If the image of the point (4, 4, 3) in the line (x - 1/2) = (y - 2/1) = (z - 1/3) is (alpha, beta, gamma), then alpha +…

Q51 jee_main_2025_08_april_evening Shortest Distance Between Lines

Let the values of

lambda

for which the shortest distance between the lines

fracx - 12 = fracy - 23 = fracz - 34quadtextandquadfracx - lambda3 = fracy - 44 = fracz - 55

is

frac1sqrt6

be

lambda_1

and

lambda_2

. Then the radius of the circle passing through the points

(0, 0)

,

(lambda_1, lambda_2)

and

(lambda_2, lambda_1)

is

A. $frac5 sqrt23$ $\frac{5 \sqrt{2}}{3}$
B. $4$ $4$
C. $fracsqrt23$ $\frac{\sqrt{2}}{3}$
D. $3$ $3$

Solution

### Related Formula

textShortest Distance = left| fracvecAB cdot (vecp times vecq)|vecp times vecq| right|

### Core Logic Identify points

A(1, 2, 3)

and

B(lambda, 4, 5)

on the lines with directions

vecp = 2hati + 3hatj + 4hatk

and

vecq = 3hati + 4hatj + 5hatk

respectively. Use the shortest distance formula to determine

lambda_1

and

lambda_2

. ### Step 1: Calculate Cross Product and Direction Vector

vecp times vecq = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 3 & 4 & 5 endvmatrix = -hati + 2hatj - hatk

|vecp times vecq| = sqrt(-1)^2 + 2^2 + (-1)^2 = sqrt6

vecAB = (lambda - 1)hati + 2hatj + 2hatk

### Step 2: Solve for Lambda

frac1sqrt6 = left| frac((lambda - 1)hati + 2hatj + 2hatk) cdot (-hati + 2hatj - hatk)sqrt6 right|

implies |-lambda + 1 + 4 - 2| = 1 implies |lambda - 3| = 1

implies lambda = 4 text or 2

### Step 3: Radius of the Passing Circle The circle passes through

(0,0)

,

(4,2)

and

(2,4)

. Using the circumradius formula

R = fracabc4Delta

:

a = sqrt20, quad b = sqrt20, quad c = sqrt8

Delta = frac12 beginvmatrix 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 endvmatrix = 6

R = fracsqrt20 times sqrt20 times sqrt84 times 6 = frac40sqrt224 = frac5sqrt23

### Pattern Recognition Shortest distance values create symmetric configurations. When finding a circle passing through

(0,0)

,

(x,y)

, and

(y,x)

, the symmetry about

y=x

simplifies radius calculations immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Circles

Q73 jee_main_2025_08_april_evening Area of Triangle Formed by Intersecting Lines

Let the area of the triangle formed by the lines

x + 2 = y - 1 = z, fracx - 35 = fracy-1 = fracz - 11

and

fracmathrmx-3 = fracmathrmy - 33 = fracmathrmz - 21

be A. Then

A^2

is equal to

Numerical Answer. Answer: 56 to 56

Solution

### Related Formula

textArea A = frac12 |vecAB times vecAC|

### Core Logic Determine the three intersection vertex positions for the matching coordinate line segments, then calculate vector cross expansions to determine face boundaries. ### Step 1: Locate Intersection Vertices Solving line pairs intersection matrices: *

L_1 cap L_2 implies A(-2, 1, 0)

*

L_2 cap L_3 implies B(3, 0, 1)

*

L_3 cap L_1 implies C(0, 3, 2)

### Step 2: Construct Vectors Cross Matrix Using vertex values to form component arrays:

vecAB = -5hati + hatj - hatk, quad vecAC = -3hati + 3hatj + hatk

vecAB times vecAC = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -1 \\ -3 & 3 & 1 endvmatrix = 4hati + 8hatj - 12hatk

### Step 3: Final Area Squared Derivation

A = frac12sqrt16 + 64 + 144 = frac12sqrt224 = sqrt56

A^2 = 56

{{SOL_IMG_73}} ### Pattern Recognition Finding the area of a triangle formed by intersecting lines involves grouping directional cross vectors once coordinates are solved. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Q56 jee_main_2025_29_jan_evening Line and Plane Intersections

Let a straight line

L

pass through the point

P(2, -1, 3)

and be perpendicular to the lines

fracx - 12 = fracy + 11 = fracz - 3-2

and \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. If the line

L

intersects the

yz

-plane at the point

Q

, then the distance between the points

P

and

Q

is:

A. $2$ $2$
B. $sqrt10$ $\sqrt{10}$
C. $3$ $3$
D. $2sqrt3$ $2\sqrt{3}$

Solution

### Related Formula The direction vector of a line perpendicular to two vectors

vecu

and

vecv

is obtained via the cross product:

vecn = vecu times vecv

### Core Logic Extract direction vectors of the given lines:

vecu = 2hati + hatj - 2hatk

vecv = hati + 3hatj + 4hatk

Compute the cross product:

vecn = beginvmatrix hati & hatj & hatk \\ 2 & 1 & -2 \\ 1 & 3 & 4 endvmatrix

= hati(4 - (-6)) - hatj(8 - (-2)) + hatk(6 - 1) = 10hati - 10hatj + 5hatk

= 5(2hati - 2hatj + hatk)

