Class 11 Maths Chapter 11: Introduction to Three Dimensional Geometry worksheet with answer key
A 20-mark worksheet on Introduction to Three Dimensional Geometry for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (d) y-axis
The plane XOY contains the x and y axes and the plane YOZ contains the y and z axes; the line common to both is Y′OY, the y-axis.
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Answer: (c) +, −, +
From Table 11.1, octant IV is XOY′Z: x positive, y negative, z positive, i.e. (+, −, +).
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Answer: (b) 4
From Table 11.1, z is positive in octants I, II, III and IV and negative in V to VIII: four octants.
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Answer: (a) (4, −3, 0)
The foot M of the perpendicular PM on the XY-plane keeps the x and y coordinates of P and has z = 0: M is (4, −3, 0), and MP = 6.
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Answer: (a) (0, −4, 0)
A point on the y-axis has the form (0, y, 0): its x and z coordinates are both zero. Only (0, −4, 0) has this form.
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Let the point be P(0, 0, z). PA2 = 1 + 25 + (z − 7)2 and PB2 = 25 + 1 + (z + 4)2. Equating, −14z + 49 = 8z + 16, so 22z = 33 and z = 32. The point is (0, 0, 32), i.e. z = 1.5.
- takes P(0, 0, z) and equates PA2 and PB2 (1)
- z = 32, the point (0, 0, 32) (1)
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xy > 0 means x and y have the same sign: (+, +) or (−, −). With z negative the sign patterns are (+, +, −) and (−, −, −), which are octants V and VII.
- x and y of the same sign, z negative: (+, +, −) or (−, −, −) (1)
- octants V and VII (1)
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OQ2 = a2 + 4a2 + 4a2 = 9a2 = 144, so a2 = 16 and a = 4 or a = −4.
- 9a2 = 144 (1)
- a = ±4 (1)
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Let P be (x, y, z). PO = PA gives x2 = (x − 4)2, i.e. 8x = 16, x = 2. PO = PB gives y2 = (y − 6)2, i.e. 12y = 36, y = 3. PO = PC gives z2 = (z − 2)2, i.e. 4z = 4, z = 1. The point is (2, 3, 1).
- x = 2 from PO = PA (1)
- y = 3 from PO = PB (1)
- z = 1, the point (2, 3, 1) (1)
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G = (4 − 1 + 33, 1 + 5 + 03, −2 + 3 + 53) = (2, 2, 2). OG = √(4 + 4 + 4) = √12 = 2√3 units.
- the centroid as the mean of the vertices' coordinates (1)
- G = (2, 2, 2) (1)
- OG = 2√3 units (1)
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D(4, 3, 0) gives OA = 4 and OB = 3; C gives OC = 6. So P = (4, 3, 6). E, above B, is (0, 3, 6) and F, above A, is (4, 0, 6). OP = √(16 + 9 + 36) = √61 units.
- P(4, 3, 6) (1)
- E(0, 3, 6) and F(4, 0, 6) (1)
- OP = √61 units (1)
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