Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : Delta ABC is an isosceles right angled triangle and (S2): the area of Delta ABC is 9sqrt(2)2.

### Related Formula 3D Cartesian distance formula: ### Core Logic Since A(x,y,z) lies in the xy-plane, its z-coordinate must be zero (z = 0). Let the reference targets be P(0,3,2), Q(2,0,3), and R(0,0,1). Setting AP2 = AR2: ### Step 1: Locating Coordinate Dimensions Setting AQ2 = AR2 with y=2: Thus, A is precisely located at (3,2,0). ### Step 2: Triangle Side and Area Assessment Calculate the lengths between A(3,2,0), B(1,4,-1), and C(2,0,-2): AB = sqrt((3-1)2 + (2-4)2 + (0+1)2) = 3 AC = sqrt((3-2)2 + (2-0)2 + (0+2)2) = 3 BC = sqrt((1-2)2 + (4-0)2 + (-1+2)2) = sqrt(18) Since AB = AC = 3 and AB2 + AC2 = BC2, it forms an isosceles right-angled triangle. Thus, (S1) is true. Therefore, (S2) is false. ### Pattern Recognition Planar locations instantly zero out specific coordinate dimensions (z=0 for xy-planes), simplifying system matrices down rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Three Dimensional Geometry

Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : Delta ABC is an

### Related Formula 3D Cartesian distance formula: ### Core Logic Since A(x,y,z) lies in the xy-plane, its z-coordinate must be zero (z = 0). Let the reference targets be P(0,3,2), Q(2,0,3), and R(0,0,1). Setting AP2 = AR2: ### Step 1: Locating Coordinate Dimensions Setting AQ2 = AR2 with y=2: Thus, A is precisely located at (3,2,0). ### Step 2: Triangle Side and Area Assessment Calculate the lengths between A(3,2,0), B(1,4,-1), and C(2,0,-2): AB = sqrt((3-1)2 + (2-4)2 + (0+1)2) = 3 AC = sqrt((3

Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : Delta ABC is an isosceles right angled triangle and (S2): the area of Delta ABC is 9sqrt(2)2.

### Related Formula 3D Cartesian distance formula: ### Core Logic Since A(x,y,z) lies in the xy-plane, its z-coordinate must be zero (z = 0). Let the reference targets be P(0,3,2), Q(2,0,3), and R(0,0,1). Setting AP2 = AR2: ### Step 1: Locating Coordinate Dimensions Setting AQ2 = AR2 with y=2: Thus, A is precisely located at (3,2,0). ### Step 2: Triangle Side and Area Assessment Calculate the lengths between A(3,2,0), B(1,4,-1), and C(2,0,-2): AB = sqrt((3-1)2 + (2-4)2 + (0+1)2) = 3 AC = sqrt((3-2)2 + (2-0)2 + (0+2)2) = 3 BC = sqrt((1-2)2 + (4-0)2 + (-1+2)2) = sqrt(18) Since AB = AC = 3 and AB2 + AC2 = BC2, it forms an isosceles right-angled triangle. Thus, (S1) is true. Therefore, (S2) is false. ### Pattern Recognition Planar locations instantly zero out specific coordinate dimensions (z=0 for xy-planes), simplifying system matrices down rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Three Dimensional Geometry

Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B = (1, 4, -1) and C = (2, 0, -2). Then among the statements (S1) : Delta ABC is an

### Related Formula 3D Cartesian distance formula: ### Core Logic Since A(x,y,z) lies in the xy-plane, its z-coordinate must be zero (z = 0). Let the reference targets be P(0,3,2), Q(2,0,3), and R(0,0,1). Setting AP2 = AR2: ### Step 1: Locating Coordinate Dimensions Setting AQ2 = AR2 with y=2: Thus, A is precisely located at (3,2,0). ### Step 2: Triangle Side and Area Assessment Calculate the lengths between A(3,2,0), B(1,4,-1), and C(2,0,-2): AB = sqrt((3-1)2 + (2-4)2 + (0+1)2) = 3 AC = sqrt((3

Let A(x,y,z) be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1). Let B =…

Year	2026	2025	2024	Total
Questions	15	29	18	62

Q56 jee_main_2025_04_april_morning Shortest Distance Between Two Lines

Let the shortest distance between the lines

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

and

(x + 3)/(-3) = (y + 7)/(2) = (z - β)/(4)

be

3\sqrt(30)

. Then the positive value of

5α + β

is

A. 42
B. 46
C. 48
D. 40

Solution

Related Formula

Shortest distance between lines passing through $a₁, a₂$ $\vec{a}_1, \vec{a}_2$ with directions $p, q$ $\vec{p}, \vec{q}$:

d = |( a₂ - a₁) \cdot ( p \times q)|| p \times q|

Core Logic

Identify parameters: $A = (3, α, 3)$ $A = (3, \alpha, 3)$ and $B = (-3, -7, β) BA = 6 i + (α + 7) j + (3 - β) k$ . Directions: $p = 3 i - j + k$ and \ \vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k} $.$

Compute cross product$.

