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Three Dimensional Geometry appeared 62 times across 3 years — 7.2% of Mathematics. This question is from Shortest Distance Between Two Lines.

Year 2026 2025 2024 Total
Questions 15 29 18 62

If the shortest distance between the lines (x - 1)/(2) = (y - 2)/(3) = (z - 3)/(4) and (x)/(1) = (y)/(α) = (z - 5)/(1) is 5√(6) , then the sum of all possible values of α is

Solution & Explanation

Related Formula

Shortest distance between two skewed lines with vector equations r = a₁ + λ b₁ and \vec{r} = \vec{a}_2 + \mu \vec{b}_2 is:

S.D. = | ( a₂ - a₁) · ( b₁ × b₂)| b₁ × b₂| |
Core Logic

From the given lines: Line 1 passes through A(1, 2, 3) with direction vector b₁ = 2 i + 3 j + 4 k. Line 2 passes through B(0, 0, 5) with direction vector b₂ = i + α j + k.

The vector connecting the two fixed points is:

BA = (1-0) i + (2-0) j + (3-5) k = i + 2 j - 2 k
Step 1: Compute Cross Product of Direction Vectors
n = b₁ × b₂ = | matrix i & j & k 2 & 3 & 4 1 & α & 1 matrix | = i(3 - 4α) - j(2 - 4) + k(2α - 3) n = (3 - 4α) i + 2 j + (2α - 3) k
Step 2: Apply Shortest Distance Formula

Shortest Distance Between Two Lines diagram for Q52 - JEE Main 2025 Morning
Shortest Distance Between Two Lines diagram for Q52 - JEE Main 2025 Morning

S.D. = | ( i + 2 j - 2 k) · n| n| | = 5√(6)

Taking dot product in numerator:

( i + 2 j - 2 k) · n = 1(3-4α) + 2(2) - 2(2α-3) = 3 - 4α + 4 - 4α + 6 = 13 - 8α

Squaring both sides:

((13 - 8α)²)/((3 - 4α)² + 4 + (2α - 3)²) = (25)/(6) 6(64α² - 208α + 169) = 25(16α² - 24α + 9 + 4 + 4α² - 12α + 9) 6(64α² - 208α + 169) = 25(20α² - 36α + 22) 384α² - 1248α + 1014 = 500α² - 900α + 550 116α² + 348α - 464 = 0 α² + 3α - 4 = 0
Step 3: Calculate the Sum of Roots

The sum of all possible values of α is given by the relation:

α₁ + α₂ = -(3)/(1) = -3
Pattern Recognition

Whenever asked for the sum of all possible parameter values resulting from a vector condition, look to directly read the linear term coefficient via Vieta's relations rather than explicitly factoring or computing the roots individually.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 4

Q51 jee_main_2025_02_april_evening Image of a Point in a Line
If the image of the point P(1, 0, 3) in the line joining the points A(4, 7, 1) and B(3, 5, 3) is Q(α, β, γ), then α + β + γ is equal to
  • A. (47)/(3)
  • B. (46)/(3)
  • C. 18
  • D. 13

Solution

Related Formula
Equation of a line through (x₁, y₁, z₁) with direction ratios (a, b, c): (x-x₁)/(a) = (y-y₁)/(b) = (z-z₁)/(c) Condition for Perpendicular Lines: a₁ a₂ + b₁ b₂ + c₁ c₂ = 0
Core Logic

Let R be the foot of the perpendicular drawn from point P(1,0,3) to the line joining A(4,7,1) and B(3,5,3). Since Q(α,β,γ) is the reflection (image) of P across the line, point R serves as the midpoint of the line segment PQ.

Step 1: Find the Equation of the Line AB

The direction ratios of the line AB are:

d = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) ≡ (1, 2, -2)

Using point B(3,5,3), the symmetric equation of the line AB is:

(x - 3)/(1) = (y - 5)/(2) = (z - 3)/(-2) = λ

Any general point R on this line can be written in terms of parameter λ:

R ≡ (λ + 3, 2λ + 5, -2λ + 3)
Step 2: Find the Foot of the Perpendicular R

The direction ratios of the line segment PR are:

PR = (λ + 3 - 1, 2λ + 5 - 0, -2λ + 3 - 3) = (λ + 2, 2λ + 5, -2λ)

Since PR is perpendicular to the line AB, the dot product of their direction vectors must equal zero:

1(λ + 2) + 2(2λ + 5) - 2(-2λ) = 0 λ + 2 + 4λ + 10 + 4λ = 0 λ = -(4)/(3)
Step 3: Coordinates of Foot of Perpendicular

Substitute λ = -(4)/(3) into the general coordinates of R:

R ≡ (-(4)/(3) + 3, 2(-(4)/(3)) + 5, -2(-(4)/(3)) + 3) R ≡ ((5)/(3), (7)/(3), (17)/(3))
Step 4: Solve for the Image Coordinates and Sum

Since R is the midpoint of PQ, where P = (1, 0, 3) and Q = (α, β, γ):

  • (α + 1)/(2) = (5)/(3) α = (10)/(3) - 1 = (7)/(3)
  • (β + 0)/(2) = (7)/(3) β = (14)/(3)
  • (γ + 3)/(2) = (17)/(3) γ = (34)/(3) - 3 = (25)/(3)
  • Now, let us calculate the sum:

α + β + γ = (7)/(3) + (14)/(3) + (25)/(3) = (46)/(3)
Pattern Recognition

Standard Midpoint reflection: The image coordinates are given directly by ximage = 2 xfoot - xₚₒᵢₙₜ, yimage = 2 yfoot - yₚₒᵢₙₜ, and zimage = 2 zfoot - zₚₒᵢₙₜ. Finding the parameter λ by using the perpendicular vector dot-product rule is the fastest and most robust method.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q56 jee_main_2025_02_april_evening Shortest Distance between Skew Lines
The line L₁ is parallel to the vector a = -3 i + 2 j + 4 k and passes through the point (7, 6, 2) and the line L₂ is parallel to the vector b = 2 i + j + 3 k and passes through the point (5, 3, 4). The shortest distance between the lines L₁ and L₂ is:
  • A. 23√(38)
  • B. 21√(57)
  • C. 23√(57)
  • D. 21√(38)

Solution

Related Formula
Shortest Distance d = | ( r₂ - r₁) · ( a × b) || a × b|
Core Logic

Shortest distance between two skew lines is the projection of the vector joining any two points of the lines onto their common normal.

Step 1: Find the vector joining the two points

Let the points be P₁(7, 6, 2) on L₁ and P₂(5, 3, 4) on L₂:

r₂ - r₁ = (5 - 7) i + (3 - 6) j + (4 - 2) k = -2 i - 3 j + 2 k
Step 2: Find the common normal vector

The direction is given by the cross product of the direction vectors:

a × b = vmatrix i & j & k -3 & 2 & 4 2 & 1 & 3 vmatrix a × b = i(6 - 4) - j(-9 - 8) + k(-3 - 4) = 2 i + 17 j - 7 k
Step 3: Calculate the distance

Calculate the dot product of the vectors:

( r₂ - r₁) · ( a × b) = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69

Calculate the magnitude of the cross product:

| a × b| = √(2² + 17² + (-7)²) = √(4 + 289 + 49) = √(342) = 3√(38)

Now, compute the shortest distance:

d = |-69|3√(38) = 23√(38)
Pattern Recognition

Matrix determinant check: In skew lines problems, the numerator can also be computed as the determinant of the 3x3 matrix composed of ( r₂- r₁), a, and b.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q74 jee_main_2025_02_april_evening Area of Triangles
Let A(4, -2), B(1, 1) and C(9, -3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ____________.
Numerical Answer. Answer: 3 to 3

Solution

Related Formula
Area of triangle ABC = (1)/(2) | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) | Maximum area of an inscribed parallelogram = (1)/(2) × Area of triangle
Core Logic

First, we compute the total area of the triangle ABC from the given coordinate points. We then apply the geometric maximum area property for inscribed parallelograms.

