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 4
Q62jee_main_2025_28_jan_morningImage 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
Q52jee_main_2025_03_april_morningIntersection of Lines
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
### 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
Q70jee_main_2025_03_april_morningShortest Distance between Skew Lines
Line L_1$L_{1}$ passes through the point (1, 2, 3)$(1, 2, 3)$ and is parallel to z-axis[cite: 685]. Line L_2$L_{2}$ passes through the point (lambda, 5, 6)$(lambda, 5, 6)$ and is parallel to y-axis[cite: 686]. Let for lambda = lambda_1, lambda_2$\lambda = \lambda_{1}, \lambda_{2}$, lambda_2 < lambda_1$\lambda_{2} < \lambda_{1}$ the shortest distance between the two lines be 3[cite: 698]. Then the square of the distance of the point (lambda_1, lambda_2, 7)$(lambda_{1}, \lambda_{2}, 7)$ from the line L_1$L_{1}$ is[cite: 698, 700]:
A. 40
B. 32
C. 25
D. 37
Solution
### Related Formula
Shortest distance between perpendicular axes vectors:
textS.D. = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$$\text{S.D.} = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$
### Core Logic
Represent equations of straight lines symmetrically using vector directions [cite: 1418]:
L_1: fracx-10 = fracy-20 = fracz-31$$L_1: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$$ [cite: 1418]
L_2: fracx-lambda0 = fracy-51 = fracz-60$$L_2: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$$ [cite: 1418]
Evaluating the standard shortest distance configuration formula[cite: 1419, 1420]:
textS.D. = |lambda - 1| = 3 implies lambda - 1 = pm 3$$\text{S.D.} = |\lambda - 1| = 3 \implies \lambda - 1 = \pm 3$$ [cite: 1420]
lambda = 4 quad textor quad lambda = -2$$\lambda = 4 \quad \text{or} \quad \lambda = -2$$ [cite: 1420]
Given the condition lambda_2 < lambda_1$\lambda_2 < \lambda_1$ [cite: 698]:
lambda_1 = 4, quad lambda_2 = -2$$\lambda_1 = 4, \quad \lambda_2 = -2$$ [cite: 1421, 1422]
### Step 1: Distance calculation from line
We need to find the square of distance from point P(4, -2, 7)$P(4, -2, 7)$ to line L_1$L_1$ [cite: 1424].
Any general matching point coordinates tracking along path L_1$L_1$ look like Q(1, 2, t+3)$Q(1, 2, t+3)$ [cite: 1424].
Form a perpendicular projection vector condition [cite: 1425]:
vecPQ = (-3, 4, t-4)$$\vec{PQ} = (-3, 4, t-4)$$ [cite: 1425]
Since vecPQ cdot hatk = 0 implies t-4 = 0 implies t=4$\vec{PQ} \cdot \hat{k} = 0 \implies t-4 = 0 \implies t=4$ [cite: 1425, 1426].
Thus, the foot of perpendicular is Q(1, 2, 7)$Q(1, 2, 7)$ [cite: 1427].
Evaluate the squared distance component magnitude [cite: 1427]:
PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$$PQ^2 = (4-1)^2 + (-2-2)^2 + (7-7)^2 = 3^2 + (-4)^2 + 0 = 9 + 16 = 25$$ [cite: 1427, 1428]
### Pattern Recognition
For lines parallel directly to independent Cartesian coordinate grid lines, the shortest paths are simply direct plane projections.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q56jee_main_2025_04_april_eveningShortest Distance Between Two Lines
Let the values of p$p$, for which the shortest distance between the lines fracx + 13 = fracy4 = fracz5$\frac{x + 1}{3} = \frac{y}{4} = \frac{z}{5}$ and vecr = (phati + 2hatj + hatk) + lambda (2hati + 3hatj + 4hatk)$\vec{r} = (p\hat{i} + 2\hat{j} + hat{k}) + \lambda (2hat{i} + 3hat{j} + 4hat{k})$ is frac1sqrt6$\frac{1}{sqrt{6}}$, be a, b, (a < b$a < b$). Then the length of the latus rectum of the ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:
A.9$9$
B.frac32$\frac{3}{2}$
C.frac23$\frac{2}{3}$
D.18$18$
Solution
### Related Formula
The shortest distance between two lines vecr = veca_1 + lambda vecb_1$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and vecr = veca_2 + mu vecb_2$\vec{r} = \vec{a}_2 + \mu \vec{b}_2$ is given by:
d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$
### Core Logic
From the given line equations:
Line 1 passes through veca_1 = -hati + 0hatj + 0hatk$\vec{a}_1 = -\hat{i} + 0\hat{j} + 0\hat{k}$ along vector vecb_1 = 3hati + 4hatj + 5hatk$\vec{b}_1 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
Line 2 passes through veca_2 = phati + 2hatj + hatk$\vec{a}_2 = p\hat{i} + 2\hat{j} + \hat{k}$ along vector vecb_2 = 2hati + 3hatj + 4hatk$\vec{b}_2 = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
Vector difference:
veca_2 - veca_1 = (p + 1)hati + 2hatj + hatk$$\vec{a}_2 - \vec{a}_1 = (p + 1)\hat{i} + 2\hat{j} + \hat{k}$$
Computing the cross product vecb_1 times vecb_2$\vec{b}_1 \times \vec{b}_2$:
vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 3 & 4 & 5 \\ 2 & 3 & 4 endvmatrix = hati(16-15) - hatj(12-10) + hatk(9-8) = hati - 2hatj + hatk$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(16-15) - \hat{j}(12-10) + \hat{k}(9-8) = \hat{i} - 2\hat{j} + \hat{k}$$
Magnitude |vecb_1 times vecb_2| = sqrt1^2 + (-2)^2 + 1^2 = sqrt6$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}$.
### Step 1: Applying the Shortest Distance Value
Substitute these into the distance equation:
d = frac|big((p + 1)hati + 2hatj + hatkbig) cdot big(hati - 2hatj + hatkbig)|sqrt6 = frac1sqrt6$$d = \frac{|\big((p + 1)\hat{i} + 2\hat{j} + \hat{k}\big) \cdot \big(\hat{i} - 2\hat{j} + \hat{k}\big)|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$$|(p + 1)(1) + 2(-2) + 1(1)| = 1$$|(p + 1)(1) + 2(-2) + 1(1)| = 1$$|p + 1 - 4 + 1| = 1 implies |p - 2| = 1$$|p + 1 - 4 + 1| = 1 \implies |p - 2| = 1$$
This yields two values for p$p$:
- p - 2 = 1 implies p = 3$p - 2 = 1 \implies p = 3$
- p - 2 = -1 implies p = 1$p - 2 = -1 \implies p = 1$
Given that a, b$a, b$ are the parameters with a < b$a < b$, we assign a = 1$a = 1$ and b = 3$b = 3$.
### Step 2: Computing Latus Rectum of the Ellipse
The ellipse equation is:
fracx^21^2 + fracy^23^2 = 1$$\frac{x^2}{1^2} + \frac{y^2}{3^2} = 1$$
Since b > a$b > a$, the formula for the length of the latus rectum is:
textLatus Rectum = frac2a^2b = frac2(1)^23 = frac23$$\text{Latus Rectum} = \frac{2a^2}{b} = \frac{2(1)^2}{3} = \frac{2}{3}$$
### Pattern Recognition
Be careful with coordinate geometry variables; when an ellipse satisfies b > a$b > a$, the major axis is along the y-axis, making the latus rectum equal to frac2a^2b$\frac{2a^2}{b}$ instead of frac2b^2a$\frac{2b^2}{a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Conic Sections
Q69jee_main_2025_04_april_eveningAngle Between Two Lines
Let A be the point of intersection of the lines L _ 1: frac x - 71 = frac y - 50 = frac z - 3- 1$L _ {1}: \frac {x - 7}{1} = \frac {y - 5}{0} = \frac {z - 3}{- 1}$ and mathrmL_2:fracmathrmx - 13 = fracmathrmy + 34 = fracmathrmz + 75$\mathrm{L}_2:\frac{\mathrm{x} - 1}{3} = \frac{\mathrm{y} + 3}{4} = \frac{\mathrm{z} + 7}{5}$. Let B and C be the point on the lines mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ respectively such that mathrmAB = mathrmAC = sqrt15$\mathrm{AB} = \mathrm{AC} = \sqrt{15}$. Then the square of the area of the triangle ABC is :
A.54$54$
B.63$63$
C.57$57$
D.60$60$
Solution
### Core Logic
First, find the point of intersection A$A$ by solving the lines. Any point on L_1$L_1$ can be written as (lambda + 7, 5, -lambda + 3)$(\lambda + 7, 5, -\lambda + 3)$.
