Let the line passing through the points (-1, 2, 1) and to the line (x - 1/2) = (y + 1/3) = (z/4) intersect the line (x…

Year	2026	2025	2024	Total
Questions	15	29	18	62

Q10 jee_main_2026_22_january_evening Distance Between Points on Line

Let

L

be the line

(x+1)/(2) = (y+1)/(3) = (z+3)/(6)

and let

S

be the set of all points

(a,b,c)

on

L

, whose distance from the line

(x+1)/(2) = (y+1)/(3) = (z-9)/(0)

along the line

L

is

7

. Then

Σ(a,b,c) in S (a+b+c)

is equal to:

A. 34
B. 28
C. 40
D. 6

Solution

Related Formula

Distance along a line from intersection point $M$ $M$ using parametric coordinates.

Core Logic

Find intersection point $M$ $M$ of $L₁$ $L_1$ and $L₂$ $L_2$:

2λ - 1 = 2μ - 1, 3λ - 1 = 3μ - 1, 6λ - 3 = 9 λ = 2, μ = 2

Point of intersection $M = (3, 5, 9)$ $M = (3, 5, 9)$.

Any parametric point $P$ $P$ on $L₁$ $L_1$ is $(2K-1, 3K-1, 6K-3)$ $(2K-1, 3K-1, 6K-3)$.

Step 1: Solve for Points P and Q

Distance $PM = 7$ $PM = 7$:

\sqrt((2K-4)² + (3K-6)² + (6K-12)²) = 7 7|K-2| = 7 K = 1 or 3

For $K=1$ $K=1$: $P = (1, 2, 3)$ $P = (1, 2, 3)$, sum $= 6$ $= 6$
For $K=3$ $K=3$: $Q = (5, 8, 15)$ $Q = (5, 8, 15)$, sum $= 28$ $= 28$

Step 2: Total Sum

Total sum $= 6 + 28 = 34$ $= 6 + 28 = 34$.

Pattern Recognition

Intersection point gives central reference; unit direction vector gives required offset points along the line.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Q15 jee_main_2026_23_january_morning Lines in 3D

The vertices

B

and

C

of a triangle

ABC

lie on the line

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

. The coordinates of

A

and

B

are

(1, 6, 3)

and

(4, 9, α)

respectively and

C

is at a distance of

10

units from

B

. The area (in sq. units) of

Δ ABC

is:

A. $5\sqrt(13)$ $5\sqrt{13}$
B. $15\sqrt(13)$ $15\sqrt{13}$
C. $20\sqrt(13)$ $20\sqrt{13}$
D. $10\sqrt(13)$ $10\sqrt{13}$

Solution

Related Formula

Area of Δ = (1)/(2) \times base \times height = (1)/(2) \times BC \times AD

Core Logic

The line is standardly written as $(x)/(1) = (y - 1)/(2) = (z - 2)/(3) = λ$ . Point $B$ $B$ $(4, 9, α)$ $(4, 9, \alpha)$ lies on this line, so we can solve for $α$ $\alpha$ by substituting it in:

(4)/(1) = (9 - 1)/(2) = (α - 2)/(3) \Rightarrow 4 = 4 = (α - 2)/(3) \Rightarrow α = 14

Thus, $B$ $B$ is $(4, 9, 14)$ $(4, 9, 14)$.

Lines in 3D diagram for Q15 - JEE Main 2026 Morning

Step 1: Find the Perpendicular Foot (D)

Let $D$ $D$ be the foot of the perpendicular from $A(1, 6, 3)$ $A(1, 6, 3)$ to the line $BC$ $BC$. The coordinates of a general point on the line are $D = (λ, 2λ + 1, 3λ + 2)$ . The direction vector of $AD$ $AD$ is $AD = (λ - 1) i + (2λ - 5) j + (3λ - 1) k$ . Since $AD$ $AD$ is perpendicular to the line whose direction ratios are $1, 2, 3$ $\langle 1, 2, 3 \rangle$:

1(λ - 1) + 2(2λ - 5) + 3(3λ - 1) = 0

λ - 1 + 4λ - 10 + 9λ - 3 = 0

14λ - 14 = 0 \Rightarrow λ = 1

Step 2: Calculate Height (AD)

Substitute $λ = 1$ $\lambda = 1$ into $D$ $D$: $D = (1, 3, 5)$ $D = (1, 3, 5)$. The perpendicular distance $AD$ $AD$ is:

AD = \sqrt((1-1)² + (6-3)² + (3-5)²) = \sqrt(0² + 3² + (-2)²) = \sqrt(9 + 4) = \sqrt(13)

Step 3: Calculate Area of Triangle

We are given that the base $BC$ $BC$ is $10$ $10$ units.

