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MathsMediumClass 12

Image of Point in Plane

3D Geometry

16

JEE Qs

8%

Hard

60

min

Master the step-by-step derivation using line and plane equations; this approach is more robust for variations than just memorizing the direct formula.

๐Ÿงฎ Key Formulas

Equation of a plane: ax + by + cz + d = 0
Equation of a line through (x1, y1, z1) with direction ratios (A, B, C): (x - x1)/A = (y - y1)/B = (z - z1)/C = ฮป
Midpoint formula: ((x1+x2)/2, (y1+y2)/2, (z1+z2)/2)
Condition for direction ratios of line perpendicular to plane: If line is perpendicular to ax+by+cz+d=0, its direction ratios are (a, b, c).
Image P'(x', y', z') of point P(x1, y1, z1) in plane ax+by+cz+d=0: (x' - x1)/a = (y' - y1)/b = (z' - z1)/c = -2 * (ax1 + by1 + cz1 + d) / (a^2 + b^2 + c^2)

โœ… Key Points for JEE

  • 1The image of a point P in a plane is a point P' such that the plane is the perpendicular bisector of the line segment PP'.
  • 2The foot of the perpendicular (M) from P to the plane is the midpoint of the line segment PP'.
  • 3The line connecting the point P and its image P' (i.e., PP') is always normal (perpendicular) to the given plane. Therefore, its direction ratios are proportional to the coefficients of x, y, z in the plane's equation.
  • 4The process to find the image involves: 1) Writing the equation of the line normal to the plane and passing through the given point. 2) Finding the intersection point (foot of perpendicular) of this line and the plane. 3) Using the midpoint formula (foot is midpoint of P and P') to find the image P'.

โš ๏ธ Common Mistakes

  • โœ•Confusing the formula for the image of a point with the formula for the foot of the perpendicular (the image formula has -2, foot of perpendicular has -1).
  • โœ•Sign errors when substituting coordinates or the constant 'd' into the formula for image or foot of perpendicular.
  • โœ•Incorrectly identifying the direction ratios of the normal vector to the plane, especially if the plane equation is not in standard form (e.g., if coefficients are zero).

๐Ÿ“ Practice Questions

See all

Q4. Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L : xโˆ’11 = y+1โˆ’1 = zโˆ’22 . Let the line โ†’r = (โˆ’^i + ^j โˆ’2^k) + ฮป(^i โˆ’^j + ^k), ฮป โˆˆR, intersect the line L at Q . Then 2(PQ)2 is equal to : (1) 25 (2) 19 (3) 29 (4) 27

2025ยทMCQMedium

Q16.Let a straight line L pass through the point P(2, โˆ’1, 3) and be perpendicular to the lines xโˆ’12 = y+11 = zโˆ’3โˆ’2 and xโˆ’3 1 = yโˆ’23 = z+24 . If the line L intersects the yz -plane at the point Q , then the distance between the points P and Q is : (1) โˆš10 (2) 2โˆš3 (3) 2 (4) 3

2025ยทMCQMedium

Q8. Let L1 : xโˆ’12 = yโˆ’23 = zโˆ’34 and L2 : xโˆ’23 = yโˆ’44 = zโˆ’55 be two lines. Then which of the following points lies on the line of the shortest distance between L1 and L2 ? (1) ( 143 , โˆ’3, 223 ) (2) (โˆ’53 , โˆ’7, 1) (3) (2, 3, 13 ) (4) ( 83 , โˆ’1, 13 )

2025ยทMCQHard

Q25.Let L1 : xโˆ’13 = yโˆ’1โˆ’1 = z+10 and L2 : xโˆ’22 = 0y = z+4ฮฑ , ฮฑ โˆˆR, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, โˆ’1) on L2 , then the value of 26ฮฑ( PB)2 is _________

2025ยทNumericalMedium

Q14.The perpendicular distance, of the line xโˆ’1 2 = โˆ’1 = z+32 from the point P(2, โˆ’10, 1), is : (1) 6 (2) 5โˆš2 (3) 4โˆš3 (4) 3โˆš5

2025ยทMCQMedium

Q3. Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be ^i + 2^j + ^k,^i + 3^j โˆ’2^k and 2^i + ^j โˆ’^k respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E . If the length of AD is โˆš110 and the volume of the 3 tetrahedron is โˆš805 , then the position vector of E is 6โˆš2 (1) 12 1 (7^i + 4^j + 3^k) (2) 12 (^i + 4^j + 7^k) (3) 1 6 (12^i + 12^j + ^k) (4) 16 (7^i + 12^j + ^k)

2025ยทMulti conceptHard

NCERT Chapters

  • Class 12 Maths Ch 11: Three Dimensional Geometry