### Step 1: Write Line Equation and Intersect with Plane Equation of line

L

through

P(2, -1, 3)

with direction

(2, -2, 1)

:

fracx - 22 = fracy + 1-2 = fracz - 31 = lambda

Any random point on this line is

Q(2lambda + 2, -2lambda - 1, lambda + 3)

. For intersection with the

yz

-plane, set

x = 0

:

2lambda + 2 = 0 implies lambda = -1

### Step 2: Find Distance Substituting

lambda = -1

into the coordinate matrix of

Q

gives:

Q(0, 1, 2)

Calculate distance

d(P, Q)

:

d = sqrt(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2 = sqrt4 + 4 + 1 = 3

### Pattern Recognition Perpendicularity to two lines always indicates using the cross-product to lock down the direction ratios. Intersection with the

yz

-plane simply forces

x = 0

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

Q63 jee_main_2025_29_jan_evening Shortest Distance and Intersection of Lines

Let

mathbfP

be the foot of the perpendicular from the point

(1,2,2)

on the line

mathrmL:fracmathrmx - 11 = fracmathrmy + 1-1 = fracmathrmz - 22

. Let the line

vecmathrmr = (-hatmathrmi +hatmathrmj -2hatmathrmk) + lambda (hatmathrmi -hatmathrmj +hatmathrmk)

,

lambda in mathbbR

, intersect the line

mathrmL

at Q. Then

2(mathrmPQ)^2

is equal to:

A. $27$ $27$
B. $25$ $25$
C. $29$ $29$
D. $19$ $19$

Solution

### Related Formula Dot product of vector projection matching orthogonal axes equals zero:

vecAP cdot vecd = 0

### Core Logic Let the target source coordinates tracking point match

A(1, 2, 2)

. General parameter points on line

L

are defined by parameter

mu

:

P(mu + 1, -mu - 1, 2mu + 2)

Shortest Distance and Intersection of Lines diagram for Q63 - JEE Main 2025 Evening

vecAP = muhati - (mu + 3)hatj + 2muhatk

Line direction vector

vecd = hati - hatj + 2hatk

. ### Step 1: Isolate Foot and Intersection Positions

(mu)cdot 1 - (-mu - 3)cdot 1 + (2mu)cdot 2 = 0 implies 6mu + 3 = 0 implies mu = -frac12

Substituting back yields coordinate positions for foot

P

:

Pleft(frac12, -frac12, 1right)

Equating general vectors between standard linear constraints tracks intersection point

Q

at

mu = -2

:

Q(-1, 1, -2)

### Step 2: Distance Formulation Compute length of line segment squared:

PQ^2 = left(frac12 - (-1)right)^2 + left(-frac12 - 1right)^2 + (1 - (-2))^2

= frac94 + frac94 + 9 = frac544

2(PQ)^2 = 2 left(frac544right) = 27

### Pattern Recognition Always separate foot evaluations from line-intersection parameter updates to ensure you do not mix up variables tracking linear metrics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

Q56 jee_main_2025_28_jan_morning Distance Formula and Properties of Triangles

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

frac9sqrt22

. (1) both are true (2) only (S1) is true (3) only (S2) is true (4) both are false

A. both are true
B. only (S1) is true
C. only (S2) is true
D. both are false

Solution

### Related Formula 3D Cartesian distance formula:

d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2

### 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

AP^2 = AR^2

:

x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 implies y = 2

### Step 1: Locating Coordinate Dimensions Setting

AQ^2 = AR^2

with

y=2

:

(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 implies x = 3

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 = sqrt18

Since

AB = AC = 3

and

AB^2 + AC^2 = BC^2

, it forms an isosceles right-angled triangle. Thus, (S1) is true.

textArea = frac12 times 3 times 3 = frac92

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

If the image of the point $(4, 4, 3)$ $(4, 4, 3)$ in the line $fracx - 12 = fracy - 21 = fracz - 13$ is $(alpha, beta, gamma)$ $(\alpha, \beta, \gamma)$, then $alpha + beta + gamma$ $\alpha + \beta + \gamma$ is equal to (1) 9 (2) 12 (3) 8 (4) 7

Solution & Explanation

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 3

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If the image of the point (4, 4, 3)$(4, 4, 3)$ in the line fracx - 12 = fracy - 21 = fracz - 13$\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is (alpha, beta, gamma)$(\alpha, \beta, \gamma)$, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to (1) 9 (2) 12 (3) 8 (4) 7

Solution & Explanation

Reference Study Guides

More Three Dimensional Geometry Previous-Year Questions — Page 3

Solution

Solution

Solution

Solution

Solution

More Three Dimensional Geometry Questions — jee_main_2025_28_jan_morning

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If the image of the point $(4, 4, 3)$ $(4, 4, 3)$ in the line $fracx - 12 = fracy - 21 = fracz - 13$ is $(alpha, beta, gamma)$ $(\alpha, \beta, \gamma)$, then $alpha + beta + gamma$ $\alpha + \beta + \gamma$ is equal to (1) 9 (2) 12 (3) 8 (4) 7