Let (alpha, beta, gamma) be the foot of perpendicular from the point (1, 2, 3) on the line (x + 3/5) = (y

Year	2026	2025	2024	Total
Questions	15	29	18	62

Q29 jee_main_2024_01_february_morning Shortest Distance Between Skew Lines

Let the line of the shortest distance between the lines

L₁: r=( i+2 j+3 k)+λ( i- j+ k)

and

L₂: r=(4 i+5 j+6 k)+μ( i+ j- k)

intersect

L₁

and

L₂

at P and Q respectively. If

$(α, β, γ)

is the midpoint of the line segment PQ, then

2(\alpha+\beta+\gamma)$ is equal to

Numerical Answer. Answer: 21 to 21

Solution

Related Formula

The vector connecting the shortest distance points $P$ $P$ and $Q$ $Q$ on two skew lines must be simultaneously perpendicular to the direction vectors $b₁$ $\vec{b}_1$ and $b₂$ $\vec{b}_2$ of both lines:

PQ ∥ ( b₁ \times b₂)

Core Logic

Let's define general points on both lines:

Point $P$ $P$ on $L₁$ $L_1$: $(1+λ, 2-λ, 3+λ)$
Point $Q$ $Q$ on $L₂$ $L_2$: $(4+μ, 5+μ, 6-μ)$ $(4+\mu, \, 5+\mu, \, 6-\mu)$

The direction ratios of vector $PQ$ $\vec{PQ}$ are:

PQ = (3+μ-λ) i + (3+μ+λ) j + (3-μ-λ) k

Step 1: Compute Perpendicular Direction Vector

Calculate the cross product of the directions of lines $L₁$ $L_1$ and $L₂$ $L_2$:

b₁ \times b₂ = vmatrix i & j & k 1 & -1 & 1 1 & 1 & -1 vmatrix = 0 i + 2 j + 2 k

Since $PQ$ $\vec{PQ}$ is parallel to $(0, 2, 2)$ $(0, 2, 2)$, we compare the coordinate ratios:

3+μ-λ = 0 λ - μ = 3 (1)

(3+μ+λ)/(2) = (3-μ-λ)/(2) 2μ + 2λ = 0 λ + μ = 0 (2)

Shortest distance foot coordinates calculation for Q29 - JEE Main 2024 01 February Morning — The graphic maps out the geometry of lines L1 and L2 intersected by their common perpendicular segment at points A and B.

Step 2: Solve for Parameters and Midpoint

Solving linear equations (1) and (2) simultaneously:

λ = (3)/(2), μ = -(3)/(2)

Substitute these values back to find the specific coordinates of points $P$ $P$ and $Q$ $Q$:

$P = ((5)/(2), (1)/(2), (9)/(2))$
$Q = ((5)/(2), (7)/(2), (15)/(2))$

The midpoint coordinates $$(α, β, γ)$ $$$(\alpha, \beta, \gamma)$$ $are:$

(α, β, γ) = ( (5/2 + 5/2)/(2), (1/2 + 7/2)/(2), (9/2 + 15/2)/(2) ) = ((5)/(2), 2, 6)

Step 3: Final Computation Calculate the required terms:

2(α+β+γ) = 2((5)/(2) + 2 + 6) = 5 + 4 + 12 = 21

Pattern Recognition Sees: Explicit endpoints of the shortest distance line vector segment. Shortcut: Since the cross product component along$ Pattern Recognition Sees: Explicit endpoints of the shortest distance line vector segment. Shortcut: Since the cross product component along $\hat{i}$ is 0, the x-coordinates of both line points are identical, providing a massive shortcut to check algebraic equations immediately. Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Q3 jee_main_2024_29_january_evening Angle Between Two Lines

Let

P(3,2,3)

,

Q(4,6,2)

and

R(7,3,2)

be the vertices of

Δ PQR

. Then, the angle

∠ QPR

is

A. $(π)/(6)$ $\frac{\pi}{6}$
B. $⁻¹((7)/(18))$ $\cos^{-1}\left(\frac{7}{18}\right)$
C. $⁻¹((1)/(18))$ $\cos^{-1}\left(\frac{1}{18}\right)$
D. $(π)/(3)$ $\frac{\pi}{3}$

Solution

Related Formula

θ = a \cdot b| a|| b|

Core Logic

Let us compute the direction ratios of the vectors $PQ$ $\overrightarrow{PQ}$ and $PR$ $\overrightarrow{PR}$ originating from vertex $P$ $P$:

Direction ratios of $PQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$ Direction ratios of $PR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$

