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![📖ఎడ్యుకేషన్✍ - Basic Algebra Formula (a+b)2 02 _ 20b +62 3 (a - b)2 200 +62 >02 'ಲ (a +b) (a-b) = a2_62] (x+a) (x+b) =22 + (+b) +ab (X+a) (X-b) = 12 + (0-b)T-0b (x-a) (x+b) =22 + (6-a) €-abl (x-a) (x-৮) = ~2_(@+৮) গ + @b = @3 +3ab (a + 6) (a + ৮)3 + 63 = @3-53_3@b (@~6) (a-b)3 (2 +y + z)2 = ೩2 _ ೈ2 +22 _ 2 2y2 | 239 + 212 _ (X+y-z)2 = 12 + 12 + 22 _ 2ல~~ 2~9 232 (x-y+2)2 = ~2 +~2_ 29 2y> + २r२ + ५२ (X-y-z)2 = 12 + 12 + 22_2Tg + 292- 22 T3 _ y3 (2+y+2)(22 + y2 _3xy2 +23_ + 22_ ೩2) ல 3~ = a " x2 +y2 = % [(c+y)2 + (v-y)2]] (x + a)(x + b)(x + c) = .3 + (a + b + c)२ + (ab + bc + ca )a + abc + y३ (2 + y)(22_ry+y2) = + (3-!) (೩2 _ ೩y + 12) x? " ~ys 2[(೩ _ !)2 _ (y _ 2)2 _ (೭ _ ೩)21 +y2 +22 Xyyz zX = ಲ + Basic Algebra Formula (a+b)2 02 _ 20b +62 3 (a - b)2 200 +62 >02 'ಲ (a +b) (a-b) = a2_62] (x+a) (x+b) =22 + (+b) +ab (X+a) (X-b) = 12 + (0-b)T-0b (x-a) (x+b) =22 + (6-a) €-abl (x-a) (x-৮) = ~2_(@+৮) গ + @b = @3 +3ab (a + 6) (a + ৮)3 + 63 = @3-53_3@b (@~6) (a-b)3 (2 +y + z)2 = ೩2 _ ೈ2 +22 _ 2 2y2 | 239 + 212 _ (X+y-z)2 = 12 + 12 + 22 _ 2ல~~ 2~9 232 (x-y+2)2 = ~2 +~2_ 29 2y> + २r२ + ५२ (X-y-z)2 = 12 + 12 + 22_2Tg + 292- 22 T3 _ y3 (2+y+2)(22 + y2 _3xy2 +23_ + 22_ ೩2) ல 3~ = a x2 +y2 = % [(c+y)2 + (v-y)2]] (x + a)(x + b)(x + c) = .3 + (a + b + c)२ + (ab + bc + ca )a + abc + y३ (2 + y)(22_ry+y2) = + (3-!) (೩2 _ ೩y + 12) x? ~ys 2[(೩ _ !)2 _ (y _ 2)2 _ (೭ _ ೩)21 +y2 +22 Xyyz zX = ಲ + - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_533131_a05db6a_1772517410548_sc.jpg?tenant=sc&referrer=user-profile-service%2FrequestType50&f=548_sc.jpg "📖ఎడ్యుకేషన్✍ - Basic Algebra Formula (a+b)2 02 _ 20b +62 3 (a - b)2 200 +62 >02 'ಲ (a +b) (a-b) = a2_62] (x+a) (x+b) =22 + (+b) +ab (X+a) (X-b) = 12 + (0-b)T-0b (x-a) (x+b) =22 + (6-a) €-abl (x-a) (x-৮) = ~2_(@+৮) গ + @b = @3 +3ab (a + 6) (a + ৮)3 + 63 = @3-53_3@b (@~6) (a-b)3 (2 +y + z)2 = ೩2 _ ೈ2 +22 _ 2 2y2 | 239 + 212 _ (X+y-z)2 = 12 + 12 + 22 _ 2ல~~ 2~9 232 (x-y+2)2 = ~2 +~2_ 29 2y> + २r२ + ५२ (X-y-z)2 = 12 + 12 + 22_2Tg + 292- 22 T3 _ y3 (2+y+2)(22 + y2 _3xy2 +23_ + 22_ ೩2) ல 3~ = a \" x2 +y2 = % [(c+y)2 + (v-y)2]] (x + a)(x + b)(x + c) = .3 + (a + b + c)२ + (ab + bc + ca )a + abc + y३ (2 + y)(22_ry+y2) = + (3-!) (೩2 _ ೩y + 12) x? \" ~ys 2[(೩ _ !)2 _ (y _ 2)2 _ (೭ _ ೩)21 +y2 +22 Xyyz zX = ಲ + Basic Algebra Formula (a+b)2 02 _ 20b +62 3 (a - b)2 200 +62 >02 'ಲ (a +b) (a-b) = a2_62] (x+a) (x+b) =22 + (+b) +ab (X+a) (X-b) = 12 + (0-b)T-0b (x-a) (x+b) =22 + (6-a) €-abl (x-a) (x-৮) = ~2_(@+৮) গ + @b = @3 +3ab (a + 6) (a + ৮)3 + 63 = @3-53_3@b (@~6) (a-b)3 (2 +y + z)2 = ೩2 _ ೈ2 +22 _ 2 2y2 | 239 + 212 _ (X+y-z)2 = 12 + 12 + 22 _ 2ல~~ 2~9 232 (x-y+2)2 = ~2 +~2_ 29 2y> + २r२ + ५२ (X-y-z)2 = 12 + 12 + 22_2Tg + 292- 22 T3 _ y3 (2+y+2)(22 + y2 _3xy2 +23_ + 22_ ೩2) ல 3~ = a x2 +y2 = % [(c+y)2 + (v-y)2]] (x + a)(x + b)(x + c) = .3 + (a + b + c)२ + (ab + bc + ca )a + abc + y३ (2 + y)(22_ry+y2) = + (3-!) (೩2 _ ೩y + 12) x? ~ys 2[(೩ _ !)2 _ (y _ 2)2 _ (೭ _ ೩)21 +y2 +22 Xyyz zX = ಲ + - ShareChat")
![👩💻టెట్/DSC ప్రత్యేకం - (ix j).k+(j > k) i [Foreign 2012] Write the value of Ex. 6 (i x j) k + (j > k) i = k k +i i =1+1 = 2 Sol. +3k If0 is the angle between two vectors i - 2j Ex 7 [CBSE 2018] and 3i - 2j + k, find sin 0. - २j + k a = i _ 2j + 3k and b = 3ಗ Sol Let Ia X b | sin 0 |q |b | ; ; a Xb = 1 -2 1 3 !(4) - j(~8) + k(4) 4i + ৪j + 4k = la X b| = 4+1+4+1 =4+6 1q| = 1] + 4 + 9 Also V14 3 19+4+1 = 114 b ~ 216 416 sin 0 1141]4 (ix j).k+(j > k) i [Foreign 2012] Write the value of Ex. 6 (i x j) k + (j > k) i = k k +i i =1+1 = 2 Sol. +3k If0 is the angle between two vectors i - 2j Ex 7 [CBSE 2018] and 3i - 2j + k, find sin 0. - २j + k a = i _ 2j + 3k and b = 3ಗ Sol Let Ia X b | sin 0 |q |b | ; ; a Xb = 1 -2 1 3 !(4) - j(~8) + k(4) 4i + ৪j + 4k = la X b| = 4+1+4+1 =4+6 1q| = 1] + 4 + 9 Also V14 3 19+4+1 = 114 b ~ 216 416 sin 0 1141]4 - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_514724_1d55b92f_1772517308770_sc.jpg?tenant=sc&referrer=user-profile-service%2FrequestType50&f=770_sc.jpg "👩💻టెట్/DSC ప్రత్యేకం - (ix j).k+(j > k) i [Foreign 2012] Write the value of Ex. 6 (i x j) k + (j > k) i = k k +i i =1+1 = 2 Sol. +3k If0 is the angle between two vectors i - 2j Ex 7 [CBSE 2018] and 3i - 2j + k, find sin 0. - २j + k a = i _ 2j + 3k and b = 3ಗ Sol Let Ia X b | sin 0 |q |b | ; ; a Xb = 1 -2 1 3 !