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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

Reference Study Guides

Three Dimensional Geometry

More Three Dimensional Geometry Previous-Year Questions

Q51 2025 Image of a Point in a Line
If the image of the point mathrmP(1, 0, 3) in the line joining the points mathrmA(4, 7, 1) and mathrmB(3, 5, 3) is mathrmQ(alpha, beta, gamma), then alpha + beta + gamma is equal to
  • A. frac473
  • B. frac463
  • C. 18
  • D. 13

Solution

### Related Formula textEquation of a line through (x_1, y_1, z_1) text with direction ratios (a, b, c): quad fracx-x_1a = fracy-y_1b = fracz-z_1c textCondition for Perpendicular Lines: quad a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 ### Core Logic Let R be the foot of the perpendicular drawn from point P(1,0,3) to the line joining A(4,7,1) and B(3,5,3). Since Q(alpha,beta,gamma) is the reflection (image) of P across the line, point R serves as the midpoint of the line segment PQ. ### Step 1: Find the Equation of the Line AB The direction ratios of the line AB are: vecd = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) equiv (1, 2, -2) Using point B(3,5,3), the symmetric equation of the line AB is: fracx - 31 = fracy - 52 = fracz - 3-2 = lambda Any general point R on this line can be written in terms of parameter lambda: R equiv (lambda + 3, 2lambda + 5, -2lambda + 3) ### Step 2: Find the Foot of the Perpendicular R The direction ratios of the line segment PR are: vecmathrmPR = (lambda + 3 - 1, \, 2lambda + 5 - 0, \, -2lambda + 3 - 3) = (lambda + 2, \, 2lambda + 5, \, -2lambda) Since PR is perpendicular to the line AB, the dot product of their direction vectors must equal zero: 1(lambda + 2) + 2(2lambda + 5) - 2(-2lambda) = 0 lambda + 2 + 4lambda + 10 + 4lambda = 0 lambda = -frac43 ### Step 3: Coordinates of Foot of Perpendicular Substitute lambda = -frac43 into the general coordinates of R: R equiv left(-frac43 + 3, \, 2left(-frac43right) + 5, \, -2left(-frac43right) + 3right) R equiv left(frac53, \, frac73, \, frac173right) ### Step 4: Solve for the Image Coordinates and Sum Since R is the midpoint of PQ, where P = (1, 0, 3) and Q = (alpha, beta, gamma): - fracalpha + 12 = frac53 implies alpha = frac103 - 1 = frac73 - fracbeta + 02 = frac73 implies beta = frac143 - fracgamma + 32 = frac173 implies gamma = frac343 - 3 = frac253 Now, let us calculate the sum: alpha + beta + gamma = frac73 + frac143 + frac253 = frac463 ### Pattern Recognition Standard Midpoint reflection: The image coordinates are given directly by x_textimage = 2 x_textfoot - x_textpoint, y_textimage = 2 y_textfoot - y_textpoint, and z_textimage = 2 z_textfoot - z_textpoint. Finding the parameter lambda by using the perpendicular vector dot-product rule is the fastest and most robust method. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry
Q56 2025 Shortest Distance between Skew Lines
The line mathrmL_1 is parallel to the vector vecmathrma = -3hatmathrmi + 2hatmathrmj + 4hatmathrmk and passes through the point (7, 6, 2) and the line mathrmL_2 is parallel to the vector vecmathrmb = 2hatmathrmi + hatmathrmj + 3hatmathrmk and passes through the point (5, 3, 4). The shortest distance between the lines mathrmL_1 and mathrmL_2 is:
  • A. frac23sqrt38
  • B. frac21sqrt57
  • C. frac23sqrt57
  • D. frac21sqrt38

Solution

### Related Formula textShortest Distance d = fracleft| (vecr_2 - vecr_1) cdot (veca times vecb) right||veca times vecb| ### Core Logic Shortest distance between two skew lines is the projection of the vector joining any two points of the lines onto their common normal. ### Step 1: Find the vector joining the two points Let the points be P_1(7, 6, 2) on L_1 and P_2(5, 3, 4) on L_2: vecr_2 - vecr_1 = (5 - 7)hatmathrmi + (3 - 6)hatmathrmj + (4 - 2)hatmathrmk = -2hatmathrmi - 3hatmathrmj + 2hatmathrmk ### Step 2: Find the common normal vector The direction is given by the cross product of the direction vectors: veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ -3 & 2 & 4 \\ 2 & 1 & 3 endvmatrix veca times vecb = hatmathrmi(6 - 4) - hatmathrmj(-9 - 8) + hatmathrmk(-3 - 4) = 2hatmathrmi + 17hatmathrmj - 7hatmathrmk ### Step 3: Calculate the distance Calculate the dot product of the vectors: (vecr_2 - vecr_1) cdot (veca times vecb) = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69 Calculate the magnitude of the cross product: |veca times vecb| = sqrt2^2 + 17^2 + (-7)^2 = sqrt4 + 289 + 49 = sqrt342 = 3sqrt38 Now, compute the shortest distance: d = frac|-69|3sqrt38 = frac23sqrt38 ### Pattern Recognition Matrix determinant check: In skew lines problems, the numerator can also be computed as the determinant of the 3x3 matrix composed of (vecr_2-vecr_1), veca, and vecb. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry
Q74 2025 Area of Triangles
Let mathrmA(4, -2), mathrmB(1, 1) and mathrmC(9, -3) be the vertices of a triangle ABC. Then the maximum area of the parallelogram AFDE, formed with vertices D, E and F on the sides BC, CA and AB of the triangle ABC respectively, is ____________.
Numerical Answer. Answer: 3 to 3

