A line with direction ratios 2, 1, 2 meets the lines x=y+2=z and x+2=2y=2z respectively at the point P and Q. if the…

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Q jee_main_2025_03_april_morning Intersection of Lines

Let a line passing through the point

(4,1,0)

intersect the line

L₁:(x-1)/(2)=(y-2)/(3)=(z-3)/(4)

at the point

A(α, β, γ)

and the line

L₂:x-6=y=-z+4

at the point

B(a, b, c)

. Then the value of the determinant

vmatrix 1 & 0 & 1 α & β & γ a & b & c vmatrix

is equal to:

A. 8
B. 16
C. 12
D. 4

Solution

Related Formula

For three points $P$ $P$, $A$ $A$, and $B$ $B$ to be collinear, their direction vectors must be proportional:

PA ∥ PB (xA - xP)/(xB - xP) = (yA - yP)/(yB - yP) = (zA - zP)/(zB - zP)

Intersection of Lines diagram for Q52 - JEE Main 2025 Morning

Core Logic

Express general coordinates for $A$ $A$ on $L₁$ $L_1$ and $B$ $B$ on $L₂$ $L_2$:

L₁: (x-1)/(2) = (y-2)/(3) = (z-3)/(4) = p A(2p+1, 3p+2, 4p+3)

L₂: (x-6)/(1) = (y)/(1) = (z-4)/(-1) = q B(q+6, q, 4-q)

Direction ratios (D.R.) from $P(4, 1, 0)$ $P(4, 1, 0)$:

D.R. of PA = (2p-3, 3p+1, 4p+3)

D.R. of PB = (q+2, q-1, 4-q)

Since $P, A, B$ $P, A, B$ lie on the same line:

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

Step 1: Solving the System of Equations

Equating the first two ratios:

2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 pq + 8p + 4q - 1 = 0 --- (1)

Equating the second and third ratios:

12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 7pq - 16p + 4q - 7 = 0 --- (2)

Subtracting (1) from (2) yields:

6pq - 24p - 6 = 0 pq = 4p + 1

Substituting $pq = 4p + 1$ $pq = 4p + 1$ into (1) gives:

12p + 4q = 0 q = -3p

Solving simultaneously yields:

p = -1, q = 3

Substituting the parameters back yields the points:

A(-1, -1, -1), B(9, 3, 1)

Step 2: Evaluating the Determinant

Substitute coordinates of $A$ $A$ and $B$ $B$ into the determinant:

vmatrix 1 & 0 & 1 -1 & -1 & -1 9 & 3 & 1 vmatrix

Applying the column operation $C₃ arrow C₃ - C₁$ $C_3 \rightarrow C_3 - C_1$:

vmatrix 1 & 0 & 0 -1 & -1 & 0 9 & 3 & -8 vmatrix = 1((-1)(-8) - 0) = 8

Pattern Recognition

Shortcut: For collinearity across two skew lines with a known external point, express coordinates parametrically and equate direction ratios. Solving the linear relation between parameters rapidly leads to coordinates of $A$ $A$ and $B$ $B$.

Evaluation Rubric / Model Answer

8

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Determinants

Q70 jee_main_2025_03_april_morning Shortest Distance between Skew Lines

Line

L₁

passes through the point

(1, 2, 3)

and is parallel to z-axis. Line

L₂

passes through the point

(λ, 5, 6)

and is parallel to y-axis. Let for

λ = λ₁, λ₂

,

λ₂ < λ₁

the shortest distance between the two lines be 3. Then the square of the distance of the point

(λ₁, λ₂, 7)

from the line

L₁

is:

A. 40
B. 32
C. 25
D. 37

Solution

Related Formula

Shortest distance between perpendicular axes vectors:

S.D. = |( a₂ - a₁) \cdot ( b₁ \times b₂)|| b₁ \times b₂|

Core Logic

Represent equations of straight lines symmetrically using vector directions:

L₁: (x-1)/(0) = (y-2)/(0) = (z-3)/(1)

L₂: (x-λ)/(0) = (y-5)/(1) = (z-6)/(0)

Evaluating the standard shortest distance configuration formula:

S.D. = |λ - 1| = 3 λ - 1 = \pm 3

λ = 4 or λ = -2

Given the condition $λ₂ < λ₁$ $\lambda_2 < \lambda_1$:

λ₁ = 4, λ₂ = -2

Step 1: Distance calculation from line

We need to find the square of distance from point $P(4, -2, 7)$ $P(4, -2, 7)$ to line $L₁$ $L_1$. Any general matching point coordinates tracking along path $L₁$ $L_1$ look like $Q(1, 2, t+3)$ $Q(1, 2, t+3)$.

Form a perpendicular projection vector condition:

PQ = (-3, 4, t-4)

Since $PQ \cdot k = 0 t-4 = 0 t=4$ .

Thus, the foot of perpendicular is $Q(1, 2, 7)$ $Q(1, 2, 7)$.

