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 10
Q14jee_main_2024_31_jan_morningDistance of a Point from a Line
The distance of the point Q(0, 2, -2)$Q(0, 2, -2)$ form the line passing through the point P(5, -4, 3)$P(5, -4, 3)$ and perpendicular to the lines vecr = (-3hati + 2hatk) + lambda(2hati + 3hatj + 5hatk), lambda in mathbbR$\vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 5\hat{k}), \lambda \in \mathbb{R}$ and vecr = (hati - 2hatj + hatk) + mu(-hati + 3hatj + 2hatk), mu in mathbbR$\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu(-\hat{i} + 3\hat{j} + 2\hat{k}), \mu \in \mathbb{R}$
A.sqrt86$\sqrt{86}$
B.sqrt20$\sqrt{20}$
C.sqrt54$\sqrt{54}$
D.sqrt74$\sqrt{74}$
Solution
### Core Logic
A vector in the direction of the required line is perpendicular to both given lines. We obtain it via cross product of their direction vectors:
vecn = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 5 \\ -1 & 3 & 2 endvmatrix = -9hati - 9hatj + 9hatk$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k}$
Taking the direction vector as hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$.
### Step 1: Required Line Equation
The line passes through P(5, -4, 3)$P(5, -4, 3)$ with direction hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$.
Equation: vecr = (5hati - 4hatj + 3hatk) + alpha(hati + hatj - hatk)$\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \alpha(\hat{i} + \hat{j} - \hat{k})$.
### Step 2: Projection & Distance
Any point on the line is M(5+alpha, -4+alpha, 3-alpha)$M(5+\alpha, -4+\alpha, 3-\alpha)$.
We need distance from Q(0, 2, -2)$Q(0, 2, -2)$.
Vector vecQM = (5+alpha)hati + (alpha-6)hatj + (5-alpha)hatk$\vec{QM} = (5+\alpha)\hat{i} + (\alpha-6)\hat{j} + (5-\alpha)\hat{k}$.
Since vecQM$\vec{QM}$ is perpendicular to the line direction (hati + hatj - hatk)$(\hat{i} + \hat{j} - \hat{k})$:
(5+alpha)(1) + (alpha-6)(1) + (5-alpha)(-1) = 0$(5+\alpha)(1) + (\alpha-6)(1) + (5-\alpha)(-1) = 0$5 + alpha + alpha - 6 - 5 + alpha = 0 implies 3alpha = 6 implies alpha = 2.$5 + \alpha + \alpha - 6 - 5 + \alpha = 0 \implies 3\alpha = 6 \implies \alpha = 2.$Distance of a Point from a Line diagram for Q14 - JEE Main 2024 Morning
### Step 3: Distance calculation
Substitute alpha = 2$\alpha = 2$ in vecQM$\vec{QM}$:
vecQM = 7hati - 4hatj + 3hatk$\vec{QM} = 7\hat{i} - 4\hat{j} + 3\hat{k}$.
Distance |vecQM| = sqrt7^2 + (-4)^2 + 3^2 = sqrt49 + 16 + 9 = sqrt74$|\vec{QM}| = \sqrt{7^2 + (-4)^2 + 3^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Class 12 Maths: Vector Algebra
Q24jee_main_2024_31_jan_morningFoot of Perpendicular and Angle
Let Q$Q$ and R$R$ be the feet of perpendiculars from the point P(a, a, a)$P(a, a, a)$ on the lines x = y, z = 1$x = y, z = 1$ and x = -y, z = -1$x = -y, z = -1$ respectively. If angle QPR$\angle QPR$ is a right angle, then 12a^2$12a^2$ is equal to
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