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

Q jee_main_2025_03_april_evening Direction Cosines and Direction Ratios

Each of the angles

β

and

γ

that a given line makes with the positive

y

- and

z

-axes, respectively, is half of the angle that this line makes with the positive

x

-axis. Then the sum of all possible values of the angle

β

is

A. $(3π)/(4)$ $\frac{3\pi}{4}$
B. $π$ $\pi$
C. $(π)/(2)$ $\frac{\pi}{2}$
D. $(3π)/(2)$ $\frac{3\pi}{2}$

Solution

Related Formula

For any line making angles $α, β, γ$ $\alpha, \beta, \gamma$ with the coordinate axes, the direction cosines satisfy:

² α + ² β + ² γ = 1

Core Logic

Given that:

β = (α)/(2) and γ = (α)/(2)

Substituting these values into the identity:

² α + 2 ²((α)/(2)) = 1

Step 1: Solving the Trigonometric Equation

Using the half-angle identity $2 ²((α)/(2)) = 1 + α$ :

² α + 1 + α = 1

² α + α = 0

α ( α + 1) = 0

This yields two possible cases:

$α = 0 α = (π)/(2)$
$α = -1 α = π$

Step 2: Finding Values of $β$ $\beta$ and their Sum

Now we find corresponding values for $β = (α)/(2)$ $\beta = \frac{\alpha}{2}$:

If $α = (π)/(2) β₁ = (π)/(4)$
If $α = π β₂ = (π)/(2)$

Sum of all possible values:

β₁ + β₂ = (π)/(4) + (π)/(2) = (3π)/(4)

Pattern Recognition

Direction cosines are bounded between $[-1, 1]$ $[-1, 1]$. Always utilize the identities relating double angles or half angles ( $2 ²θ = 1 + 2θ$ $2\cos^2\theta = 1 + \cos 2\theta$) to simplify quadratic forms involving different multiples of the coordinate angles.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Trigonometric Functions

Q69 jee_main_2025_03_april_evening Lines in 3D Space

The distance of the point

(7, 10, 11)

from the line

(x-4)/(1) = (y-4)/(0) = (z-2)/(3)

along the line

(x-9)/(2) = (y-13)/(3) = (z-17)/(6)

is

A. $18$ $18$
B. $14$ $14$
C. $12$ $12$
D. $16$ $16$

Solution

Related Formula

Distance 'along a line' means the direction vector of the line segment joining target point $P$ $P$ to intersecting point $Q$ $Q$ must be parallel to the given direction ratios:

PQ = k \cdot d

Core Logic

Let point $P = (7, 10, 11)$ $P = (7, 10, 11)$. Any general point $Q$ $Q$ on the first line $(x-4)/(1) = (y-4)/(0) = (z-2)/(3) = λ$ is:

Q = (λ + 4, 4, 3λ + 2)

Direction ratios of $PQ$ $PQ$:

PQ = (λ + 4 - 7, 4 - 10, 3λ + 2 - 11) = (λ - 3, -6, 3λ - 9)

Step 1: Equating direction vectors

Since distance is measured parallel to the line with direction vector $(2, 3, 6)$ $(2, 3, 6)$, $PQ$ $\vec{PQ}$ must be parallel to $(2, 3, 6)$ $(2, 3, 6)$:

(λ - 3)/(2) = (-6)/(3) = (3λ - 9)/(6)

From the middle term:

(λ - 3)/(2) = -2 λ - 3 = -4 λ = -1

Let's check consistency with the third term:

(3(-1) - 9)/(6) = -2 (Consistent!)

3D Lines diagram for Q69 - JEE Main 2025 Evening Shift

Step 2: Distance calculation

For $λ = -1$ $\lambda = -1$, the intersection point $Q$ $Q$ is: $Q = (3, 4, -1)$ $Q = (3, 4, -1)$

Distance $PQ$ $PQ$:

PQ = \sqrt((7 - 3)² + (10 - 4)² + (11 - (-1))²)

PQ = \sqrt(4² + 6² + 12²) = \sqrt(16 + 36 + 144) = \sqrt(196) = 14

Pattern Recognition

When solving distance parallel to a line in 3D, always write down the parametric coordinates of the general point first. Match the ratio of direction cosines directly to avoid setting up complicated systems of coordinate planes.