(4) - j(~8) + k(4) 4i + ৪j + 4k = la X b| = 4+1+4+1 =4+6 1q| = 1] + 4 + 9 Also V14 3 19+4+1 = 114 b ~ 216 416 sin 0 1141]4 (ix j).k+(j > k) i [Foreign 2012] Write the value of Ex. 6 (i x j) k + (j > k) i = k k +i i =1+1 = 2 Sol. +3k If0 is the angle between two vectors i - 2j Ex 7 [CBSE 2018] and 3i - 2j + k, find sin 0. - २j + k a = i _ 2j + 3k and b = 3ಗ Sol Let Ia X b | sin 0 |q |b | ; ; a Xb = 1 -2 1 3 !(4) - j(~8) + k(4) 4i + ৪j + 4k = la X b| = 4+1+4+1 =4+6 1q| = 1] + 4 + 9 Also V14 3 19+4+1 = 114 b ~ 216 416 sin 0 1141]4 - ShareChat")
![👩💻టెట్/DSC ప్రత్యేకం - TY T ExamSiDE] MATH/ 0 MATHS WITH | Solutions MATHEMATICS MAIHS PHANEENDRA By PHANI ByPHANI a rad sin & tan 0 cr0 coS 0 0 16+ 12 6- 12 2-13 2+~3 15 5-1 5- 215 10 V10+215 15+215 18 5 5+1 5+1 110 _ 2v5 V10 _ 215 T 36 V10-215 15 + ] 5 V10 _ 215 5+1 5+1 V10-2V5 37 54 5+1 110 _ 215 10 5~2v5 5-1 110+215 21 15+2/5 72 5 5 16 + v2 16 - v2 51 2-3 2+13 75 12 TY T ExamSiDE] MATH/ 0 MATHS WITH | Solutions MATHEMATICS MAIHS PHANEENDRA By PHANI ByPHANI a rad sin & tan 0 cr0 coS 0 0 16+ 12 6- 12 2-13 2+~3 15 5-1 5- 215 10 V10+215 15+215 18 5 5+1 5+1 110 _ 2v5 V10 _ 215 T 36 V10-215 15 + ] 5 V10 _ 215 5+1 5+1 V10-2V5 37 54 5+1 110 _ 215 10 5~2v5 5-1 110+215 21 15+2/5 72 5 5 16 + v2 16 - v2 51 2-3 2+13 75 12 - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_434232_187316f1_1772429395703_sc.jpg?tenant=sc&referrer=user-profile-service%2FrequestType50&f=703_sc.jpg "👩💻టెట్/DSC ప్రత్యేకం - TY T ExamSiDE] MATH/ 0 MATHS WITH | Solutions MATHEMATICS MAIHS PHANEENDRA By PHANI ByPHANI a rad sin & tan 0 cr0 coS 0 0 16+ 12 6- 12 2-13 2+~3 15 5-1 5- 215 10 V10+215 15+215 18 5 5+1 5+1 110 _ 2v5 V10 _ 215 T 36 V10-215 15 + ] 5 V10 _ 215 5+1 5+1 V10-2V5 37 54 5+1 110 _ 215 10 5~2v5 5-1 110+215 21 15+2/5 72 5 5 16 + v2 16 - v2 51 2-3 2+13 75 12 TY T ExamSiDE] MATH/ 0 MATHS WITH | Solutions MATHEMATICS MAIHS PHANEENDRA By PHANI ByPHANI a rad sin & tan 0 cr0 coS 0 0 16+ 12 6- 12 2-13 2+~3 15 5-1 5- 215 10 V10+215 15+215 18 5 5+1 5+1 110 _ 2v5 V10 _ 215 T 36 V10-215 15 + ] 5 V10 _ 215 5+1 5+1 V10-2V5 37 54 5+1 110 _ 215 10 5~2v5 5-1 110+215 21 15+2/5 72 5 5 16 + v2 16 - v2 51 2-3 2+13 75 12 - ShareChat")
