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 3
Q51jee_main_2025_08_april_eveningShortest Distance Between Lines
Let the values of lambda$\lambda$ for which the shortest distance between the lines
fracx - 12 = fracy - 23 = fracz - 34quadtextandquadfracx - lambda3 = fracy - 44 = fracz - 55$$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}\quad\text{and}\quad\frac{x - \lambda}{3} = \frac{y - 4}{4} = \frac{z - 5}{5}$$
is frac1sqrt6$\frac{1}{\sqrt{6}}$ be lambda_1$\lambda_{1}$ and lambda_2$\lambda_{2}$. Then the radius of the circle passing through the points (0, 0)$(0, 0)$, (lambda_1, lambda_2)$(\lambda_{1}, \lambda_{2})$ and (lambda_2, lambda_1)$(\lambda_{2}, \lambda_{1})$ is
A.frac5 sqrt23$\frac{5 \sqrt{2}}{3}$
B.4$4$
C.fracsqrt23$\frac{\sqrt{2}}{3}$
D.3$3$
Solution
### Related Formula
textShortest Distance = left| fracvecAB cdot (vecp times vecq)|vecp times vecq| right|$$\text{Shortest Distance} = \left| \frac{\vec{AB} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$$
### Core Logic
Identify points A(1, 2, 3)$A(1, 2, 3)$ and B(lambda, 4, 5)$B(\lambda, 4, 5)$ on the lines with directions vecp = 2hati + 3hatj + 4hatk$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and vecq = 3hati + 4hatj + 5hatk$\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ respectively. Use the shortest distance formula to determine lambda_1$\lambda_1$ and lambda_2$\lambda_2$.
### Step 1: Calculate Cross Product and Direction Vector
vecp times vecq = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 3 & 4 & 5 endvmatrix = -hati + 2hatj - hatk$$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$$|vecp times vecq| = sqrt(-1)^2 + 2^2 + (-1)^2 = sqrt6$$|\vec{p} \times \vec{q}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$$vecAB = (lambda - 1)hati + 2hatj + 2hatk$$\vec{AB} = (\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}$$
### Step 2: Solve for Lambda
frac1sqrt6 = left| frac((lambda - 1)hati + 2hatj + 2hatk) cdot (-hati + 2hatj - hatk)sqrt6 right|$$\frac{1}{\sqrt{6}} = \left| \frac{((\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})}{\sqrt{6}} \right|$$implies |-lambda + 1 + 4 - 2| = 1 implies |lambda - 3| = 1$$\implies |-\lambda + 1 + 4 - 2| = 1 \implies |\lambda - 3| = 1$$implies lambda = 4 text or 2$$\implies \lambda = 4 \text{ or } 2$$
### Step 3: Radius of the Passing Circle
The circle passes through (0,0)$(0,0)$, (4,2)$(4,2)$ and (2,4)$(2,4)$. Using the circumradius formula R = fracabc4Delta$R = \frac{abc}{4\Delta}$:
a = sqrt20, quad b = sqrt20, quad c = sqrt8$$a = \sqrt{20}, \quad b = \sqrt{20}, \quad c = \sqrt{8}$$Delta = frac12 beginvmatrix 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 endvmatrix = 6$$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{vmatrix} = 6$$R = fracsqrt20 times sqrt20 times sqrt84 times 6 = frac40sqrt224 = frac5sqrt23$$R = \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times 6} = \frac{40\sqrt{2}}{24} = \frac{5\sqrt{2}}{3}$$
### Pattern Recognition
Shortest distance values create symmetric configurations. When finding a circle passing through (0,0)$(0,0)$, (x,y)$(x,y)$, and (y,x)$(y,x)$, the symmetry about y=x$y=x$ simplifies radius calculations immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Class 11 Mathematics: Circles
Q73jee_main_2025_08_april_eveningArea of Triangle Formed by Intersecting Lines
Let the area of the triangle formed by the lines x + 2 = y - 1 = z, fracx - 35 = fracy-1 = fracz - 11$x + 2 = y - 1 = z, \frac{x - 3}{5} = \frac{y}{-1} = \frac{z - 1}{1}$ and fracmathrmx-3 = fracmathrmy - 33 = fracmathrmz - 21$\frac{\mathrm{x}}{-3} = \frac{\mathrm{y} - 3}{3} = \frac{\mathrm{z} - 2}{1}$ be A. Then A^2$A^2$ is equal to
Numerical Answer.Answer: 56 to 56
Solution
### Related Formula
textArea A = frac12 |vecAB times vecAC|$$\text{Area } A = \frac{1}{2} |\vec{AB} \times \vec{AC}|$$
### Core Logic
Determine the three intersection vertex positions for the matching coordinate line segments, then calculate vector cross expansions to determine face boundaries.
