Let (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ be the co-ordinates of the foot of the perpendicular drawn from the point (5, 4, 2) on the line vecr = (-hati + 3hatj + hatk) + lambda(2hati + 3hatj - hatk)$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$ .
Then the length of the projection of the vector alphahati+betahatj+gammahatk$\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ on the vector 6hati+2hatj+3hatk$6\hat{i}+2\hat{j}+3\hat{k}$ is :
A.frac157$\frac{15}{7}$
B.4
C.frac187$\frac{18}{7}$
D.3
Solution & Explanation
### Related Formula
textLength of projection of vecu text on vecw = frac|vecu cdot vecw||vecw|$$\text{Length of projection of } \vec{u} \text{ on } \vec{w} = \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|}$$
### Core Logic
Given point A(5, 4, 2)$A(5, 4, 2)$ and line (L)$(L)$:
vecr = (-hati + 3hatj + hatk) + lambda(2hati + 3hatj - hatk)$$\vec{r} = (-\hat{i} + 3\hat{j} + \hat{k}) + \lambda(2\hat{i} + 3\hat{j} - \hat{k})$$
Any general point P$P$ on this line has coordinates:
(-1 + 2lambda, 3 + 3lambda, 1 - lambda)$(-1 + 2\lambda, 3 + 3\lambda, 1 - \lambda)$
### Step 1: Finding the foot of the perpendicular
Vector vecAP = P - A = (-1 + 2lambda - 5)hati + (3 + 3lambda - 4)hatj + (1 - lambda - 2)hatk$\vec{AP} = P - A = (-1 + 2\lambda - 5)\hat{i} + (3 + 3\lambda - 4)\hat{j} + (1 - \lambda - 2)\hat{k}$vecAP = (2lambda - 6)hati + (3lambda - 1)hatj + (-lambda - 1)hatk$$\vec{AP} = (2\lambda - 6)\hat{i} + (3\lambda - 1)\hat{j} + (-\lambda - 1)\hat{k}$$
Since AP$AP$ is perpendicular to line (L)$(L)$, the dot product of vecAP$\vec{AP}$ with the direction vector of the line (2hati + 3hatj - hatk)$(2\hat{i} + 3\hat{j} - \hat{k})$ must be zero:
vecAP cdot (2hati + 3hatj - hatk) = 0$$\vec{AP} \cdot (2\hat{i} + 3\hat{j} - \hat{k}) = 0$$2(2lambda - 6) + 3(3lambda - 1) - 1(-lambda - 1) = 0$$2(2\lambda - 6) + 3(3\lambda - 1) - 1(-\lambda - 1) = 0$$4lambda - 12 + 9lambda - 3 + lambda + 1 = 0$$4\lambda - 12 + 9\lambda - 3 + \lambda + 1 = 0$$14lambda - 14 = 0 Rightarrow lambda = 1$$14\lambda - 14 = 0 \Rightarrow \lambda = 1$$Foot of perpendicular 3D diagram for Q14 - JEE Main 2026 Morning
### Step 2: Coordinates of the foot
Substitute lambda = 1$\lambda = 1$ into general point P$P$ to get (alpha, beta, gamma)$(\alpha, \beta, \gamma)$:
alpha = -1 + 2(1) = 1$\alpha = -1 + 2(1) = 1$beta = 3 + 3(1) = 6$\beta = 3 + 3(1) = 6$gamma = 1 - 1 = 0$\gamma = 1 - 1 = 0$
Foot of perpendicular is (1, 6, 0)$(1, 6, 0)$.