Step 1: Evaluating the Angle

Using the dot product formula for the angle $θ = ∠ QPR$ $\theta = \angle QPR$:

θ = 1(4) + 4(1) + (-1)(-1)\sqrt(1² + 4² + (-1)²) \cdot \sqrt(4² + 1² + (-1)²)

θ = 4 + 4 + 1\sqrt(18) \cdot \sqrt(18) = (9)/(18) = (1)/(2)

Angle Between Two Lines diagram for Q3 - JEE Main 2024 Evening

Since $θ = (1)/(2)$ $\cos \theta = \frac{1}{2}$, we have:

θ = (π)/(3)

Pattern Recognition

When asked for an angle like $∠ QPR$ $\angle QPR$, always ensure both vectors diverge from the common vertex $P$ $P$ (i.e., use $PQ$ $\overrightarrow{PQ}$ and $PR$ $\overrightarrow{PR}$) to avoid sign errors.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q26 jee_main_2024_29_january_evening Shortest Distance Between Two Lines

Let

O

be the origin, and

M

and

N

be the points on the lines

(x - 5)/(4) = (y - 4)/(1) = (z - 5)/(3)

and

(x + 8)/(12) = (y + 2)/(5) = (z + 11)/(9)

respectively such that

MN

is the shortest distance between the given lines. Then

OM\cdot ON

is equal to

Numerical Answer. Answer: 9 to 9

Solution

Related Formula

The line segment of shortest distance $MN$ $MN$ is perpendicular to both directing vectors $b₁$ $\vec{b}_1$ and $b₂$ $\vec{b}_2$.

Core Logic

Let general point coordinates be expressions of parameters $λ$ $\lambda$ and $μ$ $\mu$:

M = (4λ + 5, λ + 4, 3λ + 5)

N = (12μ - 8, 5μ - 2, 9μ - 11)

Vector direction ratios of $MN$ $\overrightarrow{MN}$:

MN = (4λ - 12μ + 13, λ - 5μ + 6, 3λ - 9μ + 16)

The cross product vector perpendicular to both lines is $b₁ \times b₂ = (-6, 0, 8)$ .

Step 1: Finding Parameters via Perpendicular Conditions

Equating directional proportional factors:

(4λ - 12μ + 13)/(-6) = (λ - 5μ + 6)/(0) = (3λ - 9μ + 16)/(8)

From the zero denominator constraint:

λ - 5μ + 6 = 0 (iii)

From the first and third components:

8(4λ - 12μ + 13) = -6(3λ - 9μ + 16) 32λ - 96μ + 104 = -18λ + 54μ - 96

50λ - 150μ + 200 = 0 λ - 3μ + 4 = 0 (iv)

Solving equations (iii) and (iv) yields $λ = -1$ $\lambda = -1$ and $μ = 1$ $\mu = 1$.

Step 2: Vector Coordinate Evaluation

Substituting parameter roots into point layout vectors:

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

N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)

Evaluating target scalar products:

OM \cdot ON = 1(4) + 3(3) + 2(-2) = 4 + 9 - 4 = 9

Pattern Recognition

Shortest distance points constrain line segments to align with the direct cross product vector. Setting proportional components yields quick parameters.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q3 jee_main_2024_27_jan_morning Distance of a Point from a Line

The distance of the point

(7, -2, 11)

from the line

(x-6)/(1)=(y-4)/(0)=(z-8)/(3)

along the line

(x-5)/(2)=(y-1)/(-3)=(z-5)/(6)

, is:

A. $12$ $12$
B. $14$ $14$
C. $18$ $18$
D. $21$ $21$

Solution

Related Formula

(x-x₁)/(a) = (y-y₁)/(b) = (z-z₁)/(c) = λ

Core Logic

Let point $A = (7, -2, 11)$ $A = (7, -2, 11)$. We want the distance from $A$ $A$ to the line $L₁: (x-6)/(1)=(y-4)/(0)=(z-8)/(3)$ measured along a line parallel to $L₂: (x-5)/(2)=(y-1)/(-3)=(z-5)/(6)$ .