### Step 1: Locate Intersection Vertices
Solving line pairs intersection matrices:
* L_1 cap L_2 implies A(-2, 1, 0)$L_1 \cap L_2 \implies A(-2, 1, 0)$
* L_2 cap L_3 implies B(3, 0, 1)$L_2 \cap L_3 \implies B(3, 0, 1)$
* L_3 cap L_1 implies C(0, 3, 2)$L_3 \cap L_1 \implies C(0, 3, 2)$
### Step 2: Construct Vectors Cross Matrix
Using vertex values to form component arrays:
vecAB = -5hati + hatj - hatk, quad vecAC = -3hati + 3hatj + hatk$$\vec{AB} = -5\hat{i} + \hat{j} - \hat{k}, \quad \vec{AC} = -3\hat{i} + 3\hat{j} + \hat{k}$$vecAB times vecAC = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -1 \\ -3 & 3 & 1 endvmatrix = 4hati + 8hatj - 12hatk$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{vmatrix} = 4\hat{i} + 8\hat{j} - 12\hat{k}$$
### Step 3: Final Area Squared Derivation
A = frac12sqrt16 + 64 + 144 = frac12sqrt224 = sqrt56$$A = \frac{1}{2}\sqrt{16 + 64 + 144} = \frac{1}{2}\sqrt{224} = \sqrt{56}$$A^2 = 56$A^2 = 56$
{{SOL_IMG_73}}
### Pattern Recognition
Finding the area of a triangle formed by intersecting lines involves grouping directional cross vectors once coordinates are solved.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q56jee_main_2025_29_jan_eveningLine and Plane Intersections
Let a straight line L$L$ pass through the point P(2, -1, 3)$P(2, -1, 3)$ and be perpendicular to the lines fracx - 12 = fracy + 11 = fracz - 3-2$\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2}$ and \frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4}. If the line L$L$ intersects the yz$yz$-plane at the point Q$Q$, then the distance between the points P$P$ and Q$Q$ is:
A.2$2$
B.sqrt10$\sqrt{10}$
C.3$3$
D.2sqrt3$2\sqrt{3}$
Solution
### Related Formula
The direction vector of a line perpendicular to two vectors vecu$\vec{u}$ and vecv$\vec{v}$ is obtained via the cross product:
vecn = vecu times vecv$$\vec{n} = \vec{u} \times \vec{v}$$
### Core Logic
Extract direction vectors of the given lines:
vecu = 2hati + hatj - 2hatk$$\vec{u} = 2\hat{i} + \hat{j} - 2\hat{k}$$vecv = hati + 3hatj + 4hatk$$\vec{v} = \hat{i} + 3\hat{j} + 4\hat{k}$$
Compute the cross product:
vecn = beginvmatrix hati & hatj & hatk \\ 2 & 1 & -2 \\ 1 & 3 & 4 endvmatrix$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$$= hati(4 - (-6)) - hatj(8 - (-2)) + hatk(6 - 1) = 10hati - 10hatj + 5hatk$$= \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$$= 5(2hati - 2hatj + hatk)$$= 5(2\hat{i} - 2\hat{j} + \hat{k})$$
### Step 1: Write Line Equation and Intersect with Plane
Equation of line L$L$ through P(2, -1, 3)$P(2, -1, 3)$ with direction (2, -2, 1)$(2, -2, 1)$:
fracx - 22 = fracy + 1-2 = fracz - 31 = lambda$$\frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda$$
Any random point on this line is Q(2lambda + 2, -2lambda - 1, lambda + 3)$Q(2\lambda + 2, -2\lambda - 1, \lambda + 3)$.