### Step 3: Calculate the projection
Let vecu = alphahati + betahatj + gammahatk = hati + 6hatj + 0hatk$\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k} = \hat{i} + 6\hat{j} + 0\hat{k}$
Let vecw = 6hati + 2hatj + 3hatk$\vec{w} = 6\hat{i} + 2\hat{j} + 3\hat{k}$textProjection = frac|vecu cdot vecw||vecw| = frac|1(6) + 6(2) + 0(3)|sqrt6^2 + 2^2 + 3^2$$\text{Projection} = \frac{|\vec{u} \cdot \vec{w}|}{|\vec{w}|} = \frac{|1(6) + 6(2) + 0(3)|}{\sqrt{6^2 + 2^2 + 3^2}}$$= frac6 + 12sqrt36 + 4 + 9 = frac18sqrt49 = frac187$$= \frac{6 + 12}{\sqrt{36 + 4 + 9}} = \frac{18}{\sqrt{49}} = \frac{18}{7}$$
### Pattern Recognition
Foot of perpendicular problems algorithm: 1) Frame general vector P(lambda)$P(\lambda)$. 2) Construct distance vector vecAP$\vec{AP}$. 3) Dot product with direction vector vecd = 0$\vec{d} = 0$. 4) Solve for lambda$\lambda$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Class 12 Maths: Vector Algebra
Keywords:#foot of the perpendicular#JEE Main 2026 Morning Q14#Three Dimensional Geometry JEE Main 2026#Foot of Perpendicular and Projection JEE Main 2026
More Three Dimensional Geometry Previous-Year Questions — Page 2
Qjee_main_2025_02_april_morningProperties of Tetrahedron
Let ABCD$ABCD$ be a tetrahedron such that the edges AB$AB$, AC$AC$ and AD$AD$ are mutually perpendicular. Let the areas of the triangles ABC$ABC$, ACD$ACD$ and ADB$ADB$ be 5, 6 and 7 square units respectively. Then the area (in square units) of the Delta BCD$\Delta BCD$ is equal to:
A.sqrt340$\sqrt{340}$
B.12$12$
C.sqrt110$\sqrt{110}$
D.7sqrt3$7\sqrt{3}$
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$$[\text{Area}(\Delta BCD)]^2 = [\text{Area}(\Delta ABC)]^2 + [\text{Area}(\Delta ACD)]^2 + [\text{Area}(\Delta ADB)]^2$$
### Core Logic
Since the three edges meeting at vertex A$A$ 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$\Delta$.
Delta^2 = 5^2 + 6^2 + 7^2$$\Delta^2 = 5^2 + 6^2 + 7^2$$Delta^2 = 25 + 36 + 49 = 110$$\Delta^2 = 25 + 36 + 49 = 110$$Delta = sqrt110$$\Delta = \sqrt{110}$$
### 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
Qjee_main_2025_03_april_eveningDirection Cosines and Direction Ratios
Each of the angles beta$\beta$ and gamma$\gamma$ that a given line makes with the positive y$y$- and z$z$-axes, respectively, is half of the angle that this line makes with the positive x$x$-axis. Then the sum of all possible values of the angle beta$\beta$ is
A.frac3pi4$\frac{3\pi}{4}$
B.pi$\pi$
C.fracpi2$\frac{\pi}{2}$
D.frac3pi2$\frac{3\pi}{2}$
Solution
### Related Formula
For any line making angles alpha, beta, gamma$\alpha, \beta, \gamma$ with the coordinate axes, the direction cosines satisfy:
cos^2 alpha + cos^2 beta + cos^2 gamma = 1$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$
### Core Logic
Given that:
beta = fracalpha2 quad textand quad gamma = fracalpha2$$\beta = \frac{\alpha}{2} \quad \text{and} \quad \gamma = \frac{\alpha}{2}$$
Substituting these values into the identity:
cos^2 alpha + 2cos^2left(fracalpha2right) = 1$$\cos^2 \alpha + 2\cos^2\left(\frac{\alpha}{2}\right) = 1$$
### Step 1: Solving the Trigonometric Equation
Using the half-angle identity 2cos^2left(fracalpha2right) = 1 + cos alpha$2\cos^2\left(\frac{\alpha}{2}\right) = 1 + \cos \alpha$:
cos^2 alpha + 1 + cos alpha = 1$$\cos^2 \alpha + 1 + \cos \alpha = 1$$cos^2 alpha + cos alpha = 0$$\cos^2 \alpha + \cos \alpha = 0$$cos alpha (cos alpha + 1) = 0$$\cos \alpha (\cos \alpha + 1) = 0$$
This yields two possible cases:
1. cos alpha = 0 implies alpha = fracpi2$\cos \alpha = 0 \implies \alpha = \frac{\pi}{2}$
2. cos alpha = -1 implies alpha = pi$\cos \alpha = -1 \implies \alpha = \pi$
### Step 2: Finding Values of beta$\beta$ and their Sum
Now we find corresponding values for beta = fracalpha2$\beta = \frac{\alpha}{2}$:
- If alpha = fracpi2 implies beta_1 = fracpi4$\alpha = \frac{\pi}{2} \implies \beta_1 = \frac{\pi}{4}$
- If alpha = pi implies beta_2 = fracpi2$\alpha = \pi \implies \beta_2 = \frac{\pi}{2}$
Sum of all possible values:
beta_1 + beta_2 = fracpi4 + fracpi2 = frac3pi4$$\beta_1 + \beta_2 = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$$