The line passing through $A$ $A$ parallel to $L₂$ $L_2$ will have the equation:

(x-7)/(2) = (y+2)/(-3) = (z-11)/(6) = λ

Any general point $B$ $B$ on this new line can be represented as:

B \equiv (2λ + 7, -3λ - 2, 6λ + 11)

Step 1: Finding Intersection Point B

Since $B$ $B$ lies on the given line $L₁$ $L_1$, its coordinates must satisfy the equation of $L₁$ $L_1$:

((2λ + 7) - 6)/(1) = ((-3λ - 2) - 4)/(0) = ((6λ + 11) - 8)/(3)

Focus on the middle term (since denominator is 0, numerator must equal 0 for intersection):

-3λ - 6 = 0 \Rightarrow λ = -2

Step 2: Coordinates of B and Distance Calculation

Substitute $λ = -2$ $\lambda = -2$ back into the coordinates of point $B$ $B$:

B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)

Now, calculate the distance $AB$ $AB$ using the 3D distance formula:

AB = \sqrt((7-3)² + (-2-4)² + (11 - (-1))²)

AB = \sqrt(4² + (-6)² + (12)²)

AB = \sqrt(16 + 36 + 144)

AB = \sqrt(196) = 14

Pattern Recognition

Distance of a point from a line along another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Q8 jee_main_2024_27_jan_morning Shortest Distance Between Two Lines

If the shortest distance between the lines

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

and

(x-λ)/(2)=(y+1)/(4)=(z-2)/(-5)

is

6\sqrt(5)

, then the sum of all possible values of

λ

is:

A. $5$ $5$
B. $8$ $8$
C. $7$ $7$
D. $10$ $10$

Solution

Related Formula

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

Core Logic

Identify the positional vectors and direction ratios for both lines: Line 1: $a₁ = (4, -1, 0)$ $\vec{a}_1 = (4, -1, 0)$, direction $b₁ = (1, 2, -3)$ $\vec{b}_1 = (1, 2, -3)$ Line 2: $a₂ = (λ, -1, 2)$ $\vec{a}_2 = (\lambda, -1, 2)$, direction $b₂ = (2, 4, -5)$ $\vec{b}_2 = (2, 4, -5)$

Vector connecting lines: $( a₂ - a₁) = (λ - 4, 0, 2)$

Step 1: Cross Product and Magnitude

Find the normal vector $n = b₁ \times b₂$ $\vec{n} = \vec{b}_1 \times \vec{b}_2$:

b₁ \times b₂ = vmatrix i & j & k 1 & 2 & -3 2 & 4 & -5 vmatrix

= i(-10 - (-12)) - j(-5 - (-6)) + k(4 - 4)

= 2 i - 1 j + 0 k = (2, -1, 0)

Magnitude of the normal vector:

| b₁ \times b₂| = \sqrt(2² + (-1)² + 0²) = \sqrt(5)

Step 2: Application of Shortest Distance Formula

Dot product of normal vector and positional difference vector:

( a₂ - a₁) \cdot ( b₁ \times b₂) = (λ - 4)(2) + (0)(-1) + (2)(0) = 2(λ - 4)

Using the shortest distance formula given as $6\sqrt(5)$ $\frac{6}{\sqrt{5}}$:

|2(λ - 4)|\sqrt(5) = 6\sqrt(5)

|2(λ - 4)| = 6

|λ - 4| = 3

Step 3: Finding Unknown values

Solve the absolute value relation:

λ - 4 = 3 \Rightarrow λ = 7

λ - 4 = -3 \Rightarrow λ = 1

Sum of possible values = $7 + 1 = 8$ $7 + 1 = 8$.

Pattern Recognition

Standard Shortest Distance methodology between skew lines. Cross product of direction vectors forms the perpendicular frame normal, and dot-producting the difference of positional anchor points yields the direct orthogonal projection.

Chapter Mix

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

Let (α, β, γ)$(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point (1, 2, 3)$(1, 2, 3)$ on the line (x + 3)/(5) = (y - 1)/(2) = (z + 4)/(3)$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$. then 19(α + β + γ)$19(\alpha + \beta + \gamma)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Finding Direction Ratios

Step 2: Using Perpendicularity Condition

Step 3: Calculating target expression

Pattern Recognition

Chapter Mix

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Compute Perpendicular Direction Vector

Step 2: Solve for Parameters and Midpoint

Step 3: Final Computation

Step 3: Final Computation

Pattern Recognition

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Evaluating the Angle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Parameters via Perpendicular Conditions

Step 2: Vector Coordinate Evaluation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Intersection Point B

Step 2: Coordinates of B and Distance Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Cross Product and Magnitude

Step 2: Application of Shortest Distance Formula

Step 3: Finding Unknown values

Pattern Recognition

Chapter Mix

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Let $(α, β, γ)$ $(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point $(1, 2, 3)$ $(1, 2, 3)$ on the line $(x + 3)/(5) = (y - 1)/(2) = (z + 4)/(3)$ . then $19(α + β + γ)$ $19(\alpha + \beta + \gamma)$ is equal to: