Q69.Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0, 1) and R be any point (2, 1, 6) .Then the image of R in the plane P is (1) (6, 5, 2) (2) (6, 5, −2) (3) (4, 3, 2) (4) (3, 4, −2)
What This Question Tests
This problem involves finding the equation of a plane passing through three given points and then determining the image of a given point in that plane, which requires using properties of reflection and normal vectors.
Concepts Tested
Formulas Used
Normal vector = Cross product of two vectors in the plane
Equation of plane: n ⋅ (r - r0) = 0
Midpoint formula: ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)
📚 NCERT Sections This Tests
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
2.2 — A Regular Hexagon Of Side 10 Cm Has A Charge 5 Mc At Each Of Its
Physics Class 11 · Chapter 2
2.2 A regular hexagon of side 10 cm has a charge 5 mC at each of its vertices. Calculate the potential at the centre of the hexagon.
5.12 — Write All The Geometrical Isomers Of [Pt(Nh3)(Br)(Cl)(Py)] And How Many Of
Chemistry Class 11 · Chapter 5
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
📋 Question Details
- Chapter
- 3D Geometry
- Topic
- Image of a Point in a Plane
- Year
- 2020
- Shift
- 07 Jan Shift 1
- Q Number
- Q69
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 11: Three Dimensional Geometry
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