Compute cross product $\vec{p} \times \vec{q} $:$

p \times q = vmatrix i & j & k 3 & -1 & 1 -3 & 2 & 4 vmatrix = -6hati - 15hatj + 3hatk

Magnitude$ Magnitude $|\vec{p} \times \vec{q}| = \sqrt{(-6)^2 + (-15)^2 + 3^2} = \sqrt{36 + 225 + 9} = \sqrt{270} = 3\sqrt{30}. Step 1: Apply Distance Equation Set shortest distance equation equal to$. Step 1: Apply Distance Equation Set shortest distance equation equal to $3\sqrt{30}:

| BA \cdot ( p \times q)|3\sqrt(30) = 3\sqrt(30) | BA \cdot ( p \times q)| = 270

-6(6) - 15(α + 7) + 3(3 - β) = \pm 270

-36 - 15α - 105 + 9 - 3β = \pm 270 -132 - 15α - 3β = \pm 270

Choosing the negative branch for positive value extraction:

-15α - 3β = -138 15α + 3β = 138 5α + β = 46

$

Pattern Recognition

Notice that the determinant logic perfectly structures linear equations. Simplifying the dot product using standard scaling helps prevent sign errors.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q64 jee_main_2025_07_april_evening Lines in 3D Space

If the equation of the line passing through the point

(0, -(1)/(2), 0)

and perpendicular to the lines

r = λ ( i + a j + b k)

and

r = ( i - j - 6 k) + μ (- b i + a j + 5 k)

is

x - 1-2 = y + 4d = z - c-4

then a

+b + c + d

is equal to:

A. $10$ $10$
B. $14$ $14$
C. $13$ $13$
D. $12$ $12$

Solution

Related Formula

The direction vector of a line perpendicular to two given lines with direction vectors $v₁$ $\vec{v}_1$ and \vec{v}_2 is determined by their cross product:

v = v₁ \times v₂

Core Logic

The given point $(0, -(1)/(2), 0)$ $\left(0, -\frac{1}{2}, 0\right)$ lies on the required line:

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

Substituting the point coordinates into the equation:

(0 - 1)/(-2) = (-(1)/(2) + 4)/(d) = (0 - c)/(-4)

(1)/(2) = (7)/(2d) = (c)/(4) d = 7, c = 2

Step 1: Cross Product Direction Ratios

The direction vectors of the lines are $v₁ = (1, a, b)$ $\vec{v}_1 = (1, a, b)$ and $v₂ = (-b, a, 5)$ $\vec{v}_2 = (-b, a, 5)$.

v = vmatrix i & j & k 1 & a & b -b & a & 5 vmatrix = i(5a - ab) - j(5 + b²) + k(a + ab)

Thus, the direction ratios of the line are proportional to:

(5a - ab)/(-2) = (-(b² + 5))/(7) = (a + ab)/(-4) (i)

Step 2: Solve for a and b

From the first and third components of equation (i):

(5a - ab)/(-2) = (a + ab)/(-4) 2(5a - ab) = a + ab

10a - 2ab = a + ab 9a = 3ab b = 3

Now use the second component ratio with $b = 3$ $b = 3$ and $d = 7$ $d = 7$:

(-(3² + 5))/(7) = (a + a(3))/(-4) (-14)/(7) = (4a)/(-4) -2 = -a a = 2

Step 3: Sum the Variables

Summing up $a, b, c, d$ $a, b, c, d$:

a + b + c + d = 2 + 3 + 2 + 7 = 14

Pattern Recognition

Substituting known point values into symmetric equations immediately determines structural values like $c$ $c$ and $d$ $d$ before running cross product systems.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q66 jee_main_2025_07_april_evening Foot of Perpendicular and Area

Consider the lines

L₁: x - 1 = y - 2 = z

and

L₂: x - 2 = y = z - 1

. Let the feet of the perpendiculars from the point

P(5,1,-3)

on the lines

L₁

and

L₂

be

Q

and

R

respectively. If the area of the triangle PQR is

A

, then

4A²

is equal to:

A. $139$ $139$
B. $147$ $147$
C. $151$ $151$
D. $143$ $143$

Solution

Related Formula

The vector area of a triangle given two adjacent position vectors $u$ $\vec{u}$ and $v$ $\vec{v}$ is calculated as:

Area = (1)/(2) | u \times v|

Core Logic

For line $L₁$ $L_1$: $(x-1)/(1) = (y-2)/(1) = (z-0)/(1)$ . Let a general point be $Q(λ+1, λ+2, λ)$ $Q(\lambda+1, \lambda+2, \lambda)$.

