base: 𝒙 = 𝟑𝟎 √ 𝟏 + 𝟏𝟎𝟎𝟎𝟎 𝟏𝟗𝟑𝟔 ≈ 𝟕𝟒 𝟒𝟕 m.; base diameter ≈ 𝟏𝟒𝟖 𝟗𝟒 m. (1) (2*) (2*) 45(b) Diagram 𝒂 ⃗ = and 𝒃 ⃗ ⃗ ⃗ = 𝒂 ⃗ × 𝒃 ⃗ = 𝐬𝐢𝐧 ( 𝜶 + 𝜷 ) 𝒌 ̂ 𝒂 ⃗ × 𝒃 ⃗ = − ( 𝐜𝐨𝐬 𝜶 𝐬𝐢𝐧 𝜷 + 𝐬𝐢𝐧 𝜶 𝐜𝐨𝐬 𝜷 ) 𝒌 ̂ 𝐬𝐢𝐧 ( 𝜶 + 𝜷 ) = 𝐜𝐨𝐬 𝜶 𝐬𝐢𝐧 𝜷 + 𝐬𝐢𝐧 𝜶 𝐜𝐨𝐬 𝜷 ∴ 𝐬𝐢𝐧 ( 𝜶 + 𝜷 ) = 𝐬𝐢𝐧 𝜶 𝐜𝐨𝐬 𝜷 + 𝐜𝐨𝐬 𝜶 𝐬𝐢𝐧 𝜷 (1) (1) (1) (1) (1) 46(a) 𝒂 ⃗ × 𝒃 ⃗ = ∣ ∣ ∣ ∣ 𝒊 ̂ 𝒋 ̂ 𝒌 ̂ 𝟐 𝟑 − 𝟏 𝟑 𝟓 𝟐 ∣ ∣ ∣ ∣ = ( 𝟏𝟏 , − 𝟕 , 𝟏 ) ( 𝒂 ⃗ × 𝒃 ⃗ ) × 𝒄 ⃗ = ( 𝟏𝟏 , − 𝟕 , 𝟏 ) × ( − 𝟏 , − 𝟐 , 𝟑 ) = ( − 𝟏𝟗 , − 𝟑𝟒 , − 𝟐𝟗 ) 𝒂 ⃗ ⋅ 𝒄 ⃗ = − 𝟏𝟏 𝒃 ⃗ ⋅ 𝒄 ⃗ = − 𝟕 ( 𝒂 ⃗ ⋅ 𝒄 ⃗ ) 𝒃 ⃗ − ( 𝒃 ⃗ ⋅ 𝒄 ⃗ ) 𝒂 ⃗ = − 𝟏𝟏 ( 𝟑 , 𝟓 , 𝟐 ) + 𝟕 ( 𝟐 , 𝟑 , − 𝟏 ) = ( − 𝟏𝟗 , − 𝟑𝟒 , − 𝟐𝟗 ) = LHS. Hence verified. (1) (1) (1) (1) (1) 46(b) Let − 𝒙 − 𝟐 𝟒 = 𝒚 + 𝟑 − 𝟐 = 𝟐 𝒛 − 𝟔 𝟑 = 𝒕 𝒙 = − 𝟐 − 𝟒 𝒕 , 𝒚 = − 𝟑 − 𝟐 𝒕 , 𝒛 = 𝟑 + 𝟑 𝟐 𝒕 A direction vector is ( − 𝟒 , − 𝟐 , 𝟑 𝟐 ) , or equivalently ( − 𝟖 , − 𝟒 , 𝟑 ) Required line passes through ( − 𝟒 , 𝟐 , − 𝟑 ) Vector form: 𝒓 ⃗ = ( − 𝟒 𝒊 ̂ + 𝟐 𝒋 ̂ − 𝟑 𝒌 ̂ ) + 𝝀 ( − 𝟖 𝒊 ̂ − 𝟒 𝒋 ̂ + 𝟑 𝒌 ̂ ) Cartesian form: 𝒙 + 𝟒 − 𝟖 = 𝒚 − 𝟐 − 𝟒 = 𝒛 + 𝟑 𝟑 (any alternate method) 47(a) Plane: 𝒙 + 𝒚 − 𝒛 − 𝟗 = 𝟎 ; normal 𝒏 ⃗ = ( 𝟏 , 𝟏 , − 𝟏 ) For 𝑷 ( 𝟓 , 𝟐 , 𝟔 ) , 𝑺 = 𝟓 + 𝟐 − 𝟔 − 𝟗 = − 𝟖 Reflection formula: 𝑷 ′ = 𝑷 − 𝟐 𝑺 𝒂 𝟐 + 𝒃 𝟐 + 𝒄 𝟐 ( 𝒂 , 𝒃 , 𝒄 ) 𝒂 𝟐 + 𝒃 𝟐 + 𝒄 𝟐 = 𝟑 𝑷 ′ = ( 𝟓 , 𝟐 , 𝟔 ) + 𝟏𝟔 𝟑 ( 𝟏 , 𝟏 , − 𝟏 ) ∴ 𝑷 ′ = ( 𝟑𝟏 𝟑 , 𝟐𝟐 𝟑 , 𝟐 𝟑 ) (1) (1) (1) (1) (1) 47(b) ( 𝒙 − 𝟒 ) ( 𝒙 + 𝟏 ) = 𝒙 𝟐 − 𝟑 𝒙 − 𝟒 ( 𝒙 − 𝟐 ) ( 𝒙 − 𝟏 ) = 𝒙 𝟐 − 𝟑 𝒙 + 𝟐 Put 𝒚 = 𝒙 𝟐 − 𝟑 𝒙 : ( 𝒚 − 𝟒 ) ( 𝒚 + 𝟐 ) + 𝟖 = 𝟎 𝒚 𝟐 − 𝟐 𝒚 = 𝟎 ⇒ 𝒚 ( 𝒚 − 𝟐 ) = 𝟎 𝒙 𝟐 − 𝟑 𝒙 = 𝟎 ⇒ 𝒙 = 𝟎 , 𝟑 𝒙 𝟐 − 𝟑 𝒙 − 𝟐 = 𝟎 ⇒ 𝒙 = 𝟑 ± √ 𝟏𝟕 𝟐 (1) (1) (1) (1) (1)