For intersection with the yz$yz$-plane, set x = 0$x = 0$:
2lambda + 2 = 0 implies lambda = -1$$2\lambda + 2 = 0 \implies \lambda = -1$$
### Step 2: Find Distance
Substituting lambda = -1$\lambda = -1$ into the coordinate matrix of Q$Q$ gives:
Q(0, 1, 2)$Q(0, 1, 2)$
Calculate distance d(P, Q)$d(P, Q)$:
d = sqrt(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2 = sqrt4 + 4 + 1 = 3$$d = \sqrt{(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 4 + 1} = 3$$
### Pattern Recognition
Perpendicularity to two lines always indicates using the cross-product to lock down the direction ratios. Intersection with the yz$yz$-plane simply forces x = 0$x = 0$ immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q63jee_main_2025_29_jan_eveningShortest Distance and Intersection of Lines
Let mathbfP$\mathbf{P}$ be the foot of the perpendicular from the point (1,2,2)$(1,2,2)$ on the line mathrmL:fracmathrmx - 11 = fracmathrmy + 1-1 = fracmathrmz - 22$\mathrm{L}:\frac{\mathrm{x} - 1}{1} = \frac{\mathrm{y} + 1}{-1} = \frac{\mathrm{z} - 2}{2}$. Let the line vecmathrmr = (-hatmathrmi +hatmathrmj -2hatmathrmk) + lambda (hatmathrmi -hatmathrmj +hatmathrmk)$\vec{\mathrm{r}} = (-\hat{\mathrm{i}} +\hat{\mathrm{j}} -2\hat{\mathrm{k}}) + \lambda (\hat{\mathrm{i}} -\hat{\mathrm{j}} +\hat{\mathrm{k}})$, lambda in mathbbR$\lambda \in \mathbb{R}$, intersect the line mathrmL$\mathrm{L}$ at Q. Then 2(mathrmPQ)^2$2(\mathrm{PQ})^2$ is equal to:
A.27$27$
B.25$25$
C.29$29$
D.19$19$
Solution
### Related Formula
Dot product of vector projection matching orthogonal axes equals zero:
vecAP cdot vecd = 0$$\vec{AP} \cdot \vec{d} = 0$$
### Core Logic
Let the target source coordinates tracking point match A(1, 2, 2)$A(1, 2, 2)$. General parameter points on line L$L$ are defined by parameter mu$\mu$:
P(mu + 1, -mu - 1, 2mu + 2)$$P(\mu + 1, -\mu - 1, 2\mu + 2)$$Shortest Distance and Intersection of Lines diagram for Q63 - JEE Main 2025 EveningvecAP = muhati - (mu + 3)hatj + 2muhatk$$\vec{AP} = \mu\hat{i} - (\mu + 3)\hat{j} + 2\mu\hat{k}$$
Line direction vector vecd = hati - hatj + 2hatk$\vec{d} = \hat{i} - \hat{j} + 2\hat{k}$.
