Let a line passing through the point (4,1,0) [cite: 520] intersect the line L_1:fracx-12=fracy-23=fracz-34 at the point A(alpha, beta, gamma) [cite: 522, 523, 524] and the line L_2:x-6=y=-z+4 at the point B(a, b, c)[cite: 525]. Then the value of the determinant beginvmatrix 1 & 0 & 1 \\ alpha & beta & gamma \\ a & b & c endvmatrix is equal to[cite: 527]:

Solution & Explanation

### Related Formula For three points P, A, and B to be collinear, their direction vectors must be proportional: vecPA parallel vecPB implies fracx_A - x_Px_B - x_P = fracy_A - y_Py_B - y_P = fracz_A - z_Pz_B - z_P
Intersection of Lines diagram for Q52 - JEE Main 2025 Morning
Intersection of Lines diagram for Q52 - JEE Main 2025 Morning
### Core Logic Express general coordinates for A on L_1 and B on L_2 [cite: 1204]: L_1: fracx-12=fracy-23=fracz-34=p implies A(2p+1, 3p+2, 4p+3) [cite: 1204, 1206] L_2: fracx-61=fracy1=fracz-4-1=q implies B(q+6, q, 4-q) [cite: 1204, 1207] Direction ratios (D.R.) from P(4,1,0) [cite: 1208, 1209]: textD.R. of PA = (2p-3, 3p+1, 4p+3) [cite: 1208] textD.R. of PB = (q+2, q-1, 4-q) [cite: 1209] Since P, A, B lie on the same line [cite: 1210]: frac2p-3q+2 = frac3p+1q-1 = frac4p+34-q [cite: 1210] ### Step 1: Solving the system of equations Equating expressions to solve for parameters p and q [cite: 1211, 1212]: From first two [cite: 1211]: 2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 implies pq + 8p + 4q - 1 = 0 [cite: 1211, 1212] From second and third components [cite: 1212]: 12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 implies 7pq - 16p + 4q - 7 = 0 [cite: 1212] Solving equations simultaneously gives [cite: 1234]: p = -1, quad q = 3 [cite: 1234] Substituting parameters back yields the points [cite: 1236]: A(-1, -1, -1), quad B(9, 3, 1) [cite: 1236] ### Step 2: Evaluating the Determinant Substitute values of A and B into the matrix grid [cite: 1244]: beginvmatrix 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 endvmatrix [cite: 1244] Applying row/column operations (C_3 rightarrow C_3 - C_1) [cite: 1244]: beginvmatrix 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 endvmatrix = 1((-1)(-8) - 0) = 8 [cite: 1244] ### Pattern Recognition Collinearity problem involving parameter matches across skew line orientations. Reducing fractions systematically prevents higher-degree calculation mistakes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Determinants

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Three Dimensional Geometry

More Three Dimensional Geometry Previous-Year Questions — Page 3

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
  • B. 4
  • C. fracsqrt23
  • D. 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
  • 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 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
  • 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 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

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