If the shortest distance between the linesfracx-41=fracy+12=fracz-3$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and fracx-lambda2=fracy+14=fracz-2-5$\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is frac6sqrt5$\frac{6}{\sqrt{5}}$, then the sum of all possible values of lambda$\lambda$ is:
A.5$5$
B.8$8$
C.7$7$
D.10$10$
Solution & Explanation
### Related Formula
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 positional vectors and direction ratios for both lines:
Line 1: veca_1 = (4, -1, 0)$\vec{a}_1 = (4, -1, 0)$, direction vecb_1 = (1, 2, -3)$\vec{b}_1 = (1, 2, -3)$
Line 2: veca_2 = (lambda, -1, 2)$\vec{a}_2 = (\lambda, -1, 2)$, direction vecb_2 = (2, 4, -5)$\vec{b}_2 = (2, 4, -5)$
Vector connecting lines: (veca_2 - veca_1) = (lambda - 4, 0, 2)$(\vec{a}_2 - \vec{a}_1) = (\lambda - 4, 0, 2)$
### Step 1: Cross Product and Magnitude
Find the normal vector vecn = vecb_1 times vecb_2$\vec{n} = \vec{b}_1 \times \vec{b}_2$:
vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 1 & 2 & -3 \\ 2 & 4 & -5 endvmatrix$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$= hati(-10 - (-12)) - hatj(-5 - (-6)) + hatk(4 - 4)$= \hat{i}(-10 - (-12)) - \hat{j}(-5 - (-6)) + \hat{k}(4 - 4)$= 2hati - 1hatj + 0hatk = (2, -1, 0)$= 2\hat{i} - 1\hat{j} + 0\hat{k} = (2, -1, 0)$
Magnitude of the normal vector:
|vecb_1 times vecb_2| = sqrt2^2 + (-1)^2 + 0^2 = sqrt5$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$
### Step 2: Application of Shortest Distance Formula
Dot product of normal vector and positional difference vector:
(veca_2 - veca_1) cdot (vecb_1 times vecb_2) = (lambda - 4)(2) + (0)(-1) + (2)(0) = 2(lambda - 4)$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\lambda - 4)(2) + (0)(-1) + (2)(0) = 2(\lambda - 4)$
Using the shortest distance formula given as frac6sqrt5$\frac{6}{\sqrt{5}}$:
frac|2(lambda - 4)|sqrt5 = frac6sqrt5$\frac{|2(\lambda - 4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$|2(lambda - 4)| = 6$|2(\lambda - 4)| = 6$|lambda - 4| = 3$|\lambda - 4| = 3$
### Step 3: Finding Unknown values
Solve the absolute value relation:
lambda - 4 = 3 Rightarrow lambda = 7$\lambda - 4 = 3 \Rightarrow \lambda = 7$lambda - 4 = -3 Rightarrow lambda = 1$\lambda - 4 = -3 \Rightarrow \lambda = 1$
Sum of possible values = 7 + 1 = 8$7 + 1 = 8$.
### Pattern Recognition
Standard Shortest Distance methodology between skew lines. Cross product of direction vectors forms the perpendicular frame normal, and dot-producting the difference of positional anchor points yields the direct orthogonal projection.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Class 12 Maths: Vector Algebra
Keywords:#shortest distance between the lines#sum of all possible values#JEE Main 2024 Morning Q8#Three Dimensional Geometry JEE Main 2024#Shortest Distance Between Two Lines JEE Main 2024
More Three Dimensional Geometry Previous-Year Questions — Page 9
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
Q28jee_main_2024_31_jan_eveningDistance of a point on a line
A line passes through A(4, -6, -2)$A(4, -6, -2)$ and B(16, -2, 4)$B(16, -2, 4)$. The point P(a, b, c)$P(a, b, c)$ where a, b, c$a, b, c$ are non-negative integers, on the line AB$AB$ lies at a distance of 21$21$ units, from the point A$A$. The distance between the points P(a, b, c)$P(a, b, c)$ and Q(4, -12, 3)$Q(4, -12, 3)$ is equal to
Numerical Answer.Answer: 22 to 22
Solution
### Related Formula
textDistance of point P text on line from A(x_1,y_1,z_1): P = (x_1 pm rd_x, y_1 pm rd_y, z_1 pm rd_z)$\text{Distance of point } P \text{ on line from } A(x_1,y_1,z_1): P = (x_1 \pm rd_x, y_1 \pm rd_y, z_1 \pm rd_z)$textwhere (d_x,d_y,d_z) text are direction cosines and r text is distance.$\text{where } (d_x,d_y,d_z) \text{ are direction cosines and } r \text{ is distance.}$
### Core Logic
Direction ratios of AB = (16-4, -2 - (-6), 4 - (-2)) = (12, 4, 6)$AB = (16-4, -2 - (-6), 4 - (-2)) = (12, 4, 6)$.
Magnitude of this vector = sqrt144 + 16 + 36 = sqrt196 = 14$= \sqrt{144 + 16 + 36} = \sqrt{196} = 14$.
Direction cosines are left(frac1214, frac414, frac614right) = left(frac67, frac27, frac37right)$\left(\frac{12}{14}, \frac{4}{14}, \frac{6}{14}\right) = \left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$.
Point P$P$ is at a distance of 21 units from A(4, -6, -2)$A(4, -6, -2)$:
P = left(4 pm 21left(frac67right), -6 pm 21left(frac27right), -2 pm 21left(frac37right)right)$P = \left(4 \pm 21\left(\frac{6}{7}\right), -6 \pm 21\left(\frac{2}{7}\right), -2 \pm 21\left(\frac{3}{7}\right)\right)$P = (4 pm 18, -6 pm 6, -2 pm 9)$P = (4 \pm 18, -6 \pm 6, -2 \pm 9)$
Since coordinates a,b,c$a,b,c$ of P$P$ are non-negative integers, we take the '+' sign:
P = (4+18, -6+6, -2+9) = (22, 0, 7)$P = (4+18, -6+6, -2+9) = (22, 0, 7)$
Calculate distance from Q(4, -12, 3)$Q(4, -12, 3)$:
PQ = sqrt(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2$PQ = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2}$PQ = sqrt18^2 + 12^2 + 4^2 = sqrt324 + 144 + 16 = sqrt484 = 22$PQ = \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
More Three Dimensional Geometry Questions — jee_main_2024_27_jan_morning
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