Evaluate the squared distance component magnitude:

PQ² = (4-1)² + (-2-2)² + (7-7)² = 3² + (-4)² + 0 = 9 + 16 = 25

Pattern Recognition

For lines parallel directly to independent Cartesian coordinate grid lines, the shortest paths are simply direct plane projections.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Q56 jee_main_2025_04_april_evening Shortest Distance Between Two Lines

Let the values of

p

, for which the shortest distance between the lines

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

and

r = (p i + 2 j + hatk) + λ (2hati + 3hatj + 4hatk)

is

1sqrt6

, be a, b, (

a < b

). Then the length of the latus rectum of the ellipse

(x²)/(a²) + (y²)/(b²) = 1

is:

A. $9$ $9$
B. $(3)/(2)$ $\frac{3}{2}$
C. $(2)/(3)$ $\frac{2}{3}$
D. $18$ $18$

Solution

Related Formula

The shortest distance between two lines $r = a₁ + λ b₁$ and $r = a₂ + μ b₂$ $\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by:

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

Core Logic

From the given line equations: Line 1 passes through $a₁ = - i + 0 j + 0 k$ along vector $b₁ = 3 i + 4 j + 5 k$ . Line 2 passes through $a₂ = p i + 2 j + k$ along vector $b₂ = 2 i + 3 j + 4 k$ .

Vector difference:

a₂ - a₁ = (p + 1) i + 2 j + k

Computing the cross product $b₁ \times b₂$ $\vec{b}_1 \times \vec{b}_2$:

b₁ \times b₂ = vmatrix i & j & k 3 & 4 & 5 2 & 3 & 4 vmatrix = i(16-15) - j(12-10) + k(9-8) = i - 2 j + k

Magnitude $| b₁ \times b₂| = \sqrt(1² + (-2)² + 1²) = \sqrt(6)$ .

Step 1: Applying the Shortest Distance Value

Substitute these into the distance equation:

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

|(p + 1)(1) + 2(-2) + 1(1)| = 1

|p + 1 - 4 + 1| = 1 |p - 2| = 1

This yields two values for $p$ $p$:

$p - 2 = 1 p = 3$ $p - 2 = 1 \implies p = 3$
$p - 2 = -1 p = 1$ $p - 2 = -1 \implies p = 1$

Given that $a, b$ $a, b$ are the parameters with $a < b$ $a < b$, we assign $a = 1$ $a = 1$ and $b = 3$ $b = 3$.

Step 2: Computing Latus Rectum of the Ellipse

The ellipse equation is:

(x²)/(1²) + (y²)/(3²) = 1

Since $b > a$ $b > a$, the formula for the length of the latus rectum is:

Latus Rectum = (2a²)/(b) = (2(1)²)/(3) = (2)/(3)

Pattern Recognition

Be careful with coordinate geometry variables; when an ellipse satisfies $b > a$ $b > a$, the major axis is along the y-axis, making the latus rectum equal to $(2a²)/(b)$ $\frac{2a^2}{b}$ instead of $(2b²)/(a)$ $\frac{2b^2}{a}$.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Conic Sections

Q69 jee_main_2025_04_april_evening Angle Between Two Lines

Let A be the point of intersection of the lines

L _ 1: (x - 7)/(1) = (y - 5)/(0) = (z - 3)/(- 1)

and

L₂: x - 13 = y + 34 = z + 75

. Let B and C be the point on the lines

L₁

and

L₂

respectively such that

AB = AC = \sqrt(15)

. Then the square of the area of the triangle ABC is :

A. $54$ $54$
B. $63$ $63$
C. $57$ $57$
D. $60$ $60$

Solution

Core Logic

First, find the point of intersection $A$ $A$ by solving the lines. Any point on $L₁$ $L_1$ can be written as $(λ + 7, 5, -λ + 3)$ $(\lambda + 7, 5, -\lambda + 3)$. Substituting this point into the equation for $L₂$ $L_2$:

((λ + 7) - 1)/(3) = (5 + 3)/(4) (λ + 6)/(3) = 2 λ = 0

Thus, the intersection point is $A = (7, 5, 3)$ $A = (7, 5, 3)$.

Step 1: Calculating the Angle between lines

The directional vectors of lines $L₁$ $L_1$ and $L₂$ $L_2$ are $u = i - k$ $\vec{u} = \hat{i} - \hat{k}$ and $v = 3hati + 4hatj + 5hatk$ $\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively.

