If the shortest distance between the lines (x - 1/2) = (y - 2/3) = (z - 3/4) and (x/1) = (y/alpha) = (z

Year	2026	2025	2024	Total
Questions	15	29	18	62

Q68 jee_main_2025_24_jan_morning Intersection of Lines in 3D Space

Let the line passing through the points

(-1, 2, 1)

and \parallel to the line

(x - 1)/(2) = (y + 1)/(3) = (z)/(4)

intersect the line

(x + 2)/(3) = (y - 3)/(2) = (z - 4)/(1)

at the point

P

. Then the distance of

P

from the point

Q(4, -5, 1)

is :

A. $5$ $5$
B. $10$ $10$
C. $5\sqrt(6)$ $5\sqrt{6}$
D. $5\sqrt(5)$ $5\sqrt{5}$

Solution

Related Formula

The line passing through $a$ $\vec{a}$ and \parallel to direction vector $v = l i + m j + n k$ is written in symmetric form as:

(x - xₐ)/(l) = (y - yₐ)/(m) = (z - zₐ)/(n)

Core Logic

Formulate the equation of the line passing through $(-1, 2, 1)$ $(-1, 2, 1)$ with direction vector components $(2, 3, 4)$ $(2, 3, 4)$:

L₁: (x + 1)/(2) = (y - 2)/(3) = (z - 1)/(4) = λ (1)

Any generic point on this line can be written as:

P = (2λ - 1, 3λ + 2, 4λ + 1)

The second given line equation is:

L₂: (x + 2)/(3) = (y - 3)/(2) = (z - 4)/(1) = μ (2)

Any generic point on line $L₂$ $L_2$ is:

P' = (3μ - 2, 2μ + 3, μ + 4)

Step 1: Compute the Point of Intersection

At the point of intersection $P$ $P$, equate the coordinates from both lines:

2λ - 1 = 3μ - 2 2λ - 3μ = -1 (3)

3λ + 2 = 2μ + 3 3λ - 2μ = 1 (4)

4λ + 1 = μ + 4 4λ - μ = 3 (5)

Solving equations (3) and (4) simultaneously gives:

λ = 1, μ = 1

Verify these values using equation (5):

4(1) - 1 = 3 (Satisfied)

Thus, substituting $λ = 1$ $\lambda = 1$ gives the coordinates of point $P$ $P$:

P = (2(1) - 1, 3(1) + 2, 4(1) + 1) = (1, 5, 5)

Step 2: Calculate Euclidean Distance PQ

Find the distance between $P(1, 5, 5)$ $P(1, 5, 5)$ and $Q(4, -5, 1)$ $Q(4, -5, 1)$ using the 3D distance formula:

PQ = \sqrt((4 - 1)² + (-5 - 5)² + (1 - 5)²)

PQ = \sqrt(3² + (-10)² + (-4)²) = \sqrt(9 + 100 + 16)

PQ = \sqrt(125) = 5\sqrt(5)

Pattern Recognition

When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q jee_main_2025_28_jan_evening Distance from a Point to a Line

The square of the distance of the point

((15)/(7),(32)/(7),7)

from the line

(x+1)/(3)=(y+3)/(5)=(z+5)/(7)

in the direction of the vector

i+4 j+7 k

is:

A. $54$ $54$
B. $41$ $41$
C. $66$ $66$
D. $44$ $44$

Solution

Related Formula

Equation of a line passing through $P(x₀, y₀, z₀)$ $P(x_0, y_0, z_0)$ along direction vector $v = a i+b j+c k$ $\vec{v} = a\hat{i}+b\hat{j}+c\hat{k}$:

(x-x₀)/(a) = (y-y₀)/(b) = (z-z₀)/(c) = λ

Core Logic

Let the given point be $P((15)/(7), (32)/(7), 7)$ . We need to find the distance measured along the line passing through $P$ $P$ parallel to the vector $v = i+4 j+7 k$ $\vec{v} = \hat{i}+4\hat{j}+7\hat{k}$.

Equation of this line $PQ$ $PQ$ is:

(x - (15)/(7))/(1) = (y - (32)/(7))/(4) = (z - 7)/(7) = λ

Any general point $Q$ $Q$ on this line can be written as:

Q(λ + (15)/(7), 4λ + (32)/(7), 7λ + 7)

Step 1: Find Intersection Point Q with Given Line

Point $Q$ $Q$ must lie on the given target line $L: (x+1)/(3) = (y+3)/(5) = (z+5)/(7)$ .

