Ex 12.2, 3 - Verify (i) (0, 7, -10), (1, 6, -6), (4, 9, -6) - Ex 12.2

  1. Chapter 12 Class 11 Introduction to Three Dimensional Geometry
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Ex 12.2, 3 Verify the following: (i) (0, 7, –10), (1, 6, –6) and (4, 9, –6) are the vertices of an isosceles triangle. Let points be A (0, 7, – 10) , B (1, 6, – 6) & C (4, 9, – 6) If any 2 sides are equal, it will be an isosceles triangle Lets calculate AB, BC, AC Calculating AB A (0, 7, – 10) B (1, 6, – 6) AB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 0, y1 = 7, z1 = – 10 x2 = 1, y2 = 6, z2 = – 6 AB = ﷐﷮﷐1−0﷯2+﷐6−7﷯2+﷐−6+10﷯2﷯ = ﷐﷮﷐1﷯2+﷐−1﷯2+﷐4﷯2﷯ = ﷐﷮1+1+16﷯ = ﷐﷮18﷯ = 3﷐﷮2﷯ Calculating BC B (1, 6, –6) C (4, 9, –6) BC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here x1 = 1, y1 = 6, z1 = – 6 x2 = 4, y2 = 9, z2 = – 6 BC = ﷐﷮﷐4−1﷯2+﷐9−6﷯2+(−6−﷐−6﷯2﷯ = ﷐﷮﷐3﷯2+﷐3﷯2+(−6+6)2﷯ = ﷐﷮9+9+0﷯ = ﷐﷮18﷯ Since AB = BC ∴ 2 Sides are equal Hence ABC is an isosceles triangle ( Note: We don’t need to calculate AC) Ex 12.2, 3 Verify the following: (ii) (0, 7, 10), (–1, 6, 6) and (–4, 9, 6) are the vertices of a right angled triangle. Let A (0, 7, – 10) B ( – 1, 6, 6) C ( – 4, 9, 6) Lets first calculate Distance AB, BC & AC and then apply Pythagoras theorem to check whether it is right angled triangle Calculating AB A (0, 7, – 10) B ( – 1, 6, 6) AB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 0, y1 = 7, z1 =10 x2 = 1, y2 = 6, z2 = 6 AB = ﷐﷮﷐−1−0﷯2+﷐6−7﷯2+﷐6−10﷯2﷯ = ﷐﷮﷐−1﷯2+﷐−1﷯2+﷐−4﷯2﷯ = ﷐﷮1+1+16﷯ = ﷐﷮18﷯ Calculating BC B ( – 1, 6, 6) C ( – 4, 9, 6) BC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = – 1, y1 = 6, z1 = 6 x2 = – 4, y2 = 9, z2 = 6 BC = ﷐﷮﷐−4−(−1)﷯2+﷐9−6﷯2+﷐6−6﷯2﷯ BC = ﷐﷮﷐−4+1﷯2+﷐3﷯2+﷐0﷯2﷯ = ﷐﷮﷐−3﷯2+9﷯ = ﷐﷮9+9﷯ = ﷐﷮18﷯ Calculating AC A (0, 7, 10) C ( – 4, 9, 6) AC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 0, y1 = 7, z1 = 10 x2 = – 4, y2 = 9, z2 = 6 AC = ﷐﷮﷐−4−0﷯2+﷐9−7﷯2+﷐6−10﷯2﷯ = ﷐﷮﷐−4﷯2+﷐2﷯2+﷐−4﷯2﷯ = ﷐﷮16+4+16﷯ = ﷐﷮36﷯ = 6 Now AB = ﷐﷮18﷯ , BC = ﷐﷮18﷯ , AC = ﷐﷮36﷯ In Right angle tringle (Hypotenuse)2 = (Height)2 + (Base)2 Since ﷐﷮36﷯ is the largest of the three sides, we take Hypotenuse = ﷐﷮36﷯ Hence we have to prove (﷐﷮36﷯)2 = (﷐﷮18﷯)2 + (﷐﷮18﷯)2 Since L.H.S = R.H.S Hence, It is a right angle tringle Ex 12.2, 3 Verify the following: (iii) (–1, 2, 1), (1, –2, 5), (4, –7, 8), and (2, –3, 4) are the vertices of a parallelogram. Let A (–1, 2, 1) , B (1, –2, 5) , C (4, –7, 8) , D (2, –3, 4) ABCD can be vertices of parallelogram only if opposite sides are equal. i.e. AB = CD & BC = AD Calculating AB A (–1, 2, 1) B (1, –2, 5) AB = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = – 1, y1 = 2, z1 = 1 x2 = 1, y2 = – 2, z2 = 5 AB = ﷐﷮﷐1−(−1)﷯2+﷐−2−2﷯2+﷐5−1﷯2﷯ = ﷐﷮﷐1+1﷯2+﷐−4﷯2+﷐4﷯2﷯ = ﷐﷮22+16+16﷯ = ﷐﷮4+16+16﷯ = ﷐﷮36﷯ = 6 Calculating BC B (1, – 2, 5) C (4, – 7, 8) BC = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 1, y1 = – 2, z1 = 5 x2 = 4, y2 = – 7, z2 = 8 BC = ﷐﷮﷐4−1﷯2+﷐−7−(2)﷯2+﷐8−5﷯2﷯ BC = ﷐﷮﷐3﷯2+﷐−7+2﷯2+﷐3﷯2﷯ = ﷐﷮9+﷐−5﷯2+9﷯ = ﷐﷮9+25+9﷯ = ﷐﷮43﷯ Calculating CD C (4, – 7, 8) D (2, – 3, 4) CD = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 4, y1 = – 7, z1 = 8 x2 = 2, y2 = – 3, z2 = 4 CD = ﷐﷮﷐2−4﷯2+﷐−3−(−7)﷯2+﷐4−8﷯2﷯ = ﷐﷮﷐−2﷯2+﷐−3+7﷯2+﷐−4﷯2﷯ = ﷐﷮4+﷐4﷯2+16﷯ = ﷐﷮4+16+16﷯ = ﷐﷮36﷯ = 6 Calculating DA D (2, – 3, 4) A ( –1, 2, 1) DA = ﷐﷮﷐x2−x1﷯2+﷐y2−y1﷯2+﷐z2 −z1﷯2﷯ Here, x1 = 2, y1 = – 3, z1 = 4 x2 = –1, y2 = 2, z2 = 1 DA = ﷐﷮﷐−1−2﷯2+﷐2−(−3)﷯2+﷐1−4﷯2﷯ = ﷐﷮﷐−3﷯2+﷐2+3﷯2+﷐−3﷯2﷯ = ﷐﷮9+﷐5﷯2+9﷯ = ﷐﷮9+25+9﷯ = ﷐﷮43﷯ Since AB = CD & BC = DA So, In ABCD both pairs of opposite sides are equal Thus, ABCD is a parallelogram.

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