Check sibling questions

Example 5 - Are A (3, 6, 9), B (10, 20, 30), C (25, -41, 5)

Example  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 2
Example  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 3
Example  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 4
Example  5 - Chapter 12 Class 11 Introduction to Three Dimensional Geometry - Part 5

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Transcript

Example 5 Are the points A (3, 6, 9), B (10, 20, 30) and C (25, – 41, 5), the vertices of a right angled triangle? Lets first calculate distances AB, BC and AC & then apply Pythagoras theorem to check whether it is right triangle Calculating AB A (3, 6, 9) B (10, 20, 30) AB = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = 3, y1 = 6, z1 = 9 x2 = 10, y2 = 20, z2 = 30 AB = √((10−3)2+(20−6)2+(30−9)2) = √((7)2+(14)2+(21)2) = √(49+196+441) = √𝟔𝟖𝟔 Calculating BC B (10, 20, 30) C (25, –41, 5) BC = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) Here, x1 = 10, y1 = 20, z1 = 30 x2 = 25, y2 = – 41, z2 = 5 BC = √((25−10)2+(−41−20)2+(5−30)2) = √((15)2+(−61)2+(−25)2) = √(225+3721+625) = √𝟒𝟓𝟕𝟏 Calculating CA C (25, –41, 5) A (3, 6, 9) CA = √((x2−x1)2+(y2−y1)2+(z2 −z1)2) x1 = 25, y1 = – 41, z1 = 5 x2 = 3, y2 = 6, z2 = 9 AB = √((3−25)2+(6−(−41))2+(9−5)2) = √((−22)2+(6+41)2+(4)2) = √(484+(47)2+16) = √(484+2209+16) = √𝟐𝟕𝟎𝟗 Now AB = √686 , BC = √4571 , CA = √2709 In Right angle tringle, (Hypotenuse)2 = (Height)2 + (Base)2 Since √4571 is the biggest of the three sides , we take Hypotenuse = √4571 Hence we have to prove (√4571)2 = (√686)2 + (√2709)2 Thus, L.H.S ≠ R.H.S Hence, it is not a right angle tringle L.H.S (√4571)2 = 4571 R.H.S (√686)2 + (√2709)2 = 686 + 2709 = 3395

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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.