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 9
Q22jee_main_2024_30_january_eveningLines in Space
Let a line passing through the point (-1, 2, 3)$(-1, 2, 3)$ intersect the lines L_1: fracx - 13 = fracy - 22 = fracz + 1-2$L_1: \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{-2}$ at M(alpha, beta, gamma)$M(\alpha, \beta, \gamma)$ and L_2: fracx + 2-3 = fracy - 2-2 = fracz - 14$L_2: \frac{x + 2}{-3} = \frac{y - 2}{-2} = \frac{z - 1}{4}$ at N(a, b, c)$N(a, b, c)$. Then the value of frac(alpha + beta + gamma)^2(a + b + c)^2$\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2}$ equals
Numerical Answer.Answer: 196 to 196
Solution
### Related Formula
textParametric form of a line: vecr = veca + lambdavecb$$\text{Parametric form of a line: } \vec{r} = \vec{a} + \lambda\vec{b}$$
### Core Logic
Let the given point be P(-1, 2, 3)$P(-1, 2, 3)$. A line passing through P$P$ intersects L_1$L_1$ at M$M$ and L_2$L_2$ at N$N$.
Write general coordinates for M$M$ on L_1$L_1$ and N$N$ on L_2$L_2$:
L_1: fracx-13 = fracy-22 = fracz+1-2 = lambda Rightarrow M(3lambda + 1, 2lambda + 2, -2lambda - 1)$L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2} = \lambda \Rightarrow M(3\lambda + 1, 2\lambda + 2, -2\lambda - 1)$L_2: fracx+2-3 = fracy-2-2 = fracz-14 = mu Rightarrow N(-3mu - 2, -2mu + 2, 4mu + 1)$L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4} = \mu \Rightarrow N(-3\mu - 2, -2\mu + 2, 4\mu + 1)$
### Step 1: Collinearity Condition
Since P, M, N$P, M, N$ lie on the same straight line, the direction vectors vecPM$\vec{PM}$ and vecPN$\vec{PN}$ must be proportional.
vecPM = (3lambda + 1 - (-1))hati + (2lambda + 2 - 2)hatj + (-2lambda - 1 - 3)hatk = (3lambda + 2)hati + (2lambda)hatj + (-2lambda - 4)hatk$\vec{PM} = (3\lambda + 1 - (-1))\hat{i} + (2\lambda + 2 - 2)\hat{j} + (-2\lambda - 1 - 3)\hat{k} = (3\lambda + 2)\hat{i} + (2\lambda)\hat{j} + (-2\lambda - 4)\hat{k}$vecPN = (-3mu - 2 - (-1))hati + (-2mu + 2 - 2)hatj + (4mu + 1 - 3)hatk = (-3mu - 1)hati + (-2mu)hatj + (4mu - 2)hatk$\vec{PN} = (-3\mu - 2 - (-1))\hat{i} + (-2\mu + 2 - 2)\hat{j} + (4\mu + 1 - 3)\hat{k} = (-3\mu - 1)\hat{i} + (-2\mu)\hat{j} + (4\mu - 2)\hat{k}$
Proportionality yields:
frac3lambda + 2-3mu - 1 = frac2lambda-2mu = frac-2lambda - 44mu - 2$$\frac{3\lambda + 2}{-3\mu - 1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda - 4}{4\mu - 2}$$
### Step 2: Solving for Lambda and Mu
Lines in Space diagram for Q22 - JEE Main 2024 Evening
From the first pair:
frac3lambda + 2-3mu - 1 = frac-lambdamu$$\frac{3\lambda + 2}{-3\mu - 1} = \frac{-\lambda}{\mu}$$mu(3lambda + 2) = -lambda(-3mu - 1)$$\mu(3\lambda + 2) = -\lambda(-3\mu - 1)$$3lambdamu + 2mu = 3lambdamu + lambda Rightarrow 2mu = lambda$$3\lambda\mu + 2\mu = 3\lambda\mu + \lambda \Rightarrow 2\mu = \lambda$$
From the second pair (using 2mu = lambda$2\mu = \lambda$):
frac-lambdamu = frac-lambda - 22mu - 1$$\frac{-\lambda}{\mu} = \frac{-\lambda - 2}{2\mu - 1}$$-2mu = frac-2mu - 22mu - 1$$-2\mu = \frac{-2\mu - 2}{2\mu - 1}$$-2mu(2mu - 1) = -2mu - 2$$-2\mu(2\mu - 1) = -2\mu - 2$$-4mu^2 + 2mu = -2mu - 2$$-4\mu^2 + 2\mu = -2\mu - 2$$-4mu^2 + 4mu + 2 = 0 quad text(Wait, alternative grouping):$$-4\mu^2 + 4\mu + 2 = 0 \quad \text{(Wait, alternative grouping)}:$$
