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Let the line passing through the points (-1, 2, 1) and \parallel to the line fracx - 12 = fracy + 13 = fracz4 intersect the line fracx + 23 = fracy - 32 = fracz - 41 at the point P. Then the distance of P from the point Q(4, -5, 1) is :

Solution & Explanation

### Related Formula The line passing through veca and \parallel to direction vector vecv = lhatmathbfi + mhatmathbfj + nhatmathbfk is written in symmetric form as: fracx - x_al = fracy - y_am = fracz - z_an ### Core Logic Formulate the equation of the line passing through (-1, 2, 1) with direction vector components (2, 3, 4): L_1: fracx + 12 = fracy - 23 = fracz - 14 = lambda quad dots (1)
Intersection of Lines in 3D Space
Intersection of Lines in 3D Space
Any generic point on this line can be written as: P = (2lambda - 1, \, 3lambda + 2, \, 4lambda + 1) The second given line equation is: L_2: fracx + 23 = fracy - 32 = fracz - 41 = mu quad dots (2) Any generic point on line L_2 is: P' = (3mu - 2, \, 2mu + 3, \, mu + 4) ### Step 1: Compute the Point of Intersection At the point of intersection P, equate the coordinates from both lines: 2lambda - 1 = 3mu - 2 implies 2lambda - 3mu = -1 quad dots (3) 3lambda + 2 = 2mu + 3 implies 3lambda - 2mu = 1 quad dots (4) 4lambda + 1 = mu + 4 implies 4lambda - mu = 3 quad dots (5) Solving equations (3) and (4) simultaneously gives: lambda = 1, quad mu = 1 Verify these values using equation (5): 4(1) - 1 = 3 quad (textSatisfied) Thus, substituting lambda = 1 gives the coordinates of point P: P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5) ### Step 2: Calculate Euclidean Distance PQ Find the distance between P(1, 5, 5) and Q(4, -5, 1) using the 3D distance formula: PQ = sqrt(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2 PQ = sqrt3^2 + (-10)^2 + (-4)^2 = sqrt9 + 100 + 16 PQ = sqrt125 = 5sqrt5 ### Pattern Recognition When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

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Three Dimensional Geometry

More Three Dimensional Geometry Previous-Year Questions — Page 3

Q73 2025 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 2025 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
  • B. sqrt10
  • C. 3
  • D. 2sqrt3

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 2025 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
  • B. 25
  • C. 29
  • D. 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
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 2025 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
Q62 2025 Image of a Point in a Line
If the image of the point (4, 4, 3) in the line fracx - 12 = fracy - 21 = fracz - 13 is (alpha, beta, gamma), then alpha + beta + gamma is equal to (1) 9 (2) 12 (3) 8 (4) 7
  • A. 9
  • B. 12
  • C. 8
  • D. 7

Solution

### Related Formula Perpendicularity condition for vectors: vecu cdot vecv = 0 ### Core Logic Let Q be the projection point on the given line parameterized by lambda: Q(2lambda + 1, lambda + 2, 3lambda + 1). The vector overrightarrowPQ from P(4,4,3) is:
Image of a Point in a Line diagram for Q62 - JEE Main 2025 Morning
Image of a Point in a Line diagram for Q62 - JEE Main 2025 Morning
overrightarrowPQ = (2lambda - 3)hati + (lambda - 2)hatj + (3lambda - 2)hatk. ### Step 1: Solving for Projected Intersection Points Since overrightarrowPQ is perpendicular to the line's direction vector (2, 1, 3): 2(2lambda - 3) + 1(lambda - 2) + 3(3lambda - 2) = 0 implies 14lambda - 14 = 0 implies lambda = 1 Thus, Q is located at (3,3,4). ### Step 2: Transforming using Midpoint Mappings The projection point Q acts as the midpoint between original point P and its target image R(alpha, beta, gamma): fracalpha + 42 = 3, quad fracbeta + 42 = 3, quad fracgamma + 32 = 4 Evaluating this gives (alpha, beta, gamma) = (2, 2, 5). textSum = 2 + 2 + 5 = 9 Wait, checking the options from the paper layout: option (2) represents the correct numerical matrix sum choice value 12? Let's verify the options mapping sequence matching. Ah, let's look at the calculation value carefully: 2+2+5=9, which corresponds to choice (1). ### Pattern Recognition Midpoint properties safely speed up spatial image transitions once you locate the perpendicular projection foot. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

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