Let a line passing through the point (4,1,0)$(4,1,0)$ [cite: 520] intersect the line L_1:fracx-12=fracy-23=fracz-34$L_{1}:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point A(alpha, beta, gamma)$A(\alpha, \beta, \gamma)$ [cite: 522, 523, 524] and the line L_2:x-6=y=-z+4$L_{2}:x-6=y=-z+4$ at the point B(a, b, c)$B(a, b, c)$[cite: 525]. Then the value of the determinant beginvmatrix 1 & 0 & 1 \\ alpha & beta & gamma \\ a & b & c endvmatrix$\begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix}$ is equal to[cite: 527]:
A.8
B.16
C.12
D.4
Solution & Explanation
### Related Formula
For three points P$P$, A$A$, and B$B$ 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$\vec{PA} \parallel \vec{PB} \implies \frac{x_A - x_P}{x_B - x_P} = \frac{y_A - y_P}{y_B - y_P} = \frac{z_A - z_P}{z_B - z_P}$Intersection of Lines diagram for Q52 - JEE Main 2025 Morning
### Core Logic
Express general coordinates for A$A$ on L_1$L_1$ and B$B$ on L_2$L_2$ [cite: 1204]:
L_1: fracx-12=fracy-23=fracz-34=p implies A(2p+1, 3p+2, 4p+3)$L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=p \implies A(2p+1, 3p+2, 4p+3)$ [cite: 1204, 1206]
L_2: fracx-61=fracy1=fracz-4-1=q implies B(q+6, q, 4-q)$L_2: \frac{x-6}{1}=\frac{y}{1}=\frac{z-4}{-1}=q \implies B(q+6, q, 4-q)$ [cite: 1204, 1207]
Direction ratios (D.R.) from P(4,1,0)$P(4,1,0)$ [cite: 1208, 1209]:
textD.R. of PA = (2p-3, 3p+1, 4p+3)$\text{D.R. of } PA = (2p-3, 3p+1, 4p+3)$ [cite: 1208]
textD.R. of PB = (q+2, q-1, 4-q)$\text{D.R. of } PB = (q+2, q-1, 4-q)$ [cite: 1209]
Since P, A, B$P, A, B$ lie on the same line [cite: 1210]:
frac2p-3q+2 = frac3p+1q-1 = frac4p+34-q$\frac{2p-3}{q+2} = \frac{3p+1}{q-1} = \frac{4p+3}{4-q}$ [cite: 1210]
### Step 1: Solving the system of equations
Equating expressions to solve for parameters p$p$ and q$q$ [cite: 1211, 1212]:
From first two [cite: 1211]:
2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 implies pq + 8p + 4q - 1 = 0$2pq - 2p - 3q + 3 = 3pq + 6p + q + 2 \implies pq + 8p + 4q - 1 = 0$ [cite: 1211, 1212]
From second and third components [cite: 1212]:
12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 implies 7pq - 16p + 4q - 7 = 0$12p - 3pq + 4 - q = 4pq + 3q - 4p - 3 \implies 7pq - 16p + 4q - 7 = 0$ [cite: 1212]
Solving equations simultaneously gives [cite: 1234]:
p = -1, quad q = 3$p = -1, \quad q = 3$ [cite: 1234]
Substituting parameters back yields the points [cite: 1236]:
A(-1, -1, -1), quad B(9, 3, 1)$A(-1, -1, -1), \quad B(9, 3, 1)$ [cite: 1236]
### Step 2: Evaluating the Determinant
Substitute values of A$A$ and B$B$ into the matrix grid [cite: 1244]:
beginvmatrix 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 endvmatrix$\begin{vmatrix} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{vmatrix}$ [cite: 1244]
Applying row/column operations (C_3 rightarrow C_3 - C_1$C_3 \rightarrow C_3 - C_1$) [cite: 1244]:
beginvmatrix 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 endvmatrix = 1((-1)(-8) - 0) = 8$\begin{vmatrix} 1 & 0 & 0 \\ -1 & -1 & 0 \\ 9 & 3 & -8 \end{vmatrix} = 1((-1)(-8) - 0) = 8$ [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
Keywords:#line passing through point (4,1,0)#JEE Main 2025 Morning Q52#Three Dimensional Geometry JEE Main 2025#Intersection of Lines
More Three Dimensional Geometry Previous-Year Questions — Page 3
Q732025Area 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$x + 2 = y - 1 = z, \frac{x - 3}{5} = \frac{y}{-1} = \frac{z - 1}{1}$ and fracmathrmx-3 = fracmathrmy - 33 = fracmathrmz - 21$\frac{\mathrm{x}}{-3} = \frac{\mathrm{y} - 3}{3} = \frac{\mathrm{z} - 2}{1}$ be A. Then A^2$A^2$ is equal to
Numerical Answer.Answer: 56 to 56
Solution
### Related Formula
textArea A = frac12 |vecAB times vecAC|$\text{Area } A = \frac{1}{2} |\vec{AB} \times \vec{AC}|$
### 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)$L_1 \cap L_2 \implies A(-2, 1, 0)$
* L_2 cap L_3 implies B(3, 0, 1)$L_2 \cap L_3 \implies B(3, 0, 1)$
* L_3 cap L_1 implies C(0, 3, 2)$L_3 \cap L_1 \implies C(0, 3, 2)$
### Step 2: Construct Vectors Cross Matrix
Using vertex values to form component arrays:
vecAB = -5hati + hatj - hatk, quad vecAC = -3hati + 3hatj + hatk$\vec{AB} = -5\hat{i} + \hat{j} - \hat{k}, \quad \vec{AC} = -3\hat{i} + 3\hat{j} + \hat{k}$vecAB times vecAC = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -1 \\ -3 & 3 & 1 endvmatrix = 4hati + 8hatj - 12hatk$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{vmatrix} = 4\hat{i} + 8\hat{j} - 12\hat{k}$
### Step 3: Final Area Squared Derivation
A = frac12sqrt16 + 64 + 144 = frac12sqrt224 = sqrt56$A = \frac{1}{2}\sqrt{16 + 64 + 144} = \frac{1}{2}\sqrt{224} = \sqrt{56}$A^2 = 56$A^2 = 56$
{{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
Q562025Line and Plane Intersections
Let a straight line L$L$ pass through the point P(2, -1, 3)$P(2, -1, 3)$ and be perpendicular to the lines fracx - 12 = fracy + 11 = fracz - 3-2$\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2}$ and \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. If the line L$L$ intersects the yz$yz$-plane at the point Q$Q$, then the distance between the points P$P$ and Q$Q$ 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$\vec{u}$ and vecv$\vec{v}$ is obtained via the cross product:
vecn = vecu times vecv$\vec{n} = \vec{u} \times \vec{v}$
### Core Logic
Extract direction vectors of the given lines:
vecu = 2hati + hatj - 2hatk$\vec{u} = 2\hat{i} + \hat{j} - 2\hat{k}$vecv = hati + 3hatj + 4hatk$\vec{v} = \hat{i} + 3\hat{j} + 4\hat{k}$
Compute the cross product:
vecn = beginvmatrix hati & hatj & hatk \\ 2 & 1 & -2 \\ 1 & 3 & 4 endvmatrix$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$= hati(4 - (-6)) - hatj(8 - (-2)) + hatk(6 - 1) = 10hati - 10hatj + 5hatk$= \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$= 5(2hati - 2hatj + hatk)$= 5(2\hat{i} - 2\hat{j} + \hat{k})$
### Step 1: Write Line Equation and Intersect with Plane
Equation of line L$L$ through P(2, -1, 3)$P(2, -1, 3)$ with direction (2, -2, 1)$(2, -2, 1)$:
fracx - 22 = fracy + 1-2 = fracz - 31 = lambda$\frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda$
Any random point on this line is Q(2lambda + 2, -2lambda - 1, lambda + 3)$Q(2\lambda + 2, -2\lambda - 1, \lambda + 3)$.