Re-checking the second pair with 2mu = lambda$2\mu = \lambda$:
frac2lambda-2mu = frac-lambdamu$$\frac{2\lambda}{-2\mu} = \frac{-\lambda}{\mu}$$frac-lambdamu = frac-2lambda - 44mu - 2$$\frac{-\lambda}{\mu} = \frac{-2\lambda - 4}{4\mu - 2}$$
Substitute lambda = 2mu$\lambda = 2\mu$:
frac-2mumu = frac-4mu - 44mu - 2$$\frac{-2\mu}{\mu} = \frac{-4\mu - 4}{4\mu - 2}$$-2(4mu - 2) = -4mu - 4$$-2(4\mu - 2) = -4\mu - 4$$-8mu + 4 = -4mu - 4$$-8\mu + 4 = -4\mu - 4$$8 = 4mu Rightarrow mu = 2$$8 = 4\mu \Rightarrow \mu = 2$$
Then lambda = 2(2) = 4$\lambda = 2(2) = 4$.
### Step 3: Finalizing Target Coordinates
For lambda = 4$\lambda = 4$:
alpha + beta + gamma = (3(4)+1) + (2(4)+2) + (-2(4)-1) = 13 + 10 - 9 = 14$\alpha + \beta + \gamma = (3(4)+1) + (2(4)+2) + (-2(4)-1) = 13 + 10 - 9 = 14$
For mu = 2$\mu = 2$:
a + b + c = (-3(2)-2) + (-2(2)+2) + (4(2)+1) = -8 - 2 + 9 = -1$a + b + c = (-3(2)-2) + (-2(2)+2) + (4(2)+1) = -8 - 2 + 9 = -1$
Finally, compute the target fraction:
frac(alpha + beta + gamma)^2(a + b + c)^2 = frac(14)^2(-1)^2 = frac1961 = 196$$\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} = \frac{(14)^2}{(-1)^2} = \frac{196}{1} = 196$$
### Pattern Recognition
A line crossing two given lines acts as a transversal holding 3 collinear points. Setting vecPM = k cdot vecPN$\vec{PM} = k \cdot \vec{PN}$ rapidly locks the parametric constants lambda$\lambda$ and mu$\mu$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Q10jee_main_2024_30_jan_morningLine in 3D
Let (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ be the foot of perpendicular from the point (1, 2, 3)$(1, 2, 3)$ on the line fracx + 35 = fracy - 12 = fracz + 43$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$. then 19(alpha + beta + gamma)$19(\alpha + \beta + \gamma)$ is equal to:
A.102$102$
B.101$101$
C.99$99$
D.100$100$
Solution
### Related Formula
a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 quad text(Condition for perpendicular vectors)$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \quad \text{(Condition for perpendicular vectors)}$$
### Core Logic
Line in 3D diagram for Q10 - JEE Main 2024 Morning
Let the given point be A(1, 2, 3)$A(1, 2, 3)$. Let the foot of the perpendicular on the line be P(alpha, beta, gamma)$P(\alpha, \beta, \gamma)$.
The equation of the line is fracx + 35 = fracy - 12 = fracz + 43 = k$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$.
So, any point on the line can be written as (5k - 3, 2k + 1, 3k - 4)$(5k - 3, 2k + 1, 3k - 4)$.
Let this point be P$P$. Thus, P(alpha, beta, gamma) equiv (5k - 3, 2k + 1, 3k - 4)$P(\alpha, \beta, \gamma) \equiv (5k - 3, 2k + 1, 3k - 4)$.
### Step 1: Finding Direction Ratios
Direction ratios (DR's) of the line segment AP$AP$ are:
((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) equiv (5k - 4, 2k - 1, 3k - 7)$$((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) \equiv (5k - 4, 2k - 1, 3k - 7)$$
Direction ratios of the given line are (5, 2, 3)$(5, 2, 3)$.
