Q78.Let the foot of perpendicular of the point P(3, –2, –9) on the plane passing through the points (–1, –2, –3), (9, 3, 4), (9, –2, 1) be Q(α, β, γ). Then the distance Q from the origin is (1) √42 (2) √38 (3) √35 (4) √29
What This Question Tests
This question requires multiple steps: finding the equation of a plane from three points, then determining the foot of the perpendicular from a given point to this plane, and finally calculating the distance of this foot from the origin. It's complex and multi-conceptual.
Concepts Tested
Formulas Used
Equation of plane (r - a) . (b - a) x (c - a) = 0
Equation of line x - x1 / a = y - y1 / b = z - z1 / c
Distance = sqrt((x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2)
📚 NCERT Sections This Tests
1.18 — A Point Charge Of 2.0 Mc Is At The Centre Of A Cubic Gaussian
Physics Class 11 · Chapter 1
1.18 A point charge of 2.0 mC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
2.1 — Two Charges 5 × 10–8 C And –3 × 10–8 C Are Located 16 Cm Apart. At
Physics Class 11 · Chapter 2
2.1 Two charges 5 × 10–8 C and –3 × 10–8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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.
📋 Question Details
- Chapter
- 3D Geometry
- Topic
- Foot of perpendicular from a point to a plane
- Year
- 2023
- Shift
- 15 Apr Shift 1
- Q Number
- Q78
- Type
- MCQ
- NCERT Ref
- Class 12 Mathematics Ch 11: 3D Geometry
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