For intersection with the yz$yz$-plane, set x = 0$x = 0$:
2lambda + 2 = 0 implies lambda = -1$2\lambda + 2 = 0 \implies \lambda = -1$
### Step 2: Find Distance
Substituting lambda = -1$\lambda = -1$ into the coordinate matrix of Q$Q$ gives:
Q(0, 1, 2)$Q(0, 1, 2)$
Calculate distance d(P, Q)$d(P, Q)$:
d = sqrt(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2 = sqrt4 + 4 + 1 = 3$d = \sqrt{(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 4 + 1} = 3$
### Pattern Recognition
Perpendicularity to two lines always indicates using the cross-product to lock down the direction ratios. Intersection with the yz$yz$-plane simply forces x = 0$x = 0$ immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q632025Shortest Distance and Intersection of Lines
Let mathbfP$\mathbf{P}$ be the foot of the perpendicular from the point (1,2,2)$(1,2,2)$ on the line mathrmL:fracmathrmx - 11 = fracmathrmy + 1-1 = fracmathrmz - 22$\mathrm{L}:\frac{\mathrm{x} - 1}{1} = \frac{\mathrm{y} + 1}{-1} = \frac{\mathrm{z} - 2}{2}$. Let the line vecmathrmr = (-hatmathrmi +hatmathrmj -2hatmathrmk) + lambda (hatmathrmi -hatmathrmj +hatmathrmk)$\vec{\mathrm{r}} = (-\hat{\mathrm{i}} +\hat{\mathrm{j}} -2\hat{\mathrm{k}}) + \lambda (\hat{\mathrm{i}} -\hat{\mathrm{j}} +\hat{\mathrm{k}})$, lambda in mathbbR$\lambda \in \mathbb{R}$, intersect the line mathrmL$\mathrm{L}$ at Q. Then 2(mathrmPQ)^2$2(\mathrm{PQ})^2$ 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$\vec{AP} \cdot \vec{d} = 0$
### Core Logic
Let the target source coordinates tracking point match A(1, 2, 2)$A(1, 2, 2)$. General parameter points on line L$L$ are defined by parameter mu$\mu$:
P(mu + 1, -mu - 1, 2mu + 2)$P(\mu + 1, -\mu - 1, 2\mu + 2)$Shortest Distance and Intersection of Lines diagram for Q63 - JEE Main 2025 EveningvecAP = muhati - (mu + 3)hatj + 2muhatk$\vec{AP} = \mu\hat{i} - (\mu + 3)\hat{j} + 2\mu\hat{k}$
Line direction vector vecd = hati - hatj + 2hatk$\vec{d} = \hat{i} - \hat{j} + 2\hat{k}$.
### 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$(\mu)\cdot 1 - (-\mu - 3)\cdot 1 + (2\mu)\cdot 2 = 0 \implies 6\mu + 3 = 0 \implies \mu = -\frac{1}{2}$
Substituting back yields coordinate positions for foot P$P$:
Pleft(frac12, -frac12, 1right)$P\left(\frac{1}{2}, -\frac{1}{2}, 1\right)$
Equating general vectors between standard linear constraints tracks intersection point Q$Q$ at mu = -2$\mu = -2$:
Q(-1, 1, -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$PQ^2 = \left(\frac{1}{2} - (-1)\right)^2 + \left(-\frac{1}{2} - 1\right)^2 + (1 - (-2))^2$= frac94 + frac94 + 9 = frac544$= \frac{9}{4} + \frac{9}{4} + 9 = \frac{54}{4}$2(PQ)^2 = 2 left(frac544right) = 27$2(PQ)^2 = 2 \left(\frac{54}{4}\right) = 27$
### 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
Q562025Distance Formula and Properties of Triangles
Let A(x,y,z)$A(x,y,z)$ 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)$B = (1, 4, -1)$ and C = (2, 0, -2)$C = (2, 0, -2)$. Then among the statements
(S1) : Delta ABC$\Delta ABC$ is an isosceles right angled triangle and
(S2): the area of Delta ABC$\Delta ABC$ is frac9sqrt22$\frac{9\sqrt{2}}{2}$.