### Step 2: Using Perpendicularity Condition
Since AP$AP$ is perpendicular to the given line, the dot product of their direction ratios is zero:
5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$$5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$$(25k - 20) + (4k - 2) + (9k - 21) = 0$$(25k - 20) + (4k - 2) + (9k - 21) = 0$$38k - 43 = 0$38k - 43 = 0$k = frac4338$$k = \frac{43}{38}$$
### Step 3: Calculating target expression
We need to find 19(alpha + beta + gamma)$19(\alpha + \beta + \gamma)$.
alpha + beta + gamma = (5k - 3) + (2k + 1) + (3k - 4)$$\alpha + \beta + \gamma = (5k - 3) + (2k + 1) + (3k - 4)$$alpha + beta + gamma = 10k - 6$$\alpha + \beta + \gamma = 10k - 6$$
Substituting k = frac4338$k = \frac{43}{38}$:
19(alpha + beta + gamma) = 19 left(10left(frac4338right) - 6right)$$19(\alpha + \beta + \gamma) = 19 \left(10\left(\frac{43}{38}\right) - 6\right)$$= 19 left( frac43038 - 6 right) = frac4302 - 114 = 215 - 114 = 101$$= 19 \left( \frac{430}{38} - 6 \right) = \frac{430}{2} - 114 = 215 - 114 = 101$$
### Pattern Recognition
To find coordinates of a foot of perpendicular, define a general point on the line using parameter k$k$, create the direction vector, and dot it with the line's direction vector set to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Q22jee_main_2024_30_jan_morningLine in 3D
If d_1$d_1$ is the shortest distance between the linesx + 1 = 2y = -12z$x + 1 = 2y = -12z$, x = y + 2 = 6z - 6$x = y + 2 = 6z - 6$ and d_2$d_2$ is the shortest distance between the linesfracx - 12 = fracy + 8-7 = fracz - 45, fracx - 12 = fracy - 21 = fracz - 6-3$\frac{x - 1}{2} = \frac{y + 8}{-7} = \frac{z - 4}{5}, \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3}$, then the value of frac32sqrt3d_1d_2$\frac{32\sqrt{3}d_1}{d_2}$ is:
Q6jee_main_2024_31_jan_eveningMirror Image of a Point
Let (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ be mirror image of the point (2, 3, 5)$(2, 3, 5)$ in the line fracx - 12 =fracy - 23 =fracz - 34$\frac{x - 1}{2} =\frac{y - 2}{3} =\frac{z - 3}{4}$. Then 2alpha + 3beta + 4gamma$2\alpha + 3\beta + 4\gamma$ is equal to
A.32$32$
B.33$33$
C.31$31$
D.34$34$
Solution
### Related Formula
textDot product constraint: vecPR cdot vecd = 0 text where vecd text is direction of the line.$$\text{Dot product constraint: } \vec{PR} \cdot \vec{d} = 0 \text{ where } \vec{d} \text{ is direction of the line.}$$
### Core Logic
Mirror Image of a Point diagram for Q6 - JEE Main 2024 Evening
Let P = (2, 3, 5)$P = (2, 3, 5)$ and its mirror image be R(alpha, beta, gamma)$R(\alpha, \beta, \gamma)$.
The vector connecting the point and its mirror image, vecPR$\vec{PR}$, is strictly perpendicular to the given line.
The direction ratios of the line are vecd = (2, 3, 4)$\vec{d} = (2, 3, 4)$.
Therefore, the dot product must be zero:
vecPR cdot vecd = 0$$\vec{PR} \cdot \vec{d} = 0$$(alpha - 2, beta - 3, gamma - 5) cdot (2, 3, 4) = 0$$(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$$2(alpha - 2) + 3(beta - 3) + 4(gamma - 5) = 0$$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$$2alpha - 4 + 3beta - 9 + 4gamma - 20 = 0$$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$$2alpha + 3beta + 4gamma = 33$$2\alpha + 3\beta + 4\gamma = 33$$
### Pattern Recognition
When asked for a linear combination of image coordinates matching the direction ratios of the line, skip finding the exact foot of perpendicular. Just use the orthogonality condition directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Q20jee_main_2024_31_jan_eveningShortest Distance between two lines
The shortest distance between lines L_1$L_1$ and L_2$L_2$, where L_1:fracx-12=fracy+1-3=fracz+42$L_1:\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+4}{2}$ and L_2$L_2$ is the line passing through the points A(-4,4,3), B(-1,6,3)$A(-4,4,3), B(-1,6,3)$ and perpendicular to the line fracx-3-2=fracy3=fracz-11$\frac{x-3}{-2}=\frac{y}{3}=\frac{z-1}{1}$, is
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