θ = | u \cdot v|| u|| v| = |1(3) + 0(4) - 1(5)|\sqrt(1²+(-1)²) \sqrt(3²+4²+5²) = |3 - 5|\sqrt(2)\sqrt(50) = (2)/(10) = (1)/(5)

Now, find $θ$ $\sin\theta$:

θ = \sqrt(1 - ²θ) = \sqrt(1 - (1)/(25)) = \sqrt(24)5

Three dimensional geometry diagram for Q69 - JEE Main 2025 Evening

Step 2: Finding Area of the Triangle

The area of $ABC$ $\triangle ABC$ given two sides and their included angle is:

Area = (1)/(2) \cdot AB \cdot AC \cdot θ

Given $AB = AC = \sqrt(15)$ $AB = AC = \sqrt{15}$:

Area = (1)/(2) \cdot \sqrt(15) \cdot \sqrt(15) \cdot \sqrt(24)5 = 15\sqrt(24)10 = 3\sqrt(24)2

Squaring the area:

Area² = ( 3\sqrt(24)2)² = (9 \times 24)/(4) = 9 \times 6 = 54

Pattern Recognition

Since $B$ $B$ and $C$ $C$ lie on lines intersecting at $A$ $A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Properties of Triangles

Q jee_main_2025_04_april_morning Line and Point Relations

Let A and B be two distinct points on the line

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

. Both A and B are at a distance

2\sqrt(17)

from the foot of perpendicular drawn from the point

(1,2,3)

on the line L. If O is the origin, then

OA\cdot OB

is equal to:

A. 49
B. 47
C. 21
D. 62

Solution

Related Formula

Dot product of two vectors:

OA \cdot OB = xA xB + yA yB + zA zB

Condition for perpendicularity of two vectors:

u \cdot v = 0

Core Logic

The symmetric equation of line $L$ $L$ is:

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

Any general point $Q$ $Q$ on line $L$ $L$ can be expressed as:

Q(3λ + 6, 2λ + 7, -2λ + 7)

The vector from $P(1, 2, 3)$ $P(1, 2, 3)$ to $Q$ $Q$ is:

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

Since $Q$ $Q$ is the foot of the perpendicular dropped from $P$ $P$ onto $L$ $L$, the vector $PQ$ $\overrightarrow{PQ}$ is orthogonal to the line's direction vector $b = 3 i + 2 j - 2 k$ :

PQ \cdot b = 0

3(3λ + 5) + 2(2λ + 5) - 2(-2λ + 4) = 0

9λ + 15 + 4λ + 10 + 4λ - 8 = 0 17λ = -17 λ = -1

Substituting $λ = -1$ $\lambda = -1$ gives the foot of perpendicular: $Q(3, 5, 9)$ $Q(3, 5, 9)$

Step 1: Locate Points A and B

Points $A$ $A$ and $B$ $B$ lie on line $L$ $L$ at a distance $d = 2\sqrt(17)$ $d = 2\sqrt{17}$ on either side of $Q$ $Q$.

Let the coordinates of points on $L$ $L$ relative to $Q$ $Q$ be $(3μ + 3, 2μ + 5, -2μ + 9)$ $(3\mu + 3,\, 2\mu + 5,\, -2\mu + 9)$. The distance squared from $Q$ $Q$ is:

(3μ)² + (2μ)² + (-2μ)² = (2\sqrt(17))²

17μ² = 68 μ² = 4 μ = \pm 2

For $μ = 2$ $\mu = 2$:

A = (3(2) + 3, 2(2) + 5, -2(2) + 9) = (9, 9, 5)

For $μ = -2$ $\mu = -2$:

B = (3(-2) + 3, 2(-2) + 5, -2(-2) + 9) = (-3, 1, 13)

Line and Point Relations diagram for Q59 - JEE Main 2025 Morning

Step 2: Vector Dot Product Evaluation

Position vectors of $A$ $A$ and $B$ $B$ from origin $O(0, 0, 0)$ $O(0, 0, 0)$:

OA = 9 i + 9 j + 5 k

OB = -3 i + j + 13 k

Compute their scalar dot product:

OA \cdot OB = (9)(-3) + (9)(1) + (5)(13) = -27 + 9 + 65 = 47

Pattern Recognition

Points lying symmetrically at equal distances along a 3D line from a known central point can be found directly using parametric displacement along the unit direction vector ( $r = rQ \pm d b$ $\vec{r} = \vec{r}_Q \pm d\,\hat{b}$).

Evaluation Rubric / Model Answer

Option B: 47

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

A line with direction ratios 2, 1, 2 meets the lines x=y+2=z$x=y+2=z$ and x+2=2y=2z$x+2=2y=2z$ respectively at the point P and Q. if the length of the perpendicular from the point (1, 2, 12)$(1, 2, 12)$ to the line PQ is l$l$, then l²$l^2$ is

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve for Line PQ

Step 2: Find Perpendicular Foot F

Step 3: Compute Final Distance Squared

Pattern Recognition

Chapter Mix

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Solving the System of Equations

Step 2: Evaluating the Determinant

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Distance calculation from line

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Applying the Shortest Distance Value

Step 2: Computing Latus Rectum of the Ellipse

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Calculating the Angle between lines

Step 2: Finding Area of the Triangle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Locate Points A and B

Step 2: Vector Dot Product Evaluation

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

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A line with direction ratios 2, 1, 2 meets the lines $x=y+2=z$ $x=y+2=z$ and $x+2=2y=2z$ $x+2=2y=2z$ respectively at the point P and Q. if the length of the perpendicular from the point $(1, 2, 12)$ $(1, 2, 12)$ to the line PQ is $l$ $l$, then $l²$ $l^2$ is