Area = (1)/(2) \times BC \times AD = (1)/(2) \times 10 \times \sqrt(13) = 5\sqrt(13)

Pattern Recognition

When a triangle area is requested given a line equation containing the base, dropping a generic lambda-perpendicular establishes height ( $AD$ $AD$), decoupling the problem from finding the exact coordinate of $C$ $C$.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Q20 jee_main_2026_23_january_morning Direction Cosines

Let the direction cosines of two lines satisfy the equations:

4 + m - n = 0

and

2mn + 10n + 3 m = 0

. Then the cosine of the acute angle between these lines is:

A. $10\sqrt(38)$ $\frac{10}{\sqrt{38}}$
B. $203\sqrt(38)$ $\frac{20}{3\sqrt{38}}$
C. $107\sqrt(38)$ $\frac{10}{7\sqrt{38}}$
D. $103\sqrt(38)$ $\frac{10}{3\sqrt{38}}$

Solution

Related Formula

θ = |a₁a₂ + b₁b₂ + c₁c₂|\sqrt(a₁²+b₁²+c₁²)\sqrt(a₂²+b₂²+c₂²)

Core Logic

From the first linear equation, isolate $n$ $n$: $n = 4 + m$ $n = 4\ell + m$ Substitute this into the second quadratic relation:

2m(4 + m) + 10 (4 + m) + 3 m = 0

8 m + 2m² + 40 ² + 10 m + 3 m = 0

40 ² + 21 m + 2m² = 0

Factorizing the quadratic equation:

(8 + m)(5 + 2m) = 0

This gives two cases, generating the direction ratios of the two lines.

Step 1: Determine Direction Ratios

Case 1: $m = -8$ $m = -8\ell$ Then $n = 4 - 8 = -4$ $n = 4\ell - 8\ell = -4\ell$. Direction Ratios (D.R.s) for line $L₁$ $L_1$: $( , -8 , -4 ) \equiv (1, -8, -4)$ .

Case 2: $m = -(5)/(2)$ $m = -\frac{5}{2}\ell$ Then $n = 4 - (5)/(2) = (3)/(2)$ . Direction Ratios (D.R.s) for line $L₂$ $L_2$: $( , -(5)/(2) , (3)/(2) ) \equiv (1, -(5)/(2), (3)/(2)) \equiv (2, -5, 3)$ .

Step 2: Calculate Angle Between Lines

Use the angle formula with D.R.s $d₁ = (1, -8, -4)$ $d_1 = (1, -8, -4)$ and $d₂ = (2, -5, 3)$ $d_2 = (2, -5, 3)$:

θ = |(1)(2) + (-8)(-5) + (-4)(3)|\sqrt(1² + (-8)² + (-4)²) \sqrt(2² + (-5)² + 3²)

θ = |2 + 40 - 12|\sqrt(1 + 64 + 16) \sqrt(4 + 25 + 9)

θ = 30\sqrt(81) \sqrt(38) = 309\sqrt(38) = 103\sqrt(38)

Pattern Recognition

When direction cosines are entangled in one linear and one quadratic equation, isolate the single-degree variable from the linear plane equation and plug it into the cone equation to split it into two generating lines.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Q22 jee_main_2026_23_january_evening Line in 3D

If the image of the point

P(a, 2, a)

in the line

(x)/(2) = (y + a)/(1) = (z)/(1)

is

Q

and the image of

Q

in the line

(x - 2b)/(2) = (y - a)/(1) = (z + 2b)/(-5)

is

P

, then

a + b

is equal to

Numerical Answer. Answer: 3 to 3

Solution

Related Formula

For a point and its image across a line, the midpoint of the segment connecting them lies exactly on the line. The segment itself is perpendicular to the line.

