Let the line L pass through (1, 1, 1) and intersect the lines (x - 1/2) = (y + 1/3) = (z - 1/4) and (x

Q56 jee_main_2025_04_april_morning Shortest Distance Between Two Lines

Let the shortest distance between the lines

fracx - 33 = fracy - alpha-1 = fracz - 31

and

fracx + 3-3 = fracy + 72 = fracz - beta4

be

3sqrt30

. Then the positive value of

5alpha + beta

is

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

Solution

### Related Formula Shortest distance between lines passing through

veca_1, veca_2

with directions

vecp, vecq

:

d = frac|(veca_2 - veca_1) cdot (vecp times vecq)||vecp times vecq|

### Core Logic Identify parameters:

A = (3, alpha, 3)

and

B = (-3, -7, beta) implies overrightarrowBA = 6hati + (alpha + 7)hatj + (3 - beta)hatk

. Directions:

vecp = 3hati - hatj + hatk

and \ \vec{q} = -3\hat{i} + 2\hat{j} + 4\hat{k}

. Compute cross product

\vec{p} \times \vec{q}

:

vecp times vecq = beginvmatrix hati & hatj & hatk \\ 3 & -1 & 1 \\ -3 & 2 & 4 endvmatrix = -6hati - 15hatj + 3hatk

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

3\sqrt{30}

:

frac|overrightarrowBA cdot (vecp times vecq)|3sqrt30 = 3sqrt30 implies |overrightarrowBA cdot (vecp times vecq)| = 270

-6(6) - 15(alpha + 7) + 3(3 - beta) = pm 270

-36 - 15alpha - 105 + 9 - 3beta = pm 270 implies -132 - 15alpha - 3beta = pm 270

Choosing the negative branch for positive value extraction:

-15alpha - 3beta = -138 implies 15alpha + 3beta = 138 implies 5alpha + beta = 46

$ ### Pattern Recognition Notice that the determinant logic perfectly structures linear equations. Simplifying the dot product using standard scaling helps prevent sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Q59 jee_main_2025_04_april_morning Line and Point Relations

Let A and B be two distinct points on the line

L: fracx - 63 = fracy - 72 = fracz - 7-2

. Both A and B are at a distance

2sqrt17

from the foot of perpendicular drawn from the point

(1,2,3)

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

overrightarrowOAcdot overrightarrowOB

is equal to:

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

Solution

### Related Formula Dot product of vectors:

overrightarrowOA cdot overrightarrowOB = x_A x_B + y_A y_B + z_A z_B

### Core Logic Let any general point

Q

on line

L

be

(3lambda + 6, 2lambda + 7, -2lambda + 7)

. Vector from

P(1,2,3)

to

Q

is

overrightarrowPQ = (3lambda + 5)hati + (2lambda + 5)hatj + (-2lambda + 4)hatk

. Since

Q

is the foot of perpendicular,

overrightarrowPQ cdot vecb = 0

, where

vecb = 3hati + 2hatj - 2hatk

:

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

9lambda + 15 + 4lambda + 10 + 4lambda - 8 = 0 implies 17lambda = -17 implies lambda = -1

Thus, the foot of perpendicular is

Q(3, 5, 9)

. ### Step 1: Locate Points A and B Points

A

and

B

are on line

L

, expressed via a distance parameter

mu

away from

Q

: General point expression from

Q

:

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

[relative tracking points]. Distance squared

= 68

:

(3mu)^2 + (2mu)^2 + (-2mu)^2 = 68 implies 17mu^2 = 68 implies mu = pm 2

For

mu = 2

:

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

For

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

$overrightarrowOA cdot overrightarrowOB = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47