### Pattern Recognition
Direction cosines are bounded between [-1, 1]$[-1, 1]$. Always utilize the identities relating double angles or half angles (2cos^2theta = 1 + cos 2theta$2\cos^2\theta = 1 + \cos 2\theta$) to simplify quadratic forms involving different multiples of the coordinate angles.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Trigonometric Functions
Q69jee_main_2025_03_april_eveningLines in 3D Space
The distance of the point (7, 10, 11)$(7, 10, 11)$ from the line fracx-41 = fracy-40 = fracz-23$\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3}$ along the line fracx-92 = fracy-133 = fracz-176$\frac{x-9}{2} = \frac{y-13}{3} = \frac{z-17}{6}$ is
A.18$18$
B.14$14$
C.12$12$
D.16$16$
Solution
### Related Formula
Distance 'along a line' means the direction vector of the line segment joining target point P$P$ to intersecting point Q$Q$ must be parallel to the given direction ratios:
vecPQ = k cdot vecd$$\vec{PQ} = k \cdot \vec{d}$$
### Core Logic
Let point P = (7, 10, 11)$P = (7, 10, 11)$.
Any general point Q$Q$ on the first line fracx-41 = fracy-40 = fracz-23 = lambda$\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3} = \lambda$ is:
Q = (lambda + 4, 4, 3lambda + 2)$$Q = (\lambda + 4, 4, 3\lambda + 2)$$
Direction ratios of PQ$PQ$:
vecPQ = (lambda + 4 - 7, 4 - 10, 3lambda + 2 - 11) = (lambda - 3, -6, 3lambda - 9)$$\vec{PQ} = (\lambda + 4 - 7, 4 - 10, 3\lambda + 2 - 11) = (\lambda - 3, -6, 3\lambda - 9)$$
### Step 1: Equating direction vectors
Since distance is measured parallel to the line with direction vector (2, 3, 6)$(2, 3, 6)$, vecPQ$\vec{PQ}$ must be parallel to (2, 3, 6)$(2, 3, 6)$:
fraclambda - 32 = frac-63 = frac3lambda - 96$$\frac{\lambda - 3}{2} = \frac{-6}{3} = \frac{3\lambda - 9}{6}$$
From the middle term:
fraclambda - 32 = -2 implies lambda - 3 = -4 implies lambda = -1$$\frac{\lambda - 3}{2} = -2 \implies \lambda - 3 = -4 \implies \lambda = -1$$
Let's check consistency with the third term:
frac3(-1) - 96 = -2 quad (textConsistent!)$$\frac{3(-1) - 9}{6} = -2 \quad (\text{Consistent!})$$3D Lines diagram for Q69 - JEE Main 2025 Evening Shift
### Step 2: Distance calculation
For lambda = -1$\lambda = -1$, the intersection point Q$Q$ is:
Q = (3, 4, -1)$Q = (3, 4, -1)$
Distance PQ$PQ$:
PQ = sqrt(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2$$PQ = \sqrt{(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2}$$PQ = sqrt4^2 + 6^2 + 12^2 = sqrt16 + 36 + 144 = sqrt196 = 14$$PQ = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$$
### Pattern Recognition
When solving distance parallel to a line in 3D, always write down the parametric coordinates of the general point first. Match the ratio of direction cosines directly to avoid setting up complicated systems of coordinate planes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 12 Mathematics: Vector Algebra
Q52jee_main_2025_07_april_morningShortest Distance Between Two Lines
If the shortest distance between the lines fracx - 12 = fracy - 23 = fracz - 34$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and fracx1 = fracyalpha = fracz - 51$\frac{x}{1} = \frac{y}{\alpha} = \frac{z - 5}{1}$ is frac5sqrt6$\frac{5}{\sqrt{6}}$ , then the sum of all possible values of alpha$\alpha$ is
A.frac32$\frac{3}{2}$
B.-frac32$-\frac{3}{2}$
C.3$3$
D.-3$-3$
Solution
### Related Formula
Shortest distance between two skewed lines with vector equations vecr = veca_1 + lambda vecb_1$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and \vec{r} = \vec{a}_2 + \mu \vec{b}_2 is:
S.D. = left| frac(veca_2 - veca_1) cdot (vecb_1 times vecb_2)|vecb_1 times vecb_2| right|$$S.D. = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$$
### Core Logic
From the given lines:
Line 1 passes through A(1, 2, 3)$A(1, 2, 3)$ with direction vector vecb_1 = 2hati + 3hatj + 4hatk$\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Line 2 passes through B(0, 0, 5)$B(0, 0, 5)$ with direction vector vecb_2 = hati + alphahatj + hatk$\vec{b}_2 = \hat{i} + \alpha\hat{j} + \hat{k}$.