Core Logic

Line in 3D diagram for Q22 - JEE Main 2026 Evening

Let the first line be

L₁

:

(x)/(2) = (y + a)/(1) = (z)/(1) = λ

. A general point on

L₁

is

(2λ, λ - a, λ)

. This is the midpoint

M₁

of

P(a, 2, a)

and

Q(x₁, y₁, z₁)

.

Thus, $Q$ $Q$ is given by:

x₁ = 4λ - a

y₁ = 2λ - 2a - 2

z₁ = 2λ - a

Now, the image of $Q$ $Q$ across $L₂$ $L_2$: $(x - 2b)/(2) = (y - a)/(1) = (z + 2b)/(-5) = μ$ is $P(a, 2, a)$ $P(a, 2, a)$. The midpoint $M₂$ $M_2$ of $PQ$ $PQ$ lies on $L₂$ $L_2$. But $M₂$ $M_2$ is the exact same set of coordinates in space as $M₁$ $M_1$ since they are both the midpoint of the segment $PQ$ $PQ$!

Step 1: Equating Midpoints

General point on $L₂$ $L_2$ is $(2μ + 2b, μ + a, -5μ - 2b)$ $(2\mu + 2b, \mu + a, -5\mu - 2b)$. Since the midpoint must be the same: $(a + (4λ - a))/(2) = 2μ + 2b 2λ = 2μ + 2b λ - μ = b$ $(2 + (2λ - 2a - 2))/(2) = μ + a λ - a = μ + a λ - μ = 2a$ $(a + (2λ - a))/(2) = -5μ - 2b λ = -5μ - 2b λ + 5μ = -2b$

Step 2: Solving System

From the first two equations, $b = 2a$ $b = 2a$. Substitute $b = 2a$ $b = 2a$ into the third: $λ + 5μ = -4a$ $\lambda + 5\mu = -4a$. Also $λ - μ = 2a$ $\lambda - \mu = 2a$. Subtracting gives $6μ = -6a μ = -a$ $6\mu = -6a \implies \mu = -a$. Thus $λ = a$ $\lambda = a$.

The direction vector of segment

PQ

is proportional to the vector from

P

to

M₁

.

PM₁ = (2λ - a) i + (λ - a - 2) j + (λ - a) k

. Since

λ = a

:

PM₁ = a i - 2 j + 0 k

.

This vector must be perpendicular to $L₁$ $L_1$ (direction vector $2 i + j + k$ $2\hat{i} + \hat{j} + \hat{k}$): $(a)(2) + (-2)(1) + (0)(1) = 0 2a - 2 = 0 a = 1$ .

Step 3: Final Values

Since $a = 1$ $a = 1$, and $b = 2a$ $b = 2a$, we have $b = 2$ $b = 2$. Therefore, $a + b = 1 + 2 = 3$ $a + b = 1 + 2 = 3$.

Pattern Recognition

If $P$ $P$ is the image of $Q$ $Q$ on $L₁$ $L_1$, and $Q$ $Q$ is the image of $P$ $P$ on $L₂$ $L_2$, then $L₁$ $L_1$ and $L₂$ $L_2$ must intersect segment $PQ$ $PQ$ at the exact same midpoint. Using the midpoint consistency across both lines creates a rapid linear system.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Q3 jee_main_2026_24_january_morning Intersection of Lines and Distances

Let the lines

L₁: r = i + 2 j + 3 k + λ(2 i + 3 j + 4 k)

,

λ in R

and

L₂: r = (4 i + j) + μ(5 i + 2 j + k)

,

μ in R

, intersect at the point

R

. Let

P

and

Q

be the points lying on lines

L₁

and

L₂

, respectively, such that

| PR| = \sqrt(29)

and

| PQ| = \sqrt((47)/(3))