Substitute coordinates of $Q$ $Q$ into the first and third fractions:

((λ + (15)/(7)) + 1)/(3) = ((7λ + 7) + 5)/(7)

(λ + (22)/(7))/(3) = (7λ + 12)/(7)

(7λ + 22)/(21) = (7λ + 12)/(7)

Multiplying by 21:

7λ + 22 = 3(7λ + 12)

7λ + 22 = 21λ + 36

14λ = -14 λ = -1

Step 2: Calculate Coordinates of Q and Distance squared

Substituting $λ = -1$ $\lambda = -1$ into coordinates of $Q$ $Q$:

Q(-1 + (15)/(7), -4 + (32)/(7), -7 + 7) = Q((8)/(7), (4)/(7), 0)

Now, compute the distance squared $(PQ)²$ $(PQ)^2$:

(PQ)² = ((15)/(7) - (8)/(7))² + ((32)/(7) - (4)/(7))² + (7 - 0)²

(PQ)² = ((7)/(7))² + ((28)/(7))² + 7² = 1² + 4² + 49 = 1 + 16 + 49 = 66

Pattern Recognition

Instead of finding coordinates of $Q$ $Q$, we could also use the vector form directly: $PQ = λ( i + 4 j + 7 k)$ . Distance squared is $PQ² = λ²(1² + 4² + 7²) = (-1)²(1 + 16 + 49) = 66$ . This saves a lot of fractional coordinate subtraction arithmetic!

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q64 jee_main_2025_29_jan_morning Intersection of Lines

Let

a = i + 2 j + k

and

b = 2 i + 7 j + 3 k

. Let

L₁: r = (- i +2 j + k) + λ a,λ in R

and

L₂: r = ( j + k) + μ b,μ in R

be two lines. If the line

L₃

passes through the point of intersection of

L₁

and

L₂

, and is \parallel to

a + b

, then

L₃

passes through the point:

A. (8, 26, 12)
B. (2, 8, 5)
C. (-1, -1, 1)
D. (5, 17, 4)

Solution

Related Formula

Equation of line passing through r₀ ∥ to v: r = r₀ + α v

Core Logic

Express general points on $L₁$ $L_1$ and $L₂$ $L_2$ vector components:

L₁: r = (λ - 1) i + 2(λ + 1) j + (λ + 1) k

L₂: r = 2μ i + (1 + 7μ) j + (1 + 3μ) k

Step 1: Equate components to find point of intersection

λ - 1 = 2μ (1)

2λ + 2 = 1 + 7μ (2)

λ + 1 = 1 + 3μ λ = 3μ (3)

Substituting (3) into (1):

3μ - 1 = 2μ μ = 1 λ = 3

Point of intersection $r₀ = 2(1) i + (1+7) j + (1+3) k = 2 i + 8 j + 4 k$ .

Step 2: Formulate line L3

Direction vector $v = a + b = 3 i + 9 j + 4 k$ .

L₃: r = (2 i + 8 j + 4 k) + α(3 i + 9 j + 4 k)

L₃(α) = (2 + 3α) i + (8 + 9α) j + (4 + 4α) k

Step 3: Match options

For $α = 2$ $\alpha = 2$:

r = (2+6) i + (8+18) j + (4+8) k = 8 i + 26 j + 12 k

This matches coordinates $(8, 26, 12)$ $(8, 26, 12)$.

Pattern Recognition

Instead of checking all lines simultaneously, match options via quick ratio checks: $(x - 2)/(3) = (y - 8)/(9) = (z - 4)/(4)$ . This eliminates false choices instantly.

Chapter Mix

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

Q67 jee_main_2025_29_jan_morning Perpendicular Lines

Let

L₁: (x - 1)/(1) = (y - 2)/(-1) = (z - 1)/(2)

and

L₂: x + 1-1 = y - 22 = z1

be two lines. Let

L₃

be a line passing through the point

(α, β, γ)

and be perpendicular to both

L₁

and

L₂

. If

L₃

intersects

L₁

, then

|5α - 11β - 8γ|

equals:

A. 18
B. 16
C. 25
D. 20

Solution

Related Formula

Direction ratios of a line perpendicular to both vectors: v = d₁ \times d₂

Core Logic

Compute direction ratios for $L₃$ $L_3$ using cross product of direction vectors of $L₁$ $L_1$ and $L₂$ $L_2$:

v = | arrayccc i & j & k 1 & -1 & 2 -1 & 2 & 1 array | = -5 i - 3 j + k

Step 1: Establish intersection constraints

Let $A$ $A$ be a general point on $L₃$ $L_3$ and $B$ $B$ be a general point on $L₁$ $L_1$:

A = (α - 5λ, β - 3λ, γ + λ)

B = (k+1, -k+2, 2k+1)

Since $L₃$ $L_3$ intersects $L₁$ $L_1$, at the point of intersection $A \equiv B$ $A \equiv B$:

α - 5λ = k + 1 α = 5λ + k + 1

β - 3λ = -k + 2 β = 3λ - k + 2

γ + λ = 2k + 1 γ = -λ + 2k + 1

Step 2: Solve final algebraic target equation

Substitute expressions into the target mod expression:

5α - 11β - 8γ = 5(5λ + k + 1) - 11(3λ - k + 2) - 8(-λ + 2k + 1)

= λ(25 - 33 + 8) + k(5 + 11 - 16) + (5 - 22 - 8)

= 0λ + 0k - 25 = -25

Taking absolute value yields $|-25| = 25$ $|-25| = 25$.