Chapter Mix

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

Q52 jee_main_2025_07_april_morning Shortest Distance Between Two Lines

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\sqrt(6)

, then the sum of all possible values of

α

is

A. $(3)/(2)$ $\frac{3}{2}$
B. $-(3)/(2)$ $-\frac{3}{2}$
C. $3$ $3$
D. $-3$ $-3$

Solution

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) \cdot ( b₁ \times b₂)| b₁ \times b₂| |

Core Logic

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

S.D. = | ( i + 2 j - 2 k) \cdot n| n| | = 5\sqrt(6)

Taking dot product in numerator:

( i + 2 j - 2 k) \cdot 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 $α$ $\alpha$ 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

Q68 jee_main_2025_07_april_morning Line Intersecting Two Lines

Let the line L pass through (1, 1, 1) and intersect the lines

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

and

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

. Then, which of the following points lies on the line L?

A. $(4,22,7)$ $(4,22,7)$
B. $(5, 4, 3)$ $(5, 4, 3)$
C. $(10, -29, -50)$ $(10, -29, -50)$
D. $(7, 15, 13)$ $(7, 15, 13)$

Solution

Related Formula

General coordinates of any variable point on a 3D line given symmetric form equations:

P₁ = (2λ + 1, 3λ - 1, 4λ + 1)

P₂ = (μ + 3, 2μ + 4, μ)

Core Logic

Let line L intersect Line 1 at point $A(2λ + 1, 3λ - 1, 4λ + 1)$ and Line 2 at point $B(μ + 3, 2μ + 4, μ)$ $B(\mu + 3, 2\mu + 4, \mu)$. Since line L passes through $C(1, 1, 1)$ $C(1, 1, 1)$, the direction ratios computed from vector segment $AC$ $AC$ must be proportional to the direction ratios computed from vector segment $BC$ $BC$.

Step 1: Determine Direction Ratio Parameters

Line Intersecting Two Lines diagram for Q68 - JEE Main 2025 Morning

Direction ratios of

AC

segment:

AC = (2λ + 1 - 1, 3λ - 1 - 1, 4λ + 1 - 1) = (2λ, 3λ - 2, 4λ)

Direction ratios of $BC$ $BC$ segment:

BC = (μ + 3 - 1, 2μ + 4 - 1, μ - 1) = (μ + 2, 2μ + 3, μ - 1)

Equating directional proportionality ratios:

(μ + 2)/(2λ) = (2μ + 3)/(3λ - 2) = (μ - 1)/(4λ)

Step 2: Solve for Parameter Intersection values

From the first and third fractional groups:

(μ + 2)/(2λ) = (μ - 1)/(4λ) 2(μ + 2) = μ - 1

2μ + 4 = μ - 1 μ = -5

Substitute $μ = -5$ $\mu = -5$ back into the second parameter group linkage to evaluate the target structural direction indicators, which gives the simplified direction ratio vector for $BC$ $BC$ as:

D.R.s = (-5 + 2, 2(-5) + 3, -5 - 1) = (-3, -7, -6) \equiv (3, 7, 6)

Step 3: Construct Line Equation and Verify Choice

Equation of line L passing through $C(1, 1, 1)$ $C(1, 1, 1)$ with direction vector $(3, 7, 6)$ $(3, 7, 6)$:

(x - 1)/(3) = (y - 1)/(7) = (z - 1)/(6)

Let's check option $(7, 15, 13)$ $(7, 15, 13)$:

(7 - 1)/(3) = (6)/(3) = 2

(15 - 1)/(7) = (14)/(7) = 2

(13 - 1)/(6) = (12)/(6) = 2

Since all values match perfectly, $(7, 15, 13)$ $(7, 15, 13)$ lies on the line L.