Substituting this point into the equation for L_2$L_2$:
frac(lambda + 7) - 13 = frac5 + 34 implies fraclambda + 63 = 2 implies lambda = 0$$\frac{(\lambda + 7) - 1}{3} = \frac{5 + 3}{4} \implies \frac{\lambda + 6}{3} = 2 \implies \lambda = 0$$
Thus, the intersection point is A = (7, 5, 3)$A = (7, 5, 3)$.
### Step 1: Calculating the Angle between lines
The directional vectors of lines L_1$L_1$ and L_2$L_2$ are vecu = hati - hatk$\vec{u} = \hat{i} - \hat{k}$ and vecv = 3hati + 4hatj + 5hatk$\vec{v} = 3hat{i} + 4hat{j} + 5hat{k}$ respectively.
costheta = frac|vecu cdot vecv||vecu||vecv| = frac|1(3) + 0(4) - 1(5)|sqrt1^2+(-1)^2 sqrt3^2+4^2+5^2 = frac|3 - 5|sqrt2sqrt50 = frac210 = frac15$$\cos\theta = \frac{|\vec{u} \cdot \vec{v}|}{|\vec{u}||\vec{v}|} = \frac{|1(3) + 0(4) - 1(5)|}{\sqrt{1^2+(-1)^2} \sqrt{3^2+4^2+5^2}} = \frac{|3 - 5|}{\sqrt{2}\sqrt{50}} = \frac{2}{10} = \frac{1}{5}$$
Now, find sintheta$\sin\theta$:
sintheta = sqrt1 - cos^2theta = sqrt1 - frac125 = fracsqrt245$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{1}{25}} = \frac{\sqrt{24}}{5}$$Three dimensional geometry diagram for Q69 - JEE Main 2025 Evening
### Step 2: Finding Area of the Triangle
The area of triangle ABC$\triangle ABC$ given two sides and their included angle is:
textArea = frac12 cdot AB cdot AC cdot sintheta$$\text{Area} = \frac{1}{2} \cdot AB \cdot AC \cdot \sin\theta$$
Given AB = AC = sqrt15$AB = AC = \sqrt{15}$:
textArea = frac12 cdot sqrt15 cdot sqrt15 cdot fracsqrt245 = frac15sqrt2410 = frac3sqrt242$$\text{Area} = \frac{1}{2} \cdot \sqrt{15} \cdot \sqrt{15} \cdot \frac{\sqrt{24}}{5} = \frac{15\sqrt{24}}{10} = \frac{3\sqrt{24}}{2}$$
Squaring the area:
textArea^2 = left(frac3sqrt242right)^2 = frac9 times 244 = 9 times 6 = 54$$\text{Area}^2 = \left(\frac{3\sqrt{24}}{2}\right)^2 = \frac{9 \times 24}{4} = 9 \times 6 = 54$$
### Pattern Recognition
Since B$B$ and C$C$ lie on lines intersecting at A$A$, you don't need to determine their exact coordinates to find the area of the triangle. The standard side-angle-side area formula works perfectly using just the directional angle.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Properties of Triangles
More Three Dimensional Geometry Questions — jee_main_2026_21_jan_morning
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