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 7
Q20jee_main_2024_01_february_morningShortest Distance Between Skew Lines
If the shortest distance between the lines fracx-lambda-2=fracy-21=fracz-11$\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and fracx-sqrt31=fracy-1-2=fracz-21$\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is 1$1$, then the sum of all possible values of lambda$\lambda$ is:
A.0$0$
B.2sqrt3$2\sqrt{3}$
C.3sqrt3$3\sqrt{3}$
D.-2sqrt3$-2\sqrt{3}$
Solution
### Related Formula
The shortest distance between two skew lines vecr = veca_1 + svecb_1$\vec{r} = \vec{a}_1 + s\vec{b}_1$ and vecr = veca_2 + tvecb_2$\vec{r} = \vec{a}_2 + t\vec{b}_2$ 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
Identify the vectors from the equations of the lines:
- Line 1 passes through veca_1 = lambdahati + 2hatj + hatk$\vec{a}_1 = \lambda\hat{i} + 2\hat{j} + \hat{k}$ with direction vecb_1 = -2hati + hatj + hatk$\vec{b}_1 = -2\hat{i} + \hat{j} + \hat{k}$.
- Line 2 passes through veca_2 = sqrt3hati + hatj + 2hatk$\vec{a}_2 = \sqrt{3}\hat{i} + \hat{j} + 2\hat{k}$ with direction vecb_2 = hati - 2hatj + hatk$\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$.
Calculate the coordinate difference vector:
veca_2 - veca_1 = (sqrt3 - lambda)hati - hatj + hatk$$\vec{a}_2 - \vec{a}_1 = (\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}$$
### Step 1: Compute the Cross Product of the Direction Vectors
Find the cross product matrix:
vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ -2 & 1 & 1 \\ 1 & -2 & 1 endvmatrix$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}$$vecb_1 times vecb_2 = hati(1 - (-2)) - hatj(-2 - 1) + hatk(4 - 1) = 3hati + 3hatj + 3hatk$$\vec{b}_1 \times \vec{b}_2 = \hat{i}(1 - (-2)) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
Compute its magnitude:
|vecb_1 times vecb_2| = sqrt3^2 + 3^2 + 3^2 = sqrt27 = 3sqrt3$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$The figure details the spatial arrangement of the two skew lines along with their common perpendicular normal vector representing the shortest path.
### Step 2: Solve the Absolute Distance Equation for lambda
Substitute these values into the shortest distance formula:
1 = frac|((sqrt3 - lambda)hati - hatj + hatk) cdot (3hati + 3hatj + 3hatk)|3sqrt3$$1 = \frac{|((\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})|}{3\sqrt{3}}$$1 = frac|3(sqrt3 - lambda) - 3 + 3|3sqrt3 = frac3|sqrt3 - lambda|3sqrt3 = frac|sqrt3 - lambda|sqrt3$$1 = \frac{|3(\sqrt{3} - \lambda) - 3 + 3|}{3\sqrt{3}} = \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} = \frac{|\sqrt{3} - \lambda|}{\sqrt{3}}$$
This gives:
|sqrt3 - lambda| = sqrt3$$|\sqrt{3} - \lambda| = \sqrt{3}$$
Unfolding the absolute value gives two cases:
- **Case A:** sqrt3 - lambda = sqrt3 implies lambda = 0$\sqrt{3} - \lambda = \sqrt{3} \implies \lambda = 0$
- **Case B:** sqrt3 - lambda = -sqrt3 implies lambda = 2sqrt3$\sqrt{3} - \lambda = -\sqrt{3} \implies \lambda = 2\sqrt{3}$
### Step 3: Calculate the Final Sum
The sum of all possible values of lambda$\lambda$ is:
textSum = 0 + 2sqrt3 = 2sqrt3$$\text{Sum} = 0 + 2\sqrt{3} = 2\sqrt{3}$$
### Pattern Recognition
Sees: Shortest distance setup between vector paths containing free variables.
Shortcut: In equations like |c - lambda| = d$|c - \lambda| = d$, the sum of the roots is simply equal to 2c$2c$, because the roots are symmetrically balanced around the center point c$c$. Thus, textSum = 2 times sqrt3 = 2sqrt3$\text{Sum} = 2 \times \sqrt{3} = 2\sqrt{3}$ holds instantly without calculating individual values.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 12 Mathematics: Vector Algebra
Q29jee_main_2024_01_february_morningShortest Distance Between Skew Lines
Let the line of the shortest distance between the lines L_1:vecr=(hati+2hatj+3hatk)+lambda(hati-hatj+hatk)$L_{1}:\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and L_2:vecr=(4hati+5hatj+6hatk)+mu(hati+hatj-hatk)$L_{2}:\vec{r}=(4\hat{i}+5\hat{j}+6\hat{k})+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect L_1$L_{1}$ and L_2$L_{2}$ at P and Q respectively. If (\alpha, \beta, \gamma) is the midpoint of the line segment PQ, then 2(alpha+beta+gamma)$2(\alpha+\beta+\gamma)$ is equal to
Numerical Answer.Answer: 21 to 21
Solution
### Related Formula
The vector connecting the shortest distance points P$P$ and Q$Q$ on two skew lines must be simultaneously perpendicular to the direction vectors vecb_1$\vec{b}_1$ and vecb_2$\vec{b}_2$ of both lines:
vecPQ parallel (vecb_1 times vecb_2)$$\vec{PQ} \parallel (\vec{b}_1 \times \vec{b}_2)$$
### Core Logic
Let's define general points on both lines:
- Point P$P$ on L_1$L_1$: (1+lambda, \, 2-lambda, \, 3+lambda)$(1+\lambda, \, 2-\lambda, \, 3+\lambda)$
- Point Q$Q$ on L_2$L_2$: (4+mu, \, 5+mu, \, 6-mu)$(4+\mu, \, 5+\mu, \, 6-\mu)$
The direction ratios of vector vecPQ$\vec{PQ}$ are:
vecPQ = (3+mu-lambda)hati + (3+mu+lambda)hatj + (3-mu-lambda)hatk$$\vec{PQ} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}$$
### Step 1: Compute Perpendicular Direction Vector
Calculate the cross product of the directions of lines L_1$L_1$ and L_2$L_2$:
vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 1 & -1 & 1 \\ 1 & 1 & -1 endvmatrix = 0hati + 2hatj + 2hatk$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 0\hat{i} + 2\hat{j} + 2\hat{k}$$
Since vecPQ$\vec{PQ}$ is parallel to (0, 2, 2)$(0, 2, 2)$, we compare the coordinate ratios:
3+mu-lambda = 0 implies lambda - mu = 3 quad implies (1)$$3+\mu-\lambda = 0 \implies \lambda - \mu = 3 \quad \implies (1)$$frac3+mu+lambda2 = frac3-mu-lambda2 implies 2mu + 2lambda = 0 implies lambda + mu = 0 quad implies (2)$$\frac{3+\mu+\lambda}{2} = \frac{3-\mu-\lambda}{2} \implies 2\mu + 2\lambda = 0 \implies \lambda + \mu = 0 \quad \implies (2)$$The graphic maps out the geometry of lines L1 and L2 intersected by their common perpendicular segment at points A and B.
### Step 2: Solve for Parameters and Midpoint
Solving linear equations (1) and (2) simultaneously:
lambda = frac32, quad mu = -frac32$$\lambda = \frac{3}{2}, \quad \mu = -\frac{3}{2}$$
Substitute these values back to find the specific coordinates of points P$P$ and Q$Q$:
- P = left(frac52, \, frac12, \, frac92
ight)$P = \left(\frac{5}{2}, \, \frac{1}{2}, \, \frac{9}{2}
ight)$
- Q = left(frac52, \, frac72, \, frac152
ight)$Q = \left(\frac{5}{2}, \, \frac{7}{2}, \, \frac{15}{2}
ight)$
The midpoint coordinates (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ are:
(alpha, beta, gamma) = left( frac5/2 + 5/22, \, frac1/2 + 7/22, \, frac9/2 + 15/22 right) = left(frac52, \, 2, \, 6
ight)$$(\alpha, \beta, \gamma) = \left( \frac{5/2 + 5/2}{2}, \, \frac{1/2 + 7/2}{2}, \, \frac{9/2 + 15/2}{2} \right) = \left(\frac{5}{2}, \, 2, \, 6
ight)$$
### Step 3: Final Computation
Calculate the required terms:
2(alpha+beta+gamma) = 2left(frac52 + 2 + 6right) = 5 + 4 + 12 = 21$$2(\alpha+\beta+\gamma) = 2\left(\frac{5}{2} + 2 + 6\right) = 5 + 4 + 12 = 21$$
### Pattern Recognition
Sees: Explicit endpoints of the shortest distance line vector segment.
Shortcut: Since the cross product component along hati$\hat{i}$ is 0, the x-coordinates of both line points are identical, providing a massive shortcut to check algebraic equations immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q3jee_main_2024_29_january_eveningAngle Between Two Lines
Let P(3,2,3)$P(3,2,3)$, Q(4,6,2)$Q(4,6,2)$ and R(7,3,2)$R(7,3,2)$ be the vertices of Delta PQR$\Delta PQR$. Then, the angle angle QPR$\angle QPR$ is
### Related Formula
cos theta = fracveca cdot vecb|veca||vecb|$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$
### Core Logic
Let us compute the direction ratios of the vectors overrightarrowPQ$\overrightarrow{PQ}$ and overrightarrowPR$\overrightarrow{PR}$ originating from vertex P$P$:
Direction ratios of overrightarrowPQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$\overrightarrow{PQ} = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$
Direction ratios of overrightarrowPR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$\overrightarrow{PR} = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$
### Step 1: Evaluating the Angle
Using the dot product formula for the angle theta = angle QPR$\theta = \angle QPR$:
cos theta = frac1(4) + 4(1) + (-1)(-1)sqrt1^2 + 4^2 + (-1)^2 cdot sqrt4^2 + 1^2 + (-1)^2$$\cos \theta = \frac{1(4) + 4(1) + (-1)(-1)}{\sqrt{1^2 + 4^2 + (-1)^2} \cdot \sqrt{4^2 + 1^2 + (-1)^2}}$$cos theta = frac4 + 4 + 1sqrt18 cdot sqrt18 = frac918 = frac12$$\cos \theta = \frac{4 + 4 + 1}{\sqrt{18} \cdot \sqrt{18}} = \frac{9}{18} = \frac{1}{2}$$Angle Between Two Lines diagram for Q3 - JEE Main 2024 Evening
Since cos theta = frac12$\cos \theta = \frac{1}{2}$, we have:
theta = fracpi3$$\theta = \frac{\pi}{3}$$
### Pattern Recognition
When asked for an angle like angle QPR$\angle QPR$, always ensure both vectors diverge from the common vertex P$P$ (i.e., use overrightarrowPQ$\overrightarrow{PQ}$ and overrightarrowPR$\overrightarrow{PR}$) to avoid sign errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q26jee_main_2024_29_january_eveningShortest Distance Between Two Lines
Let O$O$ be the origin, and M$M$ and N$N$ be the points on the lines fracx - 54 = fracy - 41 = fracz - 53$\frac{x - 5}{4} = \frac{y - 4}{1} = \frac{z - 5}{3}$ and fracx + 812 = fracy + 25 = fracz + 119$\frac{x + 8}{12} = \frac{y + 2}{5} = \frac{z + 11}{9}$ respectively such that MN$MN$ is the shortest distance between the given lines. Then overlinemathrmOMcdot overlinemathrmON$\overline{\mathrm{OM}}\cdot \overline{\mathrm{ON}}$ is equal to
Numerical Answer.Answer: 9 to 9
Solution
### Related Formula
The line segment of shortest distance MN$MN$ is perpendicular to both directing vectors vecb_1$\vec{b}_1$ and vecb_2$\vec{b}_2$.
### Core Logic
Let general point coordinates be expressions of parameters lambda$\lambda$ and mu$\mu$:
M = (4lambda + 5, lambda + 4, 3lambda + 5)$$M = (4\lambda + 5, \lambda + 4, 3\lambda + 5)$$N = (12mu - 8, 5mu - 2, 9mu - 11)$$N = (12\mu - 8, 5\mu - 2, 9\mu - 11)$$
Vector direction ratios of overrightarrowMN$\overrightarrow{MN}$:
overrightarrowMN = (4lambda - 12mu + 13, lambda - 5mu + 6, 3lambda - 9mu + 16)$$\overrightarrow{MN} = (4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16)$$
The cross product vector perpendicular to both lines is vecb_1 times vecb_2 = (-6, 0, 8)$\vec{b}_1 \times \vec{b}_2 = (-6, 0, 8)$.
### Step 1: Finding Parameters via Perpendicular Conditions
Equating directional proportional factors:
frac4lambda - 12mu + 13-6 = fraclambda - 5mu + 60 = frac3lambda - 9mu + 168$$\frac{4\lambda - 12\mu + 13}{-6} = \frac{\lambda - 5\mu + 6}{0} = \frac{3\lambda - 9\mu + 16}{8}$$
From the zero denominator constraint:
lambda - 5mu + 6 = 0 quad dots (iii)$$\lambda - 5\mu + 6 = 0 \quad \dots (iii)$$
From the first and third components:
8(4lambda - 12mu + 13) = -6(3lambda - 9mu + 16) implies 32lambda - 96mu + 104 = -18lambda + 54mu - 96$$8(4\lambda - 12\mu + 13) = -6(3\lambda - 9\mu + 16) \implies 32\lambda - 96\mu + 104 = -18\lambda + 54\mu - 96$$50lambda - 150mu + 200 = 0 implies lambda - 3mu + 4 = 0 quad dots (iv)$$50\lambda - 150\mu + 200 = 0 \implies \lambda - 3\mu + 4 = 0 \quad \dots (iv)$$
Solving equations (iii) and (iv) yields lambda = -1$\lambda = -1$ and mu = 1$\mu = 1$.
### Step 2: Vector Coordinate Evaluation
Substituting parameter roots into point layout vectors:
M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$$M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$$N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$$N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$$
Evaluating target scalar products:
overlinemathrmOM cdot overlinemathrmON = 1(4) + 3(3) + 2(-2) = 4 + 9 - 4 = 9$$\overline{\mathrm{OM}} \cdot \overline{\mathrm{ON}} = 1(4) + 3(3) + 2(-2) = 4 + 9 - 4 = 9$$
### Pattern Recognition
Shortest distance points constrain line segments to align with the direct cross product vector. Setting proportional components yields quick parameters.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q3jee_main_2024_27_jan_morningDistance of a Point from a Line
The distance of the point (7, -2, 11)$(7, -2, 11)$ from the line fracx-61=fracy-40=fracz-83$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line fracx-52=fracy-1-3=fracz-56$\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is:
A.12$12$
B.14$14$
C.18$18$
D.21$21$
Solution
### Related Formula
fracx-x_1a = fracy-y_1b = fracz-z_1c = lambda$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$$
### Core Logic
Let point A = (7, -2, 11)$A = (7, -2, 11)$. We want the distance from A$A$ to the line L_1: fracx-61=fracy-40=fracz-83$L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ measured *along* a line parallel to L_2: fracx-52=fracy-1-3=fracz-56$L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$.
The line passing through A$A$ parallel to L_2$L_2$ will have the equation:
fracx-72 = fracy+2-3 = fracz-116 = lambda$$\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$$
Any general point B$B$ on this new line can be represented as:
B equiv (2lambda + 7, -3lambda - 2, 6lambda + 11)$$B \equiv (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$$
### Step 1: Finding Intersection Point B
Since B$B$ lies on the given line L_1$L_1$, its coordinates must satisfy the equation of L_1$L_1$:
frac(2lambda + 7) - 61 = frac(-3lambda - 2) - 40 = frac(6lambda + 11) - 83$$\frac{(2\lambda + 7) - 6}{1} = \frac{(-3\lambda - 2) - 4}{0} = \frac{(6\lambda + 11) - 8}{3}$$
Focus on the middle term (since denominator is 0, numerator must equal 0 for intersection):
-3lambda - 6 = 0 Rightarrow lambda = -2$$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$$
### Step 2: Coordinates of B and Distance Calculation
Substitute lambda = -2$\lambda = -2$ back into the coordinates of point B$B$:
B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$$B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$$
Now, calculate the distance AB$AB$ using the 3D distance formula:
AB = sqrt(7-3)^2 + (-2-4)^2 + (11 - (-1))^2$$AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11 - (-1))^2}$$AB = sqrt4^2 + (-6)^2 + (12)^2$$AB = \sqrt{4^2 + (-6)^2 + (12)^2}$$AB = sqrt16 + 36 + 144$$AB = \sqrt{16 + 36 + 144}$$AB = sqrt196 = 14$$AB = \sqrt{196} = 14$$
### Pattern Recognition
Distance of a point from a line *along* another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
More Three Dimensional Geometry Questions — jee_main_2026_21_jan_morning
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