The vector connecting the two fixed points is:
vecBA = (1-0)hati + (2-0)hatj + (3-5)hatk = hati + 2hatj - 2hatk$$\vec{BA} = (1-0)\hat{i} + (2-0)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$$
### Step 1: Compute Cross Product of Direction Vectors
vecn = vecb_1 times vecb_2 = left| beginmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 1 & alpha & 1 endmatrix right| = hati(3 - 4alpha) - hatj(2 - 4) + hatk(2alpha - 3)$$\vec{n} = \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{matrix} \right| = \hat{i}(3 - 4\alpha) - \hat{j}(2 - 4) + \hat{k}(2\alpha - 3)$$vecn = (3 - 4alpha)hati + 2hatj + (2alpha - 3)hatk$$\vec{n} = (3 - 4\alpha)\hat{i} + 2\hat{j} + (2\alpha - 3)\hat{k}$$
### Step 2: Apply Shortest Distance Formula
Shortest Distance Between Two Lines diagram for Q52 - JEE Main 2025 MorningS.D. = left| frac(hati + 2hatj - 2hatk) cdot vecn|vecn| right| = frac5sqrt6$$S.D. = \left| \frac{(\hat{i} + 2\hat{j} - 2\hat{k}) \cdot \vec{n}}{|\vec{n}|} \right| = \frac{5}{\sqrt{6}}$$
Taking dot product in numerator:
(hati + 2hatj - 2hatk) cdot vecn = 1(3-4alpha) + 2(2) - 2(2alpha-3) = 3 - 4alpha + 4 - 4alpha + 6 = 13 - 8alpha$$(\hat{i} + 2\hat{j} - 2\hat{k}) \cdot \vec{n} = 1(3-4\alpha) + 2(2) - 2(2\alpha-3) = 3 - 4\alpha + 4 - 4\alpha + 6 = 13 - 8\alpha$$
Squaring both sides:
frac(13 - 8alpha)^2(3 - 4alpha)^2 + 4 + (2alpha - 3)^2 = frac256$$\frac{(13 - 8\alpha)^2}{(3 - 4\alpha)^2 + 4 + (2\alpha - 3)^2} = \frac{25}{6}$$6(64alpha^2 - 208alpha + 169) = 25(16alpha^2 - 24alpha + 9 + 4 + 4alpha^2 - 12alpha + 9)$$6(64\alpha^2 - 208\alpha + 169) = 25(16\alpha^2 - 24\alpha + 9 + 4 + 4\alpha^2 - 12\alpha + 9)$$6(64alpha^2 - 208alpha + 169) = 25(20alpha^2 - 36alpha + 22)$$6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22)$$384alpha^2 - 1248alpha + 1014 = 500alpha^2 - 900alpha + 550$$384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550$$116alpha^2 + 348alpha - 464 = 0$$116\alpha^2 + 348\alpha - 464 = 0$$alpha^2 + 3alpha - 4 = 0$$\alpha^2 + 3\alpha - 4 = 0$$
### Step 3: Calculate the Sum of Roots
The sum of all possible values of alpha$\alpha$ is given by the relation:
alpha_1 + alpha_2 = -frac31 = -3$$\alpha_1 + \alpha_2 = -\frac{3}{1} = -3$$
### Pattern Recognition
Whenever asked for the sum of all possible parameter values resulting from a vector condition, look to directly read the linear term coefficient via Vieta's relations rather than explicitly factoring or computing the roots individually.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 12 Mathematics: Vector Algebra
Q68jee_main_2025_07_april_morningLine Intersecting Two Lines
Let the line L pass through (1, 1, 1) and intersect the lines fracx - 12 = fracy + 13 = fracz - 14$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and fracx - 31 = fracy - 42 = fracz1$\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z}{1}$ . Then, which of the following points lies on the line L?