Pattern Recognition

By comparing the first and third fractional terms containing $λ$ $\lambda$ in the denominator, you can solve for $μ$ $\mu$ independently without tracking complex cross-multiplied quadratic $λμ$ $\lambda\mu$ variations.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q51 jee_main_2025_08_april_evening Shortest Distance Between Lines

Let the values of

λ

for which the shortest distance between the lines

(x - 1)/(2) = (y - 2)/(3) = (z - 3)/(4) (x - λ)/(3) = (y - 4)/(4) = (z - 5)/(5)

is

1\sqrt(6)

be

λ₁

and

λ₂

. Then the radius of the circle passing through the points

(0, 0)

,

(λ₁, λ₂)

and

(λ₂, λ₁)

is

A. $5 \sqrt(2)3$ $\frac{5 \sqrt{2}}{3}$
B. $4$ $4$
C. $\sqrt(2)3$ $\frac{\sqrt{2}}{3}$
D. $3$ $3$

Solution

Related Formula

Shortest Distance = | AB \cdot ( p \times q)| p \times q| |

Core Logic

Identify points $A(1, 2, 3)$ $A(1, 2, 3)$ and $B(λ, 4, 5)$ $B(\lambda, 4, 5)$ on the lines with directions $p = 2 i + 3 j + 4 k$ and $q = 3 i + 4 j + 5 k$ respectively. Use the shortest distance formula to determine $λ₁$ $\lambda_1$ and $λ₂$ $\lambda_2$.

Step 1: Calculate Cross Product and Direction Vector

p \times q = vmatrix i & j & k 2 & 3 & 4 3 & 4 & 5 vmatrix = - i + 2 j - k

| p \times q| = \sqrt((-1)² + 2² + (-1)²) = \sqrt(6)

AB = (λ - 1) i + 2 j + 2 k

Step 2: Solve for Lambda

1\sqrt(6) = | ((λ - 1) i + 2 j + 2 k) \cdot (- i + 2 j - k)\sqrt(6) |

|-λ + 1 + 4 - 2| = 1 |λ - 3| = 1

λ = 4 or 2

Step 3: Radius of the Passing Circle

The circle passes through $(0,0)$ $(0,0)$, $(4,2)$ $(4,2)$ and $(2,4)$ $(2,4)$. Using the circumradius formula $R = (abc)/(4Δ)$ $R = \frac{abc}{4\Delta}$:

a = \sqrt(20), b = \sqrt(20), c = \sqrt(8)

Δ = (1)/(2) vmatrix 1 & 1 & 1 0 & 4 & 2 0 & 2 & 4 vmatrix = 6

R = \sqrt(20) \times \sqrt(20) \times \sqrt(8)4 \times 6 = 40\sqrt(2)24 = 5\sqrt(2)3

Pattern Recognition

Shortest distance values create symmetric configurations. When finding a circle passing through $(0,0)$ $(0,0)$, $(x,y)$ $(x,y)$, and $(y,x)$ $(y,x)$, the symmetry about $y=x$ $y=x$ simplifies radius calculations immediately.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Circles

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 5

Solution

Related Formula

Core Logic

Step 1: Solving the Trigonometric Equation

Step 2: Finding Values of β$\beta$ and their Sum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Equating direction vectors

Step 2: Distance calculation

Pattern Recognition

Chapter Mix

Solution

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

Solution

Related Formula

Core Logic

Step 1: Determine Direction Ratio Parameters

Step 2: Solve for Parameter Intersection values

Step 3: Construct Line Equation and Verify Choice

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Cross Product and Direction Vector

Step 2: Solve for Lambda

Step 3: Radius of the Passing Circle

Pattern Recognition

Chapter Mix

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Step 2: Finding Values of $β$ $\beta$ and their Sum