. If the point

P

lies in the first octant, then

27(QR)²

is equal to

A. $340$ $340$
B. $360$ $360$
C. $320$ $320$
D. $348$ $348$

Solution

Related Formula

Distance formula: d² = (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

Core Logic

Find the point of intersection $R$ $R$ by equating general points on $L₁$ $L_1$ and $L₂$ $L_2$: General point on $L₁$ $L_1$: $(2λ + 1, 3λ + 2, 4λ + 3)$ General point on $L₂$ $L_2$: $(5μ + 4, 2μ + 1, μ)$ $(5\mu + 4, 2\mu + 1, \mu)$

Equating coordinates:

2λ + 1 = 5μ + 4

3λ + 2 = 2μ + 1

4λ + 3 = μ

Solving yields $λ = -1$ $\lambda = -1$ and $μ = -1$ $\mu = -1$. Point of intersection $R$ $R$ is $(-1, -1, -1)$ $(-1, -1, -1)$.

Step 1: Finding Point P

Let point $P$ $P$ be $(2λ + 1, 3λ + 2, 4λ + 3)$ . $PR² = 29$ $PR^2 = 29$

(2λ + 1 + 1)² + (3λ + 2 + 1)² + (4λ + 3 + 1)² = 29

(2λ + 2)² + (3λ + 3)² + (4λ + 4)² = 29

4(λ + 1)² + 9(λ + 1)² + 16(λ + 1)² = 29

29(λ + 1)² = 29 \Rightarrow (λ + 1)² = 1

λ + 1 = \pm 1 \Rightarrow λ = 0 or λ = -2

If $λ = -2$ $\lambda = -2$, $P(-3, -4, -5)$ $P(-3, -4, -5)$ (not in first octant, rejected). If $λ = 0$ $\lambda = 0$, $P(1, 2, 3)$ $P(1, 2, 3)$ (in first octant, accepted).

Step 2: Finding Point Q

Let $Q$ $Q$ be $(5μ + 4, 2μ + 1, μ)$ $(5\mu + 4, 2\mu + 1, \mu)$.

PQ² = (47)/(3)

(5μ + 4 - 1)² + (2μ + 1 - 2)² + (μ - 3)² = (47)/(3)

(5μ + 3)² + (2μ - 1)² + (μ - 3)² = (47)/(3)

25μ² + 30μ + 9 + 4μ² - 4μ + 1 + μ² - 6μ + 9 = (47)/(3)

30μ² + 20μ + 19 = (47)/(3)

90μ² + 60μ + 57 = 47 \Rightarrow 90μ² + 60μ + 10 = 0

9μ² + 6μ + 1 = 0 \Rightarrow (3μ + 1)² = 0

μ = -(1)/(3)

Step 3: Calculating QR^2

Point $Q = ( 5(-(1)/(3))+4, 2(-(1)/(3))+1, -(1)/(3) ) = ( (7)/(3), (1)/(3), -(1)/(3) )$ . $R(-1, -1, -1)$ $R(-1, -1, -1)$.

(QR)² = ( (7)/(3) + 1 )² + ( (1)/(3) + 1 )² + ( -(1)/(3) + 1 )²

= ( (10)/(3) )² + ( (4)/(3) )² + ( (2)/(3) )² = (100 + 16 + 4)/(9) = (120)/(9)

27 \times (QR)² = 27 \times (120)/(9) = 3 \times 120 = 360

Pattern Recognition

When given fixed line parameters and distances, expressing all points in terms of their single parameter ( $λ$ $\lambda$ or $μ$ $\mu$) converts 3D distance problems into simple single-variable quadratic equations.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry

Solution & Explanation

Related Formula

Core Logic

Step 1: Compute the Point of Intersection

Step 2: Calculate Euclidean Distance PQ

Pattern Recognition

Chapter Mix

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Solve for Points P and Q

Step 2: Total Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find the Perpendicular Foot (D)

Step 2: Calculate Height (AD)

Step 3: Calculate Area of Triangle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Direction Ratios

Step 2: Calculate Angle Between Lines

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating Midpoints

Step 2: Solving System

Step 3: Final Values

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Point P

Step 2: Finding Point Q

Step 3: Calculating QR^2

Pattern Recognition

Chapter Mix

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