Step 1: Calculate the area of triangle ABC

The coordinate points are A(4, -2), B(1, 1), and C(9, -3):

Area(Δ ABC) = (1)/(2) vmatrix 4 & -2 & 1 1 & 1 & 1 9 & -3 & 1 vmatrix Area(Δ ABC) = (1)/(2) | 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) | Area(Δ ABC) = (1)/(2) | 4(4) + 2(-8) + 1(-12) | Area(Δ ABC) = (1)/(2) | 16 - 16 - 12 | = 6 square units
Step 2: Apply the maximum area theorem

The maximum area of a parallelogram inscribed in a triangle of area Δ is always exactly half the area of the triangle:

Maximum Area = (1)/(2) × Area(Δ ABC) = (1)/(2) × 6 = 3 square units
Pattern Recognition

Inscribed shapes maximization: The maximum area of any inscribed parallelogram AFDE on the sides of a triangle ABC occurs when the vertices D, E, F are exactly the midpoints of the respective sides of the triangle.

Chapter Mix

Class 11 Mathematics: Straight Lines

Q jee_main_2025_02_april_morning Area of Triangle in 3D Space
Let the vertices Q and R of the triangle PQR lie on the line (x+3)/(5)=(y-1)/(2)=(z+4)/(3), QR=5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is (m)/(n) then:
  • A. m-5√(21)n=0
  • B. 2m-5√(21)n=0
  • C. 5m-2√(21)n=0
  • D. 5m-21√(2)n=0

Solution

Related Formula

Area of a triangle given base length b and perpendicular height h:

Area = (1)/(2) · b · h
Core Logic

Since Q and R lie on the line, the length QR=5 forms the base of the triangle. The perpendicular distance from point P to the line represents the height h.

Area of Triangle in 3D Space diagram for Q67 - JEE Main 2025 Morning
Area of Triangle in 3D Space diagram for Q67 - JEE Main 2025 Morning

Step 1: Define Perpendicular Foot coordinates

Let M be the foot of the perpendicular from P(0,2,3) to the line. Parametric form of any point on the line:

M(5λ - 3, 2λ + 1, 3λ - 4)

Direction ratios of line segment PM:

DRs = (5λ - 3 - 0, 2λ + 1 - 2, 3λ - 4 - 3) = (5λ - 3, 2λ - 1, 3λ - 7)
Step 2: Solve for Parameter using Perpendicularity

Since PM is perpendicular to the given line (DRs: 5, 2, 3):

5(5λ - 3) + 2(2λ - 1) + 3(3λ - 7) = 0 25λ - 15 + 4λ - 2 + 9λ - 21 = 0 38λ = 38 λ = 1
Step 3: Compute Perpendicular Distance

For λ = 1, the coordinates of M are (2, 3, -1). Calculate length PM:

PM = √((2-0)² + (3-2)² + (-1-3)²) = √(4 + 1 + 16) = √(21)
Step 4: Formulate the Area Equation
Area = (1)/(2) · QR · PM = (1)/(2) · 5 · √(21) = (m)/(n) 5√(21)2 = (m)/(n) 2m = 5√(21)n 2m - 5√(21)n = 0
Pattern Recognition

Finding the foot of a perpendicular via parametric variables is a guaranteed shortcut for 3D area problems instead of cross-product calculations, keeping computation times minimal.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q jee_main_2025_02_april_morning Properties of Tetrahedron
Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the Δ BCD is equal to:
  • A. √(340)
  • B. 12
  • C. √(110)
  • D. 7√(3)

Solution

Related Formula

De Gua's Theorem (3D extension of Pythagorean theorem) for right corner tetrahedron states:

[Area(Δ BCD)]² = [Area(Δ ABC)]² + [Area(Δ ACD)]² + [Area(Δ ADB)]²
Core Logic

Since the three edges meeting at vertex A are mutually perpendicular, we can align them directly with orthogonal coordinate axes to compute the slanted face area.

Step 1: Apply Square Area Summation

Let the target area be Δ.

Δ² = 5² + 6² + 7² Δ² = 25 + 36 + 49 = 110 Δ = √(110)
Pattern Recognition

This is exactly analogous to finding the length of a vector given three perpendicular component lengths. Squaring, adding, and taking the square root yields the value immediately.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

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