(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$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$
### Core Logic
Since A(x,y,z)$A(x,y,z)$ lies in the xy-plane, its z-coordinate must be zero (z = 0$z = 0$). Let the reference targets be P(0,3,2)$P(0,3,2)$, Q(2,0,3)$Q(2,0,3)$, and R(0,0,1)$R(0,0,1)$.
Setting AP^2 = AR^2$AP^2 = AR^2$:
x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 implies y = 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$AQ^2 = AR^2$ with y=2$y=2$:
(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 implies x = 3$(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 \implies x = 3$
Thus, A$A$ is precisely located at (3,2,0)$(3,2,0)$.
### Step 2: Triangle Side and Area Assessment
Calculate the lengths between A(3,2,0)$A(3,2,0)$, B(1,4,-1)$B(1,4,-1)$, and C(2,0,-2)$C(2,0,-2)$:
AB = sqrt(3-1)^2 + (2-4)^2 + (0+1)^2 = 3$AB = \sqrt{(3-1)^2 + (2-4)^2 + (0+1)^2} = 3$AC = sqrt(3-2)^2 + (2-0)^2 + (0+2)^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$BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{18}$
Since AB = AC = 3$AB = AC = 3$ and AB^2 + AC^2 = BC^2$AB^2 + AC^2 = BC^2$, it forms an isosceles right-angled triangle. Thus, (S1) is true.
textArea = frac12 times 3 times 3 = frac92$\text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$
Therefore, (S2) is false.
### Pattern Recognition
Planar locations instantly zero out specific coordinate dimensions (z=0$z=0$ for xy-planes), simplifying system matrices down rapidly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Three Dimensional Geometry
Q622025Image of a Point in a Line
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
A. 9
B. 12
C. 8
D. 7
Solution
### Related Formula
Perpendicularity condition for vectors:
vecu cdot vecv = 0$\vec{u} \cdot \vec{v} = 0$
### Core Logic
Let Q$Q$ be the projection point on the given line parameterized by lambda$\lambda$:
Q(2lambda + 1, lambda + 2, 3lambda + 1)$Q(2\lambda + 1, \lambda + 2, 3\lambda + 1)$.
The vector overrightarrowPQ$\overrightarrow{PQ}$ from P(4,4,3)$P(4,4,3)$ is: Image of a Point in a Line diagram for Q62 - JEE Main 2025 MorningoverrightarrowPQ = (2lambda - 3)hati + (lambda - 2)hatj + (3lambda - 2)hatk$\overrightarrow{PQ} = (2\lambda - 3)\hat{i} + (\lambda - 2)\hat{j} + (3\lambda - 2)\hat{k}$.
### Step 1: Solving for Projected Intersection Points
Since overrightarrowPQ$\overrightarrow{PQ}$ is perpendicular to the line's direction vector (2, 1, 3)$(2, 1, 3)$:
2(2lambda - 3) + 1(lambda - 2) + 3(3lambda - 2) = 0 implies 14lambda - 14 = 0 implies lambda = 1$2(2\lambda - 3) + 1(\lambda - 2) + 3(3\lambda - 2) = 0 \implies 14\lambda - 14 = 0 \implies \lambda = 1$
Thus, Q$Q$ is located at (3,3,4)$(3,3,4)$.
### Step 2: Transforming using Midpoint Mappings
The projection point Q$Q$ acts as the midpoint between original point P$P$ and its target image R(alpha, beta, gamma)$R(\alpha, \beta, \gamma)$:
fracalpha + 42 = 3, quad fracbeta + 42 = 3, quad fracgamma + 32 = 4$\frac{\alpha + 4}{2} = 3, \quad \frac{\beta + 4}{2} = 3, \quad \frac{\gamma + 3}{2} = 4$
Evaluating this gives (alpha, beta, gamma) = (2, 2, 5)$(\alpha, \beta, \gamma) = (2, 2, 5)$.
textSum = 2 + 2 + 5 = 9$\text{Sum} = 2 + 2 + 5 = 9$
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$2+2+5=9$, 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
More Three Dimensional Geometry Questions — jee_main_2025_03_april_morning
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