Let P be the image of the point Q(7,-2,5)$Q(7,-2,5)$ in the line L: fracx-12=fracy+13=fracz4$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and R(5,p,q)$R(5,p,q)$ be a point on L. Then the square of the area of triangle PQR$\triangle PQR$ is \_\_\_\_. [cite: 3405, 3406]
Numerical Answer Type:
Enter a numerical valueAnswer: 957+4 marks
Solution & Explanation
### Related Formula
1. Area of a \triangle with perpendicular height h$h$ and base b$b$:
textArea = frac12 times b times h$\text{Area} = \frac{1}{2} \times b \times h$
2. Since P$P$ is the reflection image of Q$Q$ across line L$L$, the line acts as a perpendicular bisector. For any point R$R$ lying on the line, the height from R$R$ to the line is RT$RT$, and the base QP = 2QT$QP = 2QT$.
### Core Logic
Determine parameters for point R$R$ lying directly on line L [cite: 3405, 4071]:
frac5-12 = fracp+13 = fracq4 Rightarrow 2 = fracp+13 = fracq4$\frac{5-1}{2} = \frac{p+1}{3} = \frac{q}{4} \Rightarrow 2 = \frac{p+1}{3} = \frac{q}{4}$p+1 = 6 Rightarrow p = 5, quad q = 8 Rightarrow R = (5, 5, 8)$p+1 = 6 \Rightarrow p = 5, \quad q = 8 \Rightarrow R = (5, 5, 8)$ [cite: 4071, 4073]
3D line reflection \triangle diagram for Q72 - JEE Main 2025 Evening
### Step 1: Locate Foot of Perpendicular (T$T$)
Let the foot of the perpendicular from Q(7, -2, 5)$Q(7, -2, 5)$ on line L$L$ be T(2lambda+1, 3lambda-1, 4lambda)$T(2\lambda+1, 3\lambda-1, 4\lambda)$ .
The directional direction of L$L$ is vecb = 2hati + 3hatj + 4hatk$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ .
Vector overrightarrowQT = (2lambda - 6)hati + (3lambda + 1)hatj + (4lambda - 5)hatk$\overrightarrow{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k}$ .
Apply orthogonality condition overrightarrowQT cdot vecb = 0$\overrightarrow{QT} \cdot \vec{b} = 0$ :
2(2lambda - 6) + 3(3lambda + 1) + 4(4lambda - 5) = 0$2(2\lambda - 6) + 3(3\lambda + 1) + 4(4\lambda - 5) = 0$4lambda - 12 + 9lambda + 3 + 16lambda - 20 = 0 Rightarrow 29lambda - 29 = 0 Rightarrow lambda = 1$4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \Rightarrow 29\lambda - 29 = 0 \Rightarrow \lambda = 1$ [cite: 4077, 4078]
Thus, T = (3, 2, 4)$T = (3, 2, 4)$.
### Step 2: Measure Geometric Distances
Compute length QT$QT$ using distance metrics :
QT = sqrt(3-7)^2 + (2 - (-2))^2 + (4-5)^2 = sqrt16 + 16 + 1 = sqrt33$QT = \sqrt{(3-7)^2 + (2 - (-2))^2 + (4-5)^2} = \sqrt{16 + 16 + 1} = \sqrt{33}$
Since P$P$ is the symmetrical image, base QP = 2QT = 2sqrt33$QP = 2QT = 2\sqrt{33}$.
Compute length RT$RT$ representing height from vertex R(5, 5, 8)$R(5, 5, 8)$ to base line at T(3, 2, 4)$T(3, 2, 4)$ :
RT = sqrt(5-3)^2 + (5-2)^2 + (8-4)^2 = sqrt4 + 9 + 16 = sqrt29$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$
### Step 3: Calculate Squared Area
Compute the \triangle area squared value [cite: 3406, 4081]:
textArea = frac12 times QP times RT = frac12 times left(2sqrt33right) times sqrt29 = sqrt957$\text{Area} = \frac{1}{2} \times QP \times RT = \frac{1}{2} \times \left(2\sqrt{33}\right) \times \sqrt{29} = \sqrt{957}$left(textArearight)^2 = 957$\left(\text{Area}\right)^2 = 957$ [cite: 4081, 4083]
### Pattern Recognition
Because the image geometry creates an isosceles pairing from any point on the mirror line to the object and image point, the area reduces beautifully to 2 times textArea(triangle QTR) = QT times RT$2 \times \text{Area}(\triangle QTR) = QT \times RT$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Keywords:#image of point line 3D geometry#JEE Main 2025 Evening Q72#foot of perpendicular line \lambda method#\triangle area in 3D space
More Three Dimensional Geometry Previous-Year Questions — Page 6
Q682025Intersection of Lines in 3D Space
Let the line passing through the points (-1, 2, 1)$(-1, 2, 1)$ and \parallel to the line fracx - 12 = fracy + 13 = fracz4$\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4}$ intersect the line fracx + 23 = fracy - 32 = fracz - 41$\frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1}$ at the point P$P$. Then the distance of P$P$ from the point Q(4, -5, 1)$Q(4, -5, 1)$ is :
A.5$5$
B.10$10$
C.5sqrt6$5\sqrt{6}$
D.5sqrt5$5\sqrt{5}$
Solution
### Related Formula
The line passing through veca$\vec{a}$ and \parallel to direction vector vecv = lhatmathbfi + mhatmathbfj + nhatmathbfk$\vec{v} = l\hat{\mathbf{i}} + m\hat{\mathbf{j}} + n\hat{\mathbf{k}}$ is written in symmetric form as:
fracx - x_al = fracy - y_am = fracz - z_an$\frac{x - x_a}{l} = \frac{y - y_a}{m} = \frac{z - z_a}{n}$
### Core Logic
Formulate the equation of the line passing through (-1, 2, 1)$(-1, 2, 1)$ with direction vector components (2, 3, 4)$(2, 3, 4)$:
L_1: fracx + 12 = fracy - 23 = fracz - 14 = lambda quad dots (1)$L_1: \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4} = \lambda \quad \dots (1)$Intersection of Lines in 3D Space
Any generic point on this line can be written as:
P = (2lambda - 1, \, 3lambda + 2, \, 4lambda + 1)$P = (2\lambda - 1, \, 3\lambda + 2, \, 4\lambda + 1)$
The second given line equation is:
L_2: fracx + 23 = fracy - 32 = fracz - 41 = mu quad dots (2)$L_2: \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = \mu \quad \dots (2)$
Any generic point on line L_2$L_2$ is:
P' = (3mu - 2, \, 2mu + 3, \, mu + 4)$P' = (3\mu - 2, \, 2\mu + 3, \, \mu + 4)$
### Step 1: Compute the Point of Intersection
At the point of intersection P$P$, equate the coordinates from both lines:
2lambda - 1 = 3mu - 2 implies 2lambda - 3mu = -1 quad dots (3)$2\lambda - 1 = 3\mu - 2 \implies 2\lambda - 3\mu = -1 \quad \dots (3)$3lambda + 2 = 2mu + 3 implies 3lambda - 2mu = 1 quad dots (4)$3\lambda + 2 = 2\mu + 3 \implies 3\lambda - 2\mu = 1 \quad \dots (4)$4lambda + 1 = mu + 4 implies 4lambda - mu = 3 quad dots (5)$4\lambda + 1 = \mu + 4 \implies 4\lambda - \mu = 3 \quad \dots (5)$
Solving equations (3) and (4) simultaneously gives:
lambda = 1, quad mu = 1$\lambda = 1, \quad \mu = 1$
Verify these values using equation (5):
4(1) - 1 = 3 quad (textSatisfied)$4(1) - 1 = 3 \quad (\text{Satisfied})$
Thus, substituting lambda = 1$\lambda = 1$ gives the coordinates of point P$P$:
P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$
### Step 2: Calculate Euclidean Distance PQ
Find the distance between P(1, 5, 5)$P(1, 5, 5)$ and Q(4, -5, 1)$Q(4, -5, 1)$ using the 3D distance formula:
PQ = sqrt(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2$PQ = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}$PQ = sqrt3^2 + (-10)^2 + (-4)^2 = sqrt9 + 100 + 16$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16}$PQ = sqrt125 = 5sqrt5$PQ = \sqrt{125} = 5\sqrt{5}$
### Pattern Recognition
When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q582025Distance from a Point to a Line
The square of the distance of the point (frac157,frac327,7)$(\frac{15}{7},\frac{32}{7},7)$ from the line fracx+13=fracy+35=fracz+57$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector hati+4hatj+7hatk$\hat{i}+4\hat{j}+7\hat{k}$ is:
A.54$54$
B.41$41$
C.66$66$
D.44$44$
Solution
### Related Formula
Equation of a line passing through P(x_0, y_0, z_0)$P(x_0, y_0, z_0)$ along direction vector vecv = ahati+bhatj+chatk$\vec{v} = a\hat{i}+b\hat{j}+c\hat{k}$:
fracx-x_0a = fracy-y_0b = fracz-z_0c = lambda$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$
### Core Logic
Let the given point be Pleft(frac157, frac327, 7right)$P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$.
We need to find the distance measured along the line passing through P$P$ parallel to the vector vecv = hati+4hatj+7hatk$\vec{v} = \hat{i}+4\hat{j}+7\hat{k}$.
Equation of this line PQ$PQ$ is:
fracx - frac1571 = fracy - frac3274 = fracz - 77 = lambda$\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = \lambda$
Any general point Q$Q$ on this line can be written as:
Qleft(lambda + frac157, 4lambda + frac327, 7lambda + 7right)$Q\left(\lambda + \frac{15}{7}, 4\lambda + \frac{32}{7}, 7\lambda + 7\right)$
### Step 1: Find Intersection Point Q with Given Line
Point Q$Q$ must lie on the given target line L: fracx+13 = fracy+35 = fracz+57$L: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$.
Substitute coordinates of Q$Q$ into the first and third fractions:
fracleft(lambda + frac157right) + 13 = frac(7lambda + 7) + 57$\frac{\left(\lambda + \frac{15}{7}\right) + 1}{3} = \frac{(7\lambda + 7) + 5}{7}$fraclambda + frac2273 = frac7lambda + 127$\frac{\lambda + \frac{22}{7}}{3} = \frac{7\lambda + 12}{7}$frac7lambda + 2221 = frac7lambda + 127$\frac{7\lambda + 22}{21} = \frac{7\lambda + 12}{7}$
Multiplying by 21:
7lambda + 22 = 3(7lambda + 12)$7\lambda + 22 = 3(7\lambda + 12)$7lambda + 22 = 21lambda + 36$7\lambda + 22 = 21\lambda + 36$14lambda = -14 implies lambda = -1$14\lambda = -14 \implies \lambda = -1$
### Step 2: Calculate Coordinates of Q and Distance squared
Substituting lambda = -1$\lambda = -1$ into coordinates of Q$Q$:
Qleft(-1 + frac157, -4 + frac327, -7 + 7right) = Qleft(frac87, frac47, 0right)$Q\left(-1 + \frac{15}{7}, -4 + \frac{32}{7}, -7 + 7\right) = Q\left(\frac{8}{7}, \frac{4}{7}, 0\right)$
Now, compute the distance squared (PQ)^2$(PQ)^2$:
(PQ)^2 = left(frac157 - frac87right)^2 + left(frac327 - frac47right)^2 + (7 - 0)^2$(PQ)^2 = \left(\frac{15}{7} - \frac{8}{7}\right)^2 + \left(\frac{32}{7} - \frac{4}{7}\right)^2 + (7 - 0)^2$(PQ)^2 = left(frac77right)^2 + left(frac287
ight)^2 + 7^2 = 1^2 + 4^2 + 49 = 1 + 16 + 49 = 66$(PQ)^2 = \left(\frac{7}{7}\right)^2 + \left(\frac{28}{7}
ight)^2 + 7^2 = 1^2 + 4^2 + 49 = 1 + 16 + 49 = 66$
### Pattern Recognition
Instead of finding coordinates of Q$Q$, we could also use the vector form directly: vecPQ = lambda(hati + 4hatj + 7hatk)$\vec{PQ} = \lambda(\hat{i} + 4\hat{j} + 7\hat{k})$. Distance squared is PQ^2 = lambda^2(1^2 + 4^2 + 7^2) = (-1)^2(1 + 16 + 49) = 66$PQ^2 = \lambda^2(1^2 + 4^2 + 7^2) = (-1)^2(1 + 16 + 49) = 66$. This saves a lot of fractional coordinate subtraction arithmetic!