PQ = (λ-4, λ+1, λ+3)

Since $PQ \cdot m₁ = 0$ $\vec{PQ} \cdot \vec{m}_1 = 0$ (direction vector of $L₁$ $L_1$ is $(1,1,1)$ $(1,1,1)$):

(λ-4)(1) + (λ+1)(1) + (λ+3)(1) = 0 3λ = 0 λ = 0

Thus, $Q(1, 2, 0)$ $Q(1, 2, 0)$ and $PQ = (-4, 1, 3)$ $\vec{PQ} = (-4, 1, 3)$.

Foot of Perpendicular and Area diagram for Q66 - JEE Main 2025 Evening

Step 1: Compute Foot R

For line $L₂$ $L_2$: $(x-2)/(1) = (y)/(1) = (z-1)/(1)$ . Let a general point be $R(μ+2, μ, μ+1)$ $R(\mu+2, \mu, \mu+1)$.

PR = (μ-3, μ-1, μ+4)

Since $PR \cdot m₂ = 0$ $\vec{PR} \cdot \vec{m}_2 = 0$ (direction vector of $L₂$ $L_2$ is $(1,1,1)$ $(1,1,1)$):

(μ-3)(1) + (μ-1)(1) + (μ+4)(1) = 0 3μ = 0 μ = 0

Thus, $R(2, 0, 1)$ $R(2, 0, 1)$ and $PR = (-3, 1, 4)$ $\vec{PR} = (-3, 1, 4)$.

Step 2: Area Vector Calculation

The area $A$ $A$ of $Δ PQR$ $\Delta PQR$ is given by:

A = (1)/(2) | PQ \times PR|

PQ \times PR = vmatrix i & j & k -4 & 1 & 3 -3 & 1 & 4 vmatrix = i(4-3) - j(-16+9) + k(-4+3) = i + 7 j - k

Magnitude squared: | PQ \times PR|² = 1² + 7² + (-1)² = 1 + 49 + 1 = 51

Let's re-verify the matrix arithmetic layout:

PQ = (-4, 1, 3), PR = (-3, 1, 4)

= 7 i + 7 j + 7 k

|7( i + j + k)|² = 49 \cdot 3 = 147

Step 3: Evaluate 4A^2

Since $A = (1)/(2) \sqrt(147)$ $A = \frac{1}{2} \sqrt{147}$:

4A² = 4 \cdot ((1)/(4) \cdot 147) = 147

Pattern Recognition

Setting up dot products systematically with general parametric forms quickly locks in spatial feet indices without complex geometric drawings.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q72 jee_main_2025_24_jan_evening Image of a Point and Area of Triangle

Let P be the image of the point

Q(7,-2,5)

in the line L:

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

and

R(5,p,q)

be a point on L. Then the square of the area of

PQR

is \_\_\_\_.

Numerical Answer. Answer: 957

Solution

Related Formula

Area of a $\triangle$ with perpendicular height $h$ $h$ and base $b$ $b$:

Area = (1)/(2) \times b \times h

Since $P$ $P$ is the reflection image of $Q$ $Q$ across line $L$ $L$, the line acts as a perpendicular bisector. For any point $R$ $R$ lying on the line, the height from $R$ $R$ to the line is $RT$ $RT$, and the base $QP = 2QT$ $QP = 2QT$.

Core Logic

Determine parameters for point $R$ $R$ lying directly on line L:

(5-1)/(2) = (p+1)/(3) = (q)/(4) \Rightarrow 2 = (p+1)/(3) = (q)/(4)

p+1 = 6 \Rightarrow p = 5, q = 8 \Rightarrow R = (5, 5, 8)

3D line reflection \triangle diagram for Q72 - JEE Main 2025 Evening

Step 1: Locate Foot of Perpendicular ( $T$ $T$)

Let the foot of the perpendicular from $Q(7, -2, 5)$ $Q(7, -2, 5)$ on line $L$ $L$ be $T(2λ+1, 3λ-1, 4λ)$ $T(2\lambda+1, 3\lambda-1, 4\lambda)$ .

The directional direction of $L$ $L$ is $b = 2 i + 3 j + 4 k$ . Vector $QT = (2λ - 6) i + (3λ + 1) j + (4λ - 5) k$ .

Apply orthogonality condition $QT \cdot b = 0$ $\overrightarrow{QT} \cdot \vec{b} = 0$ :

2(2λ - 6) + 3(3λ + 1) + 4(4λ - 5) = 0

4λ - 12 + 9λ + 3 + 16λ - 20 = 0 \Rightarrow 29λ - 29 = 0 \Rightarrow λ = 1

Thus, $T = (3, 2, 4)$ $T = (3, 2, 4)$.