### Pattern Recognition When symmetrical points on a line are equidistant from a central foot position, utilizing standard parametric tracking simplifies vector resolution instantly. Double check calculations sequentially via parallel distance components. ### Evaluation Rubric / Model Answer null ### 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

left(0, -frac12, 0right)

and perpendicular to the lines

vec mathrm r = lambda (hat mathrm i + mathrm a hat mathrm j + mathrm b hat mathrm k)

and

vec mathrm r = left(hat mathrm i - hat mathrm j - 6 hat mathrm kright) + mu left(- b hat mathrm i + a hat mathrm j + 5 hat mathrm kright)

is

fracmathrmx - 1-2 = fracmathrmy + 4mathrmd = fracmathrmz - mathrmc-4

then a

+mathrmb + mathrmc + mathrmd

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

vecv_1

and \vec{v}_2 is determined by their cross product:

vecv = vecv_1 times vecv_2

### Core Logic The given point

left(0, -frac12, 0right)

lies on the required line:

fracx - 1-2 = fracy + 4d = fracz - c-4

Substituting the point coordinates into the equation:

frac0 - 1-2 = frac-frac12 + 4d = frac0 - c-4

frac12 = frac72d = fracc4 implies d = 7, quad c = 2

### Step 1: Cross Product Direction Ratios The direction vectors of the lines are

vecv_1 = (1, a, b)

and

vecv_2 = (-b, a, 5)

.

vecv = beginvmatrix hati & hatj & hatk \\ 1 & a & b \\ -b & a & 5 endvmatrix = hati(5a - ab) - hatj(5 + b^2) + hatk(a + ab)

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

frac5a - ab-2 = frac-(b^2 + 5)7 = fraca + ab-4 quad dots text(i)

### Step 2: Solve for a and b From the first and third components of equation (i):

frac5a - ab-2 = fraca + ab-4 implies 2(5a - ab) = a + ab

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

Now use the second component ratio with

b = 3

and

d = 7

:

frac-(3^2 + 5)7 = fraca + a(3)-4 implies frac-147 = frac4a-4 implies -2 = -a implies a = 2

### Step 3: Sum the Variables Summing up

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

and

d

before running cross product systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Q66 jee_main_2025_07_april_evening Foot of Perpendicular and Area

Consider the lines

mathrmL_1: mathrmx - 1 = mathrmy - 2 = mathrmz

and

mathrmL_2: mathrmx - 2 = mathrmy = mathrmz - 1

. Let the feet of the perpendiculars from the point

mathrmP(5,1,-3)

on the lines

mathrmL_1

and

mathrmL_2

be

mathrmQ

and

mathrmR

respectively. If the area of the triangle PQR is

mathrmA

, then

4mathrmA^2

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

vecu

and

vecv

is calculated as:

textArea = frac12 |vecu times vecv|

### Core Logic For line

L_1

:

fracx-11 = fracy-21 = fracz-01

. Let a general point be

Q(lambda+1, lambda+2, lambda)

.

vecPQ = (lambda-4, lambda+1, lambda+3)

Since

vecPQ cdot vecm_1 = 0

(direction vector of

L_1

is

(1,1,1)

):

(lambda-4)(1) + (lambda+1)(1) + (lambda+3)(1) = 0 implies 3lambda = 0 implies lambda = 0

Thus,

Q(1, 2, 0)

and

vecPQ = (-4, 1, 3)

.

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

### Step 1: Compute Foot R For line

L_2

:

fracx-21 = fracy1 = fracz-11

. Let a general point be

R(mu+2, mu, mu+1)

.

vecPR = (mu-3, mu-1, mu+4)

Since

vecPR cdot vecm_2 = 0

(direction vector of

L_2

is

(1,1,1)

):

(mu-3)(1) + (mu-1)(1) + (mu+4)(1) = 0 implies 3mu = 0 implies mu = 0

Thus,

R(2, 0, 1)

and

vecPR = (-3, 1, 4)

. ### Step 2: Area Vector Calculation The area

A

of

Delta PQR

is given by:

A = frac12 |vecPQ times vecPR|

vecPQ times vecPR = beginvmatrix hati & hatj & hatk \\ -4 & 1 & 3 \\ -3 & 1 & 4 endvmatrix = hati(4-3) - hatj(-16+9) + hatk(-4+3) = hati + 7hatj - hatk

textMagnitude squared: |vecPQ times vecPR|^2 = 1^2 + 7^2 + (-1)^2 = 1 + 49 + 1 = 51

Let's re-verify the matrix arithmetic layout:

vecPQ = (-4, 1, 3), vecPR = (-3, 1, 4)

= 7hati + 7hatj + 7hatk

|7(hati + hatj + hatk)|^2 = 49 cdot 3 = 147

### Step 3: Evaluate 4A^2 Since

A = frac12 sqrt147

:

4A^2 = 4 cdot left(frac14 cdot 147right) = 147

### Pattern Recognition Setting up dot products systematically with general parametric forms quickly locks in spatial feet indices without complex geometric drawings. ### Evaluation Rubric / Model Answer null ### 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:

fracx-12=fracy+13=fracz4

and

R(5,p,q)

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

triangle PQR

is \_\_\_\_. [cite: 3405, 3406]

Numerical Answer. Answer: 957

Solution

### Related Formula 1. Area of a \triangle with perpendicular height

h

and base

b

:

textArea = frac12 times b times h

2. Since

P

is the reflection image of

Q

across line

L

, the line acts as a perpendicular bisector. For any point

R

lying on the line, the height from

R

to the line is

RT

, and the base

QP = 2QT

. ### Core Logic Determine parameters for point

R

lying directly on line L [cite: 3405, 4071]:

frac5-12 = fracp+13 = fracq4 Rightarrow 2 = fracp+13 = fracq4

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

[cite: 4071, 4073]

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

### Step 1: Locate Foot of Perpendicular (

T

) Let the foot of the perpendicular from

Q(7, -2, 5)

on line

L

be

T(2lambda+1, 3lambda-1, 4lambda)

. The directional direction of

L

is

vecb = 2hati + 3hatj + 4hatk

. Vector

overrightarrowQT = (2lambda - 6)hati + (3lambda + 1)hatj + (4lambda - 5)hatk

. Apply orthogonality condition

overrightarrowQT cdot vecb = 0

:

2(2lambda - 6) + 3(3lambda + 1) + 4(4lambda - 5) = 0

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

[cite: 4077, 4078] Thus,

T = (3, 2, 4)

. ### Step 2: Measure Geometric Distances Compute length

QT

using distance metrics :

QT = sqrt(3-7)^2 + (2 - (-2))^2 + (4-5)^2 = sqrt16 + 16 + 1 = sqrt33

Since

P

is the symmetrical image, base

QP = 2QT = 2sqrt33

. Compute length

RT

representing height from vertex

R(5, 5, 8)

to base line at

T(3, 2, 4)

:

RT = sqrt(5-3)^2 + (5-2)^2 + (8-4)^2 = sqrt4 + 9 + 16 = sqrt29

### Step 3: Calculate Squared Area Compute the \triangle area squared value [cite: 3406, 4081]:

textArea = frac12 times QP times RT = frac12 times left(2sqrt33right) times sqrt29 = sqrt957

left(textArearight)^2 = 957

[cite: 4081, 4083] ### 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 textArea(triangle QTR) = QT times RT

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Let the line L pass through (1, 1, 1) and intersect the lines $fracx - 12 = fracy + 13 = fracz - 14$ and $fracx - 31 = fracy - 42 = fracz1$ . Then, which of the following points lies on the line L?

Solution & Explanation

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 5

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Let the line L pass through (1, 1, 1) and intersect the lines fracx - 12 = fracy + 13 = fracz - 14$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and fracx - 31 = fracy - 42 = fracz1$\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then, which of the following points lies on the line L?

Solution & Explanation

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 5

Solution

Solution

Solution

Solution

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Let the line L pass through (1, 1, 1) and intersect the lines $fracx - 12 = fracy + 13 = fracz - 14$ and $fracx - 31 = fracy - 42 = fracz1$ . Then, which of the following points lies on the line L?