If the image of the point mathrmP(1, 0, 3)$\mathrm{P}(1, 0, 3)$ in the line joining the points mathrmA(4, 7, 1)$\mathrm{A}(4, 7, 1)$ and mathrmB(3, 5, 3)$\mathrm{B}(3, 5, 3)$ is mathrmQ(alpha, beta, gamma)$\mathrm{Q}(\alpha, \beta, \gamma)$, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to
A.frac473$\frac{47}{3}$
B.frac463$\frac{46}{3}$
C.18$18$
D.13$13$
Solution & Explanation
### 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$$\text{Equation of a line through } (x_1, y_1, z_1) \text{ with direction ratios } (a, b, c): \quad \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$$textCondition for Perpendicular Lines: quad a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$\text{Condition for Perpendicular Lines:} \quad a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$$
### Core Logic
Let R$R$ be the foot of the perpendicular drawn from point P(1,0,3)$P(1,0,3)$ to the line joining A(4,7,1)$A(4,7,1)$ and B(3,5,3)$B(3,5,3)$. Since Q(alpha,beta,gamma)$Q(\alpha,\beta,\gamma)$ is the reflection (image) of P$P$ across the line, point R$R$ serves as the midpoint of the line segment PQ$PQ$.
### 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)$$\vec{d} = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) \equiv (1, 2, -2)$$
Using point B(3,5,3)$B(3,5,3)$, the symmetric equation of the line AB is:
fracx - 31 = fracy - 52 = fracz - 3-2 = lambda$$\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 3}{-2} = \lambda$$
Any general point R$R$ on this line can be written in terms of parameter lambda$\lambda$:
R equiv (lambda + 3, 2lambda + 5, -2lambda + 3)$$R \equiv (\lambda + 3, 2\lambda + 5, -2\lambda + 3)$$
### 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)$$\vec{\mathrm{PR}} = (\lambda + 3 - 1, \, 2\lambda + 5 - 0, \, -2\lambda + 3 - 3) = (\lambda + 2, \, 2\lambda + 5, \, -2\lambda)$$
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$$1(\lambda + 2) + 2(2\lambda + 5) - 2(-2\lambda) = 0$$lambda + 2 + 4lambda + 10 + 4lambda = 0$$\lambda + 2 + 4\lambda + 10 + 4\lambda = 0$$lambda = -frac43$$\lambda = -\frac{4}{3}$$
### Step 3: Coordinates of Foot of Perpendicular
Substitute lambda = -frac43$\lambda = -\frac{4}{3}$ into the general coordinates of R$R$:
R equiv left(-frac43 + 3, \, 2left(-frac43right) + 5, \, -2left(-frac43right) + 3right)$$R \equiv \left(-\frac{4}{3} + 3, \, 2\left(-\frac{4}{3}\right) + 5, \, -2\left(-\frac{4}{3}\right) + 3\right)$$R equiv left(frac53, \, frac73, \, frac173right)$$R \equiv \left(\frac{5}{3}, \, \frac{7}{3}, \, \frac{17}{3}\right)$$
### Step 4: Solve for the Image Coordinates and Sum
Since R$R$ is the midpoint of PQ$PQ$, where P = (1, 0, 3)$P = (1, 0, 3)$ and Q = (alpha, beta, gamma)$Q = (\alpha, \beta, \gamma)$:
- fracalpha + 12 = frac53 implies alpha = frac103 - 1 = frac73$\frac{\alpha + 1}{2} = \frac{5}{3} \implies \alpha = \frac{10}{3} - 1 = \frac{7}{3}$
- fracbeta + 02 = frac73 implies beta = frac143$\frac{\beta + 0}{2} = \frac{7}{3} \implies \beta = \frac{14}{3}$
- fracgamma + 32 = frac173 implies gamma = frac343 - 3 = frac253$\frac{\gamma + 3}{2} = \frac{17}{3} \implies \gamma = \frac{34}{3} - 3 = \frac{25}{3}$
Now, let us calculate the sum:
alpha + beta + gamma = frac73 + frac143 + frac253 = frac463$$\alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$$
### Pattern Recognition
Standard Midpoint reflection: The image coordinates are given directly by x_textimage = 2 x_textfoot - x_textpoint$x_{\text{image}} = 2 x_{\text{foot}} - x_{\text{point}}$, y_textimage = 2 y_textfoot - y_textpoint$y_{\text{image}} = 2 y_{\text{foot}} - y_{\text{point}}$, and z_textimage = 2 z_textfoot - z_textpoint$z_{\text{image}} = 2 z_{\text{foot}} - z_{\text{point}}$. Finding the parameter lambda$\lambda$ 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