Pattern Recognition

The cancellation of internal variables ( $λ$ $\lambda$ and $k$ $k$) shows that the absolute expression describes a invariant geometric plane coordinate constraint, allowing simple baseline parameter values (like $k=0, λ=0$ $k=0, \lambda=0$) to evaluate the question instantly.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q20 jee_main_2024_01_february_morning Shortest Distance Between Skew Lines

If the shortest distance between the lines

(x-λ)/(-2)=(y-2)/(1)=(z-1)/(1)

and

x-\sqrt(3)1=(y-1)/(-2)=(z-2)/(1)

is

1

, then the sum of all possible values of

λ

is:

A. $0$ $0$
B. $2\sqrt(3)$ $2\sqrt{3}$
C. $3\sqrt(3)$ $3\sqrt{3}$
D. $-2\sqrt(3)$ $-2\sqrt{3}$

Solution

Related Formula

The shortest distance between two skew lines $r = a₁ + s b₁$ $\vec{r} = \vec{a}_1 + s\vec{b}_1$ and $r = a₂ + t b₂$ $\vec{r} = \vec{a}_2 + t\vec{b}_2$ is given by:

d = |( a₂ - a₁) \cdot ( b₁ \times b₂)|| b₁ \times b₂|

Core Logic

Identify the vectors from the equations of the lines:

Line 1 passes through $a₁ = λ i + 2 j + k$ with direction $b₁ = -2 i + j + k$ .
Line 2 passes through $a₂ = \sqrt(3) i + j + 2 k$ with direction $b₂ = i - 2 j + k$ .

Calculate the coordinate difference vector:

a₂ - a₁ = (\sqrt(3) - λ) i - j + k

Step 1: Compute the Cross Product of the Direction Vectors

Find the cross product matrix:

b₁ \times b₂ = vmatrix i & j & k -2 & 1 & 1 1 & -2 & 1 vmatrix

b₁ \times b₂ = i(1 - (-2)) - j(-2 - 1) + k(4 - 1) = 3 i + 3 j + 3 k

Compute its magnitude:

| b₁ \times b₂| = \sqrt(3² + 3² + 3²) = \sqrt(27) = 3\sqrt(3)

Shortest distance between skew lines vector representation for Q20 - JEE Main 2024 01 February Morning — The figure details the spatial arrangement of the two skew lines along with their common perpendicular normal vector representing the shortest path.

Step 2: Solve the Absolute Distance Equation for lambda

Substitute these values into the shortest distance formula:

1 = |((\sqrt(3) - λ) i - j + k) \cdot (3 i + 3 j + 3 k)|3\sqrt(3)

1 = |3(\sqrt(3) - λ) - 3 + 3|3\sqrt(3) = 3|\sqrt(3) - λ|3\sqrt(3) = |\sqrt(3) - λ|\sqrt(3)

This gives:

|\sqrt(3) - λ| = \sqrt(3)

Unfolding the absolute value gives two cases:

Case A: $\sqrt(3) - λ = \sqrt(3) λ = 0$
Case B: $\sqrt(3) - λ = -\sqrt(3) λ = 2\sqrt(3)$

Step 3: Calculate the Final Sum

The sum of all possible values of $λ$ $\lambda$ is:

Sum = 0 + 2\sqrt(3) = 2\sqrt(3)

Pattern Recognition

Sees: Shortest distance setup between vector paths containing free variables. Shortcut: In equations like $|c - λ| = d$ $|c - \lambda| = d$, the sum of the roots is simply equal to $2c$ $2c$, because the roots are symmetrically balanced around the center point $c$ $c$. Thus, $Sum = 2 \times \sqrt(3) = 2\sqrt(3)$ holds instantly without calculating individual values.

Chapter Mix

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

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute Cross Product of Direction Vectors

Step 2: Apply Shortest Distance Formula

Step 3: Calculate the Sum of Roots

Pattern Recognition

Chapter Mix

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Compute the Point of Intersection

Step 2: Calculate Euclidean Distance PQ

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find Intersection Point Q with Given Line

Step 2: Calculate Coordinates of Q and Distance squared

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equate components to find point of intersection

Step 2: Formulate line L3

Step 3: Match options

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Establish intersection constraints

Step 2: Solve final algebraic target equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute the Cross Product of the Direction Vectors

Step 2: Solve the Absolute Distance Equation for lambda

Step 3: Calculate the Final Sum

Pattern Recognition

Chapter Mix

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