![👩💻టెట్/DSC ప్రత్యేకం - Show that the points A, B, C with position vectors 19 2i - j + k, i - 3j - Sk 4j _ 4k and 3i _ respectively, are the vertices of a right-angled triangle Hence find the area ofthe triangle [AI2017] Given points are -j+1), B(i - 3j 501 5k) A(2i and _ (31 4j - 4k) (i - 3j - 51) - (2i _ j + k) = -i - 2j - 6k AB = BC 4j - 4k) - ({ - 3j - 5k) = 21 - j + k (3i _ (2i _ j + k) - (3i _ 4j - 4k) = -i + 3j + 5k CA Me notice BC.CA = _ 2 _ 3 + $ = 0 BC LCA triangle ABC is right-angled at C. K i j 1 1 CB | 3 5 Area of CA -1 X २ 2 2 _1 2/8i 11j - 5k | [ + v210 sq units 64 + 121 + 25 2 2 Show that the points A, B, C with position vectors 19 2i - j + k, i - 3j - Sk 4j _ 4k and 3i _ respectively, are the vertices of a right-angled triangle Hence find the area ofthe triangle [AI2017] Given points are -j+1), B(i - 3j 501 5k) A(2i and _ (31 4j - 4k) (i - 3j - 51) - (2i _ j + k) = -i - 2j - 6k AB = BC 4j - 4k) - ({ - 3j - 5k) = 21 - j + k (3i _ (2i _ j + k) - (3i _ 4j - 4k) = -i + 3j + 5k CA Me notice BC.CA = _ 2 _ 3 + $ = 0 BC LCA triangle ABC is right-angled at C. K i j 1 1 CB | 3 5 Area of CA -1 X २ 2 2 _1 2/8i 11j - 5k | [ + v210 sq units 64 + 121 + 25 2 2 - ShareChat](https://cdn4.sharechat.com/bd5223f_s1w/compressed_gm_40_img_534004_a68e7a7_1772381532717_sc.jpg?tenant=sc&referrer=user-profile-service%2FrequestType50&f=717_sc.jpg "👩💻టెట్/DSC ప్రత్యేకం - Show that the points A, B, C with position vectors 19 2i - j + k, i - 3j - Sk 4j _ 4k and 3i _ respectively, are the vertices of a right-angled triangle Hence find the area ofthe triangle [AI2017] Given points are -j+1), B(i - 3j 501 5k) A(2i and _ (31 4j - 4k) (i - 3j - 51) - (2i _ j + k) = -i - 2j - 6k AB = BC 4j - 4k) - ({ - 3j - 5k) = 21 - j + k (3i _ (2i _ j + k) - (3i _ 4j - 4k) = -i + 3j + 5k CA Me notice BC.CA = _ 2 _ 3 + $ = 0 BC LCA triangle ABC is right-angled at C. K i j 1 1 CB | 3 5 Area of CA -1 X २ 2 2 _1 2/8i 11j - 5k | [ + v210 sq units 64 + 121 + 25 2 2 Show that the points A, B, C with position vectors 19 2i - j + k, i - 3j - Sk 4j _ 4k and 3i _ respectively, are the vertices of a right-angled triangle Hence find the area ofthe triangle [AI2017] Given points are -j+1), B(i - 3j 501 5k) A(2i and _ (31 4j - 4k) (i - 3j - 51) - (2i _ j + k) = -i - 2j - 6k AB = BC 4j - 4k) - ({ - 3j - 5k) = 21 - j + k (3i _ (2i _ j + k) - (3i _ 4j - 4k) = -i + 3j + 5k CA Me notice BC.CA = _ 2 _ 3 + $ = 0 BC LCA triangle ABC is right-angled at C. K i j 1 1 CB | 3 5 Area of CA -1 X २ 2 2 _1 2/8i 11j - 5k | [ + v210 sq units 64 + 121 + 25 2 2 - ShareChat")