Keywords:#reflection of point line 3D geometry#JEE Main 2025 Evening Q51#foot of perpendicular parameter lambda#direction ratios dot product perpendicular
More Three Dimensional Geometry Previous-Year Questions
Q14jee_main_2026_21_jan_morningFoot of Perpendicular and Projection
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
### 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
Q56jee_main_2025_02_april_eveningShortest Distance between Skew Lines
The line mathrmL_1$\mathrm{L}_1$ is parallel to the vector vecmathrma = -3hatmathrmi + 2hatmathrmj + 4hatmathrmk$\vec{\mathrm{a}} = -3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 4\hat{\mathrm{k}}$ and passes through the point (7, 6, 2)$(7, 6, 2)$ and the line mathrmL_2$\mathrm{L}_2$ is parallel to the vector vecmathrmb = 2hatmathrmi + hatmathrmj + 3hatmathrmk$\vec{\mathrm{b}} = 2\hat{\mathrm{i}} + \hat{\mathrm{j}} + 3\hat{\mathrm{k}}$ and passes through the point (5, 3, 4)$(5, 3, 4)$. The shortest distance between the lines mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ is:
A.frac23sqrt38$\frac{23}{\sqrt{38}}$
B.frac21sqrt57$\frac{21}{\sqrt{57}}$
C.frac23sqrt57$\frac{23}{\sqrt{57}}$
D.frac21sqrt38$\frac{21}{\sqrt{38}}$
Solution
### Related Formula
textShortest Distance d = fracleft| (vecr_2 - vecr_1) cdot (veca times vecb) right||veca times vecb|$$\text{Shortest Distance } d = \frac{\left| (\vec{r}_2 - \vec{r}_1) \cdot (\vec{a} \times \vec{b}) \right|}{|\vec{a} \times \vec{b}|}$$
### 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)$P_1(7, 6, 2)$ on L_1$L_1$ and P_2(5, 3, 4)$P_2(5, 3, 4)$ on L_2$L_2$:
vecr_2 - vecr_1 = (5 - 7)hatmathrmi + (3 - 6)hatmathrmj + (4 - 2)hatmathrmk = -2hatmathrmi - 3hatmathrmj + 2hatmathrmk$$\vec{r}_2 - \vec{r}_1 = (5 - 7)\hat{\mathrm{i}} + (3 - 6)\hat{\mathrm{j}} + (4 - 2)\hat{\mathrm{k}} = -2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$$
### 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$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix}$$veca times vecb = hatmathrmi(6 - 4) - hatmathrmj(-9 - 8) + hatmathrmk(-3 - 4) = 2hatmathrmi + 17hatmathrmj - 7hatmathrmk$$\vec{a} \times \vec{b} = \hat{\mathrm{i}}(6 - 4) - \hat{\mathrm{j}}(-9 - 8) + \hat{\mathrm{k}}(-3 - 4) = 2\hat{\mathrm{i}} + 17\hat{\mathrm{j}} - 7\hat{\mathrm{k}}$$
### 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$$(\vec{r}_2 - \vec{r}_1) \cdot (\vec{a} \times \vec{b}) = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69$$
Calculate the magnitude of the cross product:
|veca times vecb| = sqrt2^2 + 17^2 + (-7)^2 = sqrt4 + 289 + 49 = sqrt342 = 3sqrt38$$|\vec{a} \times \vec{b}| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{4 + 289 + 49} = \sqrt{342} = 3\sqrt{38}$$
Now, compute the shortest distance:
d = frac|-69|3sqrt38 = frac23sqrt38$$d = \frac{|-69|}{3\sqrt{38}} = \frac{23}{\sqrt{38}}$$
### 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)$(\vec{r}_2-\vec{r}_1)$, veca$\vec{a}$, and vecb$\vec{b}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q74jee_main_2025_02_april_eveningArea of Triangles
Let mathrmA(4, -2)$\mathrm{A}(4, -2)$, mathrmB(1, 1)$\mathrm{B}(1, 1)$ and mathrmC(9, -3)$\mathrm{C}(9, -3)$ 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|$$\text{Area of triangle ABC} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$textMaximum area of an inscribed parallelogram = frac12 times textArea of triangle$$\text{Maximum area of an inscribed parallelogram} = \frac{1}{2} \times \text{Area of triangle}$$
### 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)$A(4, -2)$, B(1, 1)$B(1, 1)$, and C(9, -3)$C(9, -3)$:
textArea(Delta textABC) = frac12 beginvmatrix 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 endvmatrix$$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \begin{vmatrix} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 \end{vmatrix}$$textArea(Delta textABC) = frac12 left| 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) right|$$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) \right|$$textArea(Delta textABC) = frac12 left| 4(4) + 2(-8) + 1(-12) right|$$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 4(4) + 2(-8) + 1(-12) \right|$$textArea(Delta textABC) = frac12 left| 16 - 16 - 12 right| = 6 text square units$$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 16 - 16 - 12 \right| = 6 \text{ square units}$$