Solution

### Related Formula textArea of triangle ABC = frac12 left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right| textMaximum area of an inscribed parallelogram = frac12 times textArea of triangle ### Core Logic First, we compute the total area of the triangle ABC from the given coordinate points. We then apply the geometric maximum area property for inscribed parallelograms. ### Step 1: Calculate the area of triangle ABC The coordinate points are A(4, -2), B(1, 1), and C(9, -3): textArea(Delta textABC) = frac12 beginvmatrix 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 endvmatrix textArea(Delta textABC) = frac12 left| 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) right| textArea(Delta textABC) = frac12 left| 4(4) + 2(-8) + 1(-12) right| textArea(Delta textABC) = frac12 left| 16 - 16 - 12 right| = 6 text square units ### Step 2: Apply the maximum area theorem The maximum area of a parallelogram inscribed in a triangle of area Delta is always exactly half the area of the triangle: textMaximum Area = frac12 times textArea(Delta textABC) = frac12 times 6 = 3 text square units ### Pattern Recognition Inscribed shapes maximization: The maximum area of any inscribed parallelogram AFDE on the sides of a triangle ABC occurs when the vertices D, E, F are exactly the midpoints of the respective sides of the triangle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines
Q 2025 Area of Triangle in 3D Space
Let the vertices Q and R of the triangle PQR lie on the line fracx+35=fracy-12=fracz+43, QR=5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is fracmn then:
  • A. m-5sqrt21n=0
  • B. 2m-5sqrt21n=0
  • C. 5m-2sqrt21n=0
  • D. 5m-21sqrt2n=0

Solution

### Related Formula Area of a triangle given base length b and perpendicular height h: textArea = frac12 cdot b cdot h ### Core Logic Since Q and R lie on the line, the length QR=5 forms the base of the triangle. The perpendicular distance from point P to the line represents the height h.
Area of Triangle in 3D Space diagram for Q67 - JEE Main 2025 Morning
Area of Triangle in 3D Space diagram for Q67 - JEE Main 2025 Morning
### Step 1: Define Perpendicular Foot coordinates Let M be the foot of the perpendicular from P(0,2,3) to the line. Parametric form of any point on the line: M(5lambda - 3, 2lambda + 1, 3lambda - 4) Direction ratios of line segment PM: textDRs = (5lambda - 3 - 0, 2lambda + 1 - 2, 3lambda - 4 - 3) = (5lambda - 3, 2lambda - 1, 3lambda - 7) ### Step 2: Solve for Parameter using Perpendicularity Since PM is perpendicular to the given line (DRs: 5, 2, 3): 5(5lambda - 3) + 2(2lambda - 1) + 3(3lambda - 7) = 0 25lambda - 15 + 4lambda - 2 + 9lambda - 21 = 0 implies 38lambda = 38 implies lambda = 1 ### Step 3: Compute Perpendicular Distance For lambda = 1, the coordinates of M are (2, 3, -1). Calculate length PM: PM = sqrt(2-0)^2 + (3-2)^2 + (-1-3)^2 = sqrt4 + 1 + 16 = sqrt21 ### Step 4: Formulate the Area Equation textArea = frac12 cdot QR cdot PM = frac12 cdot 5 cdot sqrt21 = fracmn frac5sqrt212 = fracmn implies 2m = 5sqrt21n implies 2m - 5sqrt21n = 0 ### Pattern Recognition Finding the foot of a perpendicular via parametric variables is a guaranteed shortcut for 3D area problems instead of cross-product calculations, keeping computation times minimal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry
Q 2025 Properties of Tetrahedron
Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the Delta BCD is equal to:
  • A. sqrt340
  • B. 12
  • C. sqrt110
  • D. 7sqrt3

Solution

### Related Formula De Gua's Theorem (3D extension of Pythagorean theorem) for right corner tetrahedron states: [textArea(Delta BCD)]^2 = [textArea(Delta ABC)]^2 + [textArea(Delta ACD)]^2 + [textArea(Delta ADB)]^2 ### Core Logic Since the three edges meeting at vertex A are mutually perpendicular, we can align them directly with orthogonal coordinate axes to compute the slanted face area. ### Step 1: Apply Square Area Summation Let the target area be Delta. Delta^2 = 5^2 + 6^2 + 7^2 Delta^2 = 25 + 36 + 49 = 110 Delta = sqrt110 ### Pattern Recognition This is exactly analogous to finding the length of a vector given three perpendicular component lengths. Squaring, adding, and taking the square root yields the value immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

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