A.(4,22,7)$(4,22,7)$
B.(5, 4, 3)$(5, 4, 3)$
C.(10, -29, -50)$(10, -29, -50)$
D.(7, 15, 13)$(7, 15, 13)$
Solution
### Related Formula
General coordinates of any variable point on a 3D line given symmetric form equations:
P_1 = (2lambda + 1, \, 3lambda - 1, \, 4lambda + 1)$$P_1 = (2\lambda + 1, \, 3\lambda - 1, \, 4\lambda + 1)$$P_2 = (mu + 3, \, 2mu + 4, \, mu)$$P_2 = (\mu + 3, \, 2\mu + 4, \, \mu)$$
### Core Logic
Let line L intersect Line 1 at point A(2lambda + 1, 3lambda - 1, 4lambda + 1)$A(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and Line 2 at point B(mu + 3, 2mu + 4, mu)$B(\mu + 3, 2\mu + 4, \mu)$.
Since line L passes through C(1, 1, 1)$C(1, 1, 1)$, the direction ratios computed from vector segment AC$AC$ must be proportional to the direction ratios computed from vector segment BC$BC$.
### Step 1: Determine Direction Ratio Parameters
Line Intersecting Two Lines diagram for Q68 - JEE Main 2025 Morning
Direction ratios of AC$AC$ segment:
vecAC = (2lambda + 1 - 1, \, 3lambda - 1 - 1, \, 4lambda + 1 - 1) = (2lambda, \, 3lambda - 2, \, 4lambda)$$\vec{AC} = (2\lambda + 1 - 1, \, 3\lambda - 1 - 1, \, 4\lambda + 1 - 1) = (2\lambda, \, 3\lambda - 2, \, 4\lambda)$$
Direction ratios of BC$BC$ segment:
vecBC = (mu + 3 - 1, \, 2mu + 4 - 1, \, mu - 1) = (mu + 2, \, 2mu + 3, \, mu - 1)$$\vec{BC} = (\mu + 3 - 1, \, 2\mu + 4 - 1, \, \mu - 1) = (\mu + 2, \, 2\mu + 3, \, \mu - 1)$$
Equating directional proportionality ratios:
fracmu + 22lambda = frac2mu + 33lambda - 2 = fracmu - 14lambda$$\frac{\mu + 2}{2\lambda} = \frac{2\mu + 3}{3\lambda - 2} = \frac{\mu - 1}{4\lambda}$$
### Step 2: Solve for Parameter Intersection values
From the first and third fractional groups:
fracmu + 22lambda = fracmu - 14lambda implies 2(mu + 2) = mu - 1$$\frac{\mu + 2}{2\lambda} = \frac{\mu - 1}{4\lambda} \implies 2(\mu + 2) = \mu - 1$$2mu + 4 = mu - 1 implies mu = -5$$2\mu + 4 = \mu - 1 \implies \mu = -5$$
Substitute mu = -5$\mu = -5$ back into the second parameter group linkage to evaluate the target structural direction indicators, which gives the simplified direction ratio vector for BC$BC$ as:
textD.R.s = (-5 + 2, \, 2(-5) + 3, \, -5 - 1) = (-3, \, -7, \, -6) equiv (3, \, 7, \, 6)$$\text{D.R.s} = (-5 + 2, \, 2(-5) + 3, \, -5 - 1) = (-3, \, -7, \, -6) \equiv (3, \, 7, \, 6)$$
### Step 3: Construct Line Equation and Verify Choice
Equation of line L passing through C(1, 1, 1)$C(1, 1, 1)$ with direction vector (3, 7, 6)$(3, 7, 6)$:
fracx - 13 = fracy - 17 = fracz - 16$$\frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}$$
Let's check option (7, 15, 13)$(7, 15, 13)$:
frac7 - 13 = frac63 = 2$$\frac{7 - 1}{3} = \frac{6}{3} = 2$$frac15 - 17 = frac147 = 2$$\frac{15 - 1}{7} = \frac{14}{7} = 2$$frac13 - 16 = frac126 = 2$$\frac{13 - 1}{6} = \frac{12}{6} = 2$$
Since all values match perfectly, (7, 15, 13)$(7, 15, 13)$ lies on the line L.
### Pattern Recognition
By comparing the first and third fractional terms containing lambda$\lambda$ in the denominator, you can solve for mu$\mu$ independently without tracking complex cross-multiplied quadratic lambdamu$\lambda\mu$ variations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
More Three Dimensional Geometry Questions — jee_main_2026_21_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