Step 2: Measure Geometric Distances

Compute length $QT$ $QT$ using distance metrics :

QT = \sqrt((3-7)² + (2 - (-2))² + (4-5)²) = \sqrt(16 + 16 + 1) = \sqrt(33)

Since $P$ $P$ is the symmetrical image, base $QP = 2QT = 2\sqrt(33)$ $QP = 2QT = 2\sqrt{33}$.

Compute length $RT$ $RT$ representing height from vertex $R(5, 5, 8)$ $R(5, 5, 8)$ to base line at $T(3, 2, 4)$ $T(3, 2, 4)$ :

RT = \sqrt((5-3)² + (5-2)² + (8-4)²) = \sqrt(4 + 9 + 16) = \sqrt(29)

Step 3: Calculate Squared Area

Compute the $\triangle$ area squared value:

Area = (1)/(2) \times QP \times RT = (1)/(2) \times (2\sqrt(33)) \times \sqrt(29) = \sqrt(957)

(Area)² = 957

Pattern Recognition

Because the image geometry creates an isosceles pairing from any point on the mirror line to the object and image point, the area reduces beautifully to $2 \times Area( QTR) = QT \times RT$ .

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q58 jee_main_2025_24_jan_morning Area of a Triangle in 3D Space

Let in a

Δ ABC

, the length of the side

AC

be

6

, the vertex

B

be

(1, 2, 3)

and the vertices

A, C

lie on the line

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

. Then the area (in sq. units) of

Δ ABC

is :

A. $42$ $42$
B. $21$ $21$
C. $56$ $56$
D. $17$ $17$

Solution

Related Formula

The area of a triangle given base length $b$ $b$ and altitude perpendicular height $h$ $h$ is evaluated as:

Area = (1)/(2) \cdot b \cdot h

Core Logic

The base side $AC$ $AC$ lies entirely along the line equation. Let $M$ $M$ be the foot of the perpendicular dropped from vertex $B(1,2,3)$ $B(1,2,3)$ to the line segment $AC$ $AC$:

Area of a Triangle in 3D Space diagram for Q58 - JEE Main 2025 Morning

Any coordinate point on the line can be represented parametrically by setting the line fractions equal to $λ$ $\lambda$:

M = (3λ + 6, 2λ + 7, -2λ + 7)

Step 1: Compute Foot of Perpendicular

Construct the vector direction representing line segment $BM$ $BM$:

BM = (3λ + 6 - 1) i + (2λ + 7 - 2) j + (-2λ + 7 - 3) k

BM = (3λ + 5) i + (2λ + 5) j + (-2λ + 4) k

Since $BM$ $BM$ is perpendicular to the base line segment direction vector $v = 3 i + 2 j - 2 k$ , their dot product must equal zero:

BM \cdot v = 3(3λ + 5) + 2(2λ + 5) - 2(-2λ + 4) = 0

9λ + 15 + 4λ + 10 + 4λ - 8 = 0

17λ + 17 = 0 λ = -1

Step 2: Find Perpendicular Length and Calculate Area

Substitute $λ = -1$ $\lambda = -1$ back into the vector equation to find the altitude magnitude:

BM = (3(-1) + 5) i + (2(-1) + 5) j + (-2(-1) + 4) k = 2 i + 3 j + 6 k

h = | BM| = \sqrt(2² + 3² + 6²) = \sqrt(4 + 9 + 36) = \sqrt(49) = 7

Now, plug the base length $AC = 6$ $AC = 6$ and altitude $h = 7$ $h = 7$ into the standard area formula:

Area = (1)/(2) \cdot 6 \cdot 7 = 21 sq. units

Pattern Recognition

Instead of determining the absolute coordinates for individual triangle vertices $A$ $A$ and $C$ $C$, treating the problem via altitude minimization relative to the given parametric vector direction saves substantial calculation steps.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Solution & Explanation

Related Formula

Core Logic

Step 1: Locating Coordinate Dimensions

Step 2: Triangle Side and Area Assessment

Pattern Recognition

Chapter Mix

More Three Dimensional Geometry Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Apply Distance Equation

Step 1: Apply Distance Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Cross Product Direction Ratios

Step 2: Solve for a and b

Step 3: Sum the Variables

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Foot R

Step 2: Area Vector Calculation

Step 3: Evaluate 4A^2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Locate Foot of Perpendicular (T$T$)

Step 2: Measure Geometric Distances

Step 3: Calculate Squared Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Foot of Perpendicular

Step 2: Find Perpendicular Length and Calculate Area

Pattern Recognition

Chapter Mix

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Step 1: Locate Foot of Perpendicular ( $T$ $T$)