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q642025Intersection of Lines
Let veca = hati + 2hatj + hatk$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and vecb = 2hati + 7hatj + 3hatk$\vec{b} = 2\hat{i} + 7\hat{j} + 3\hat{k}$ . Let mathrmL_1:vecmathrmr = left(-hatmathrmi +2hatmathrmj +hatmathrmkright) + lambda vecmathrma,lambda in mathbbR$\mathrm{L}_1:\vec{\mathrm{r}} = \left(-\hat{\mathrm{i}} +2\hat{\mathrm{j}} +\hat{\mathrm{k}}\right) + \lambda \vec{\mathrm{a}},\lambda \in \mathbb{R}$ and mathrmL_2:vecmathrmr = left(hatmathrmj +hatmathrmkright) + mu vecmathrmb,mu in mathbbR$\mathrm{L}_2:\vec{\mathrm{r}} = \left(\hat{\mathrm{j}} +\hat{\mathrm{k}}\right) + \mu \vec{\mathrm{b}},\mu \in \mathbb{R}$ be two lines. If the line mathrmL_3$\mathrm{L}_3$ passes through the point of intersection of mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ , and is \parallel to vecmathrma +vecb$\vec{\mathrm{a}} +\vec{\b}$, then mathrmL_3$\mathrm{L}_3$ passes through the point:
Let L_1: fracx - 11 = fracy - 2-1 = fracz - 12$L_1: \frac{x - 1}{1} = \frac{y - 2}{-1} = \frac{z - 1}{2}$ and mathrmL_2:fracmathrmx + 1-1 = fracmathrmy - 22 = fracmathrmz1$\mathrm{L}_2:\frac{\mathrm{x} + 1}{-1} = \frac{\mathrm{y} - 2}{2} = \frac{\mathrm{z}}{1}$ be two lines. Let mathrmL_3$\mathrm{L}_3$ be a line passing through the point (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ and be perpendicular to both mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ . If mathrmL_3$\mathrm{L}_3$ intersects mathrmL_1$\mathrm{L}_1$ , then |5alpha - 11beta - 8gamma|$|5\alpha - 11\beta - 8\gamma|$ equals:
A. 18
B. 16
C. 25
D. 20
Solution
### Related Formula
textDirection ratios of a line perpendicular to both vectors: vecv = vecd_1 times vecd_2$\text{Direction ratios of a line perpendicular to both vectors: } \vec{v} = \vec{d}_1 \times \vec{d}_2$
### Core Logic
Compute direction ratios for L_3$L_3$ using cross product of direction vectors of L_1$L_1$ and L_2$L_2$:
vecv = left| beginarrayccc hati & hatj & hatk \\ 1 & -1 & 2 \\ -1 & 2 & 1 endarray right| = -5hati - 3hatj + hatk$\vec{v} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{array} \right| = -5\hat{i} - 3\hat{j} + \hat{k}$
### Step 1: Establish intersection constraints
Let A$A$ be a general point on L_3$L_3$ and B$B$ be a general point on L_1$L_1$:
A = (alpha - 5lambda, beta - 3lambda, gamma + lambda)$A = (\alpha - 5\lambda, \beta - 3\lambda, \gamma + \lambda)$B = (k+1, -k+2, 2k+1)$B = (k+1, -k+2, 2k+1)$
Since L_3$L_3$ intersects L_1$L_1$, at the point of intersection A equiv B$A \equiv B$:
alpha - 5lambda = k + 1 implies alpha = 5lambda + k + 1$\alpha - 5\lambda = k + 1 \implies \alpha = 5\lambda + k + 1$beta - 3lambda = -k + 2 implies beta = 3lambda - k + 2$\beta - 3\lambda = -k + 2 \implies \beta = 3\lambda - k + 2$gamma + lambda = 2k + 1 implies gamma = -lambda + 2k + 1$\gamma + \lambda = 2k + 1 \implies \gamma = -\lambda + 2k + 1$
### Step 2: Solve final algebraic target equation
Substitute expressions into the target mod expression:
5alpha - 11beta - 8gamma = 5(5lambda + k + 1) - 11(3lambda - k + 2) - 8(-lambda + 2k + 1)$5\alpha - 11\beta - 8\gamma = 5(5\lambda + k + 1) - 11(3\lambda - k + 2) - 8(-\lambda + 2k + 1)$= lambda(25 - 33 + 8) + k(5 + 11 - 16) + (5 - 22 - 8)$= \lambda(25 - 33 + 8) + k(5 + 11 - 16) + (5 - 22 - 8)$= 0lambda + 0k - 25 = -25$= 0\lambda + 0k - 25 = -25$
Taking absolute value yields |-25| = 25$|-25| = 25$.
### Pattern Recognition
The cancellation of internal variables (lambda$\lambda$ and k$k$) shows that the absolute expression describes a invariant geometric plane coordinate constraint, allowing simple baseline parameter values (like k=0, lambda=0$k=0, \lambda=0$) to evaluate the question instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
More Three Dimensional Geometry Questions — jee_main_2025_24_jan_evening
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