### Step 1: Isolate Foot and Intersection Positions
(mu)cdot 1 - (-mu - 3)cdot 1 + (2mu)cdot 2 = 0 implies 6mu + 3 = 0 implies mu = -frac12$$(\mu)\cdot 1 - (-\mu - 3)\cdot 1 + (2\mu)\cdot 2 = 0 \implies 6\mu + 3 = 0 \implies \mu = -\frac{1}{2}$$
Substituting back yields coordinate positions for foot P$P$:
Pleft(frac12, -frac12, 1right)$$P\left(\frac{1}{2}, -\frac{1}{2}, 1\right)$$
Equating general vectors between standard linear constraints tracks intersection point Q$Q$ at mu = -2$\mu = -2$:
Q(-1, 1, -2)$Q(-1, 1, -2)$
### Step 2: Distance Formulation
Compute length of line segment squared:
PQ^2 = left(frac12 - (-1)right)^2 + left(-frac12 - 1right)^2 + (1 - (-2))^2$$PQ^2 = \left(\frac{1}{2} - (-1)\right)^2 + \left(-\frac{1}{2} - 1\right)^2 + (1 - (-2))^2$$= frac94 + frac94 + 9 = frac544$$= \frac{9}{4} + \frac{9}{4} + 9 = \frac{54}{4}$$2(PQ)^2 = 2 left(frac544right) = 27$$2(PQ)^2 = 2 \left(\frac{54}{4}\right) = 27$$
### Pattern Recognition
Always separate foot evaluations from line-intersection parameter updates to ensure you do not mix up variables tracking linear metrics.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Three Dimensional Geometry
Q56jee_main_2025_28_jan_morningDistance Formula and Properties of Triangles
Let A(x,y,z)$A(x,y,z)$ be a point in xy-plane, which is equidistant from three points (0, 3, 2), (2, 0, 3) and (0, 0, 1).
Let B = (1, 4, -1)$B = (1, 4, -1)$ and C = (2, 0, -2)$C = (2, 0, -2)$. Then among the statements
(S1) : Delta ABC$\Delta ABC$ is an isosceles right angled triangle and
(S2): the area of Delta ABC$\Delta ABC$ is frac9sqrt22$\frac{9\sqrt{2}}{2}$.
(1) both are true
(2) only (S1) is true
(3) only (S2) is true
(4) both are false
A. both are true
B. only (S1) is true
C. only (S2) is true
D. both are false
Solution
### Related Formula
3D Cartesian distance formula:
d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
### Core Logic
Since A(x,y,z)$A(x,y,z)$ lies in the xy-plane, its z-coordinate must be zero (z = 0$z = 0$). Let the reference targets be P(0,3,2)$P(0,3,2)$, Q(2,0,3)$Q(2,0,3)$, and R(0,0,1)$R(0,0,1)$.
Setting AP^2 = AR^2$AP^2 = AR^2$:
x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 implies y = 2$$x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 \implies y = 2$$
### Step 1: Locating Coordinate Dimensions
Setting AQ^2 = AR^2$AQ^2 = AR^2$ with y=2$y=2$:
(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 implies x = 3$$(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 \implies x = 3$$
Thus, A$A$ is precisely located at (3,2,0)$(3,2,0)$.
### Step 2: Triangle Side and Area Assessment
Calculate the lengths between A(3,2,0)$A(3,2,0)$, B(1,4,-1)$B(1,4,-1)$, and C(2,0,-2)$C(2,0,-2)$:
AB = sqrt(3-1)^2 + (2-4)^2 + (0+1)^2 = 3$AB = \sqrt{(3-1)^2 + (2-4)^2 + (0+1)^2} = 3$AC = sqrt(3-2)^2 + (2-0)^2 + (0+2)^2 = 3$AC = \sqrt{(3-2)^2 + (2-0)^2 + (0+2)^2} = 3$BC = sqrt(1-2)^2 + (4-0)^2 + (-1+2)^2 = sqrt18$BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{18}$
Since AB = AC = 3$AB = AC = 3$ and AB^2 + AC^2 = BC^2$AB^2 + AC^2 = BC^2$, it forms an isosceles right-angled triangle. Thus, (S1) is true.
textArea = frac12 times 3 times 3 = frac92$$\text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$$
Therefore, (S2) is false.
### Pattern Recognition
Planar locations instantly zero out specific coordinate dimensions (z=0$z=0$ for xy-planes), simplifying system matrices down rapidly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Three Dimensional Geometry
More Three Dimensional Geometry Questions — jee_main_2026_21_jan_morning
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