### Step 2: Apply the maximum area theorem
The maximum area of a parallelogram inscribed in a triangle of area Delta$\Delta$ is always exactly half the area of the triangle:
textMaximum Area = frac12 times textArea(Delta textABC) = frac12 times 6 = 3 text square units$$\text{Maximum Area} = \frac{1}{2} \times \text{Area}(\Delta \text{ABC}) = \frac{1}{2} \times 6 = 3 \text{ square units}$$
### Pattern Recognition
Inscribed shapes maximization: The maximum area of any inscribed parallelogram AFDE$AFDE$ on the sides of a triangle ABC$ABC$ occurs when the vertices D, E, F$D, E, F$ are exactly the midpoints of the respective sides of the triangle.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Qjee_main_2025_02_april_morningArea of Triangle in 3D Space
Let the vertices Q$Q$ and R$R$ of the triangle PQR$PQR$ lie on the line fracx+35=fracy-12=fracz+43$\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$, QR=5$QR=5$ and the coordinates of the point P$P$ be (0, 2, 3)$(0, 2, 3)$. If the area of the triangle PQR$PQR$ is fracmn$\frac{m}{n}$ then:
A.m-5sqrt21n=0$m-5\sqrt{21}n=0$
B.2m-5sqrt21n=0$2m-5\sqrt{21}n=0$
C.5m-2sqrt21n=0$5m-2\sqrt{21}n=0$
D.5m-21sqrt2n=0$5m-21\sqrt{2}n=0$
Solution
### Related Formula
Area of a triangle given base length b$b$ and perpendicular height h$h$:
textArea = frac12 cdot b cdot h$$\text{Area} = \frac{1}{2} \cdot b \cdot h$$
### Core Logic
Since Q$Q$ and R$R$ lie on the line, the length QR=5$QR=5$ forms the base of the triangle. The perpendicular distance from point P$P$ to the line represents the height h$h$. Area of Triangle in 3D Space diagram for Q67 - JEE Main 2025 Morning
### Step 1: Define Perpendicular Foot coordinates
Let M$M$ be the foot of the perpendicular from P(0,2,3)$P(0,2,3)$ to the line. Parametric form of any point on the line:
M(5lambda - 3, 2lambda + 1, 3lambda - 4)$$M(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$$
Direction ratios of line segment PM$PM$:
textDRs = (5lambda - 3 - 0, 2lambda + 1 - 2, 3lambda - 4 - 3) = (5lambda - 3, 2lambda - 1, 3lambda - 7)$$\text{DRs} = (5\lambda - 3 - 0, 2\lambda + 1 - 2, 3\lambda - 4 - 3) = (5\lambda - 3, 2\lambda - 1, 3\lambda - 7)$$
### Step 2: Solve for Parameter using Perpendicularity
Since PM$PM$ is perpendicular to the given line (DRs: 5, 2, 3$5, 2, 3$):
5(5lambda - 3) + 2(2lambda - 1) + 3(3lambda - 7) = 0$$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$$25lambda - 15 + 4lambda - 2 + 9lambda - 21 = 0 implies 38lambda = 38 implies lambda = 1$$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \implies 38\lambda = 38 \implies \lambda = 1$$
### Step 3: Compute Perpendicular Distance
For lambda = 1$\lambda = 1$, the coordinates of M$M$ are (2, 3, -1)$(2, 3, -1)$. Calculate length PM$PM$:
PM = sqrt(2-0)^2 + (3-2)^2 + (-1-3)^2 = sqrt4 + 1 + 16 = sqrt21$$PM = \sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$$
### Step 4: Formulate the Area Equation
textArea = frac12 cdot QR cdot PM = frac12 cdot 5 cdot sqrt21 = fracmn$$\text{Area} = \frac{1}{2} \cdot QR \cdot PM = \frac{1}{2} \cdot 5 \cdot \sqrt{21} = \frac{m}{n}$$frac5sqrt212 = fracmn implies 2m = 5sqrt21n implies 2m - 5sqrt21n = 0$$\frac{5\sqrt{21}}{2} = \frac{m}{n} \implies 2m = 5\sqrt{21}n \implies 2m - 5\sqrt{21}n = 0$$
### 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
More Three Dimensional Geometry Questions — jee_main_2025_02_april_evening
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