Let the line of the shortest distance between the lines $L_1 : \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $L_2 : \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(\hat{i} + \hat{j} - \hat{k})$ intersect $L_1$ and $L_2$ at $P$ and $Q$ respectively. If $(\alpha, \beta, \gamma)$ is the midpoint of the line segment $PQ$, then $2(\alpha + \beta + \gamma)$ is equal to

### Analyzing the Setup We are exploring the geometry of two skew lines, $L_1$ and $L_2$, which are neither parallel nor intersecting. Our goal is to find the unique line segment $PQ$ that represents the shortest distance between these two lines. To define our positions on these lines, we utilize parameters $\lambda$ and $\mu$. Any point $P$ on $L_1$ is defined by the coordinates: $$P = (1 + \lambda, 2 - \lambda, 3 + \lambda)$$ Similarly, any point $Q$ on $L_2$ is defined by: $$Q = (4 + \mu, 5 + \mu, 6 - \mu)$$ ### The Vector Approach The vector $\vec{PQ}$ is determined by the difference $Q - P$. Calculating this, we obtain: $$\vec{PQ} = (3 + \mu - \lambda, 3 + \mu + \lambda, 3 - \mu - \lambda)$$ Because $PQ$ represents the shortest distance, it must be perpendicular to the direction vectors of both lines, denoted as $\vec{b_1}$ and $\vec{b_2}$. This geometric constraint requires that the dot products satisfy: $$\vec{PQ} \cdot \vec{b_1} = 0 \quad \text{and} \quad \vec{PQ} \cdot \vec{b_2} = 0$$ ### Solving the System By calculating these dot products, we transform the 3D geometric problem into a system of two linear equations: $$\mu + 3\lambda = 3$$ $$3\mu + \lambda = -3$$ Solving this system of equations yields the specific parameters: $$\lambda = \frac{3}{2}, \quad \mu = -\frac{3}{2}$$ ### Final Calculation Substituting these values back into our expressions for $P$ and $Q$ allows us to find the exact coordinates of the endpoints of the shortest path. The midpoint $M$ is then calculated as the average of $P$ and $Q$. Following this rigorous process, we arrive at the final result. The culmination of our geometric journey is: **21**

JEE Main 2024 (01 Feb Shift 1)

LEVELJEE Advanced

Animated Solution for Mathematics - Three Dimensional Geometry: Let the line of the shortest distance between the lines $L_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $L_{2} : r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$ intersect $L_{1}$ and $L_{2}$ at $P$ and $Q$ respectively. If $(α, β, γ)$ is the midpoint of the line segment $P Q$ , then $2 (α + β + γ)$ is equal to

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S k e w L in es L_{1} an d L_{2}$

Given line: $L_{1} : r = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$
Given line: $L_{2} : r = (4 \hat{i} + 5 \hat{j} + 6 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$
Objective: Find midpoint $(α, β, γ)$ of shortest distance segment $P Q$ .

$G e n er a l P o in t P o n L_{1}$

General point $P$ on $L_{1}$ :
$P \equiv (1 + λ, 2 - λ, 3 + λ)$

$G e n er a l P o in t Q o n L_{2}$

General point $Q$ on $L_{2}$ :
$Q \equiv (4 + μ, 5 + μ, 6 - μ)$

$D e f inin g t h e V ec t or P Q$

Vector $P Q = Q - P$
$P Q = (3 + μ - λ) \hat{i} + (3 + μ + λ) \hat{j} + (3 - μ - λ) \hat{k}$

$T h e C o n d i t i o n f or S h or t es t D i s t an ce$

Condition: $P Q ⊥ L_{1}$ and $P Q ⊥ L_{2}$
Direction of $L_{1}$ : $b = \hat{i} - \hat{j} + \hat{k}$
Direction of $L_{2}$ : $d = \hat{i} + \hat{j} - \hat{k}$

$A ppl y in g P Q \cdot b = 0$

$P Q \cdot (\hat{i} - \hat{j} + \hat{k}) = 0$
$(3 + μ - λ) - (3 + μ + λ) + (3 - μ - λ) = 0$
Simplifying: $3 - μ - 3 λ = 0 ⟹ μ + 3 λ = 3$

$A ppl y in g P Q \cdot d = 0$

$P Q \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$
$(3 + μ - λ) + (3 + μ + λ) - (3 - μ - λ) = 0$
Simplifying: $3 + 3 μ + λ = 0 ⟹ 3 μ + λ = - 3$

$S o l v in g f or λ an d μ$

System:
1) $μ + 3 λ = 3$
2) $3 μ + λ = - 3$
Solving gives: $λ = \frac{3}{2}$ and $μ = - \frac{3}{2}$

$F in d in g C oor d ina t es o f P o in t P$

Substitute $λ = \frac{3}{2}$ into $P$ :
$P = (1 + \frac{3}{2}, 2 - \frac{3}{2}, 3 + \frac{3}{2})$
$P = (\frac{5}{2}, \frac{1}{2}, \frac{9}{2})$

$F in d in g C oor d ina t es o f P o in t Q$

Substitute $μ = - \frac{3}{2}$ into $Q$ :
$Q = (4 - \frac{3}{2}, 5 - \frac{3}{2}, 6 - (- \frac{3}{2}))$
$Q = (\frac{5}{2}, \frac{7}{2}, \frac{15}{2})$

$C a l c u l a t in g t h e M i d p o in t M (α, β, γ)$

Midpoint $M (α, β, γ) = \frac{P + Q}{2}$
$α = \frac{\frac{5}{2} + \frac{5}{2}}{2} = \frac{5}{2}$
$β = \frac{\frac{1}{2} + \frac{7}{2}}{2} = 2$
$γ = \frac{\frac{9}{2} + \frac{15}{2}}{2} = 6$

$F ina l C a l c u l a t i o n : 2 (α + β + γ)$

Target: $2 (α + β + γ)$
Substitute values: $2 (\frac{5}{2} + 2 + 6)$
$2 (\frac{5}{2} + 8) = 5 + 16$
Final Answer: $21$

00:00 / 00:00

The Sigma Insight: Shortest Distance Between Two Skew Lines

Solution Diagram

Analyzing the Setup

We are exploring the geometry of two skew lines,

L_{1}

and

L_{2}

, which are neither parallel nor intersecting. Our goal is to find the unique line segment

P Q

that represents the shortest distance between these two lines.

To define our positions on these lines, we utilize parameters

λ

and

μ

. Any point

P

L_{1}

is defined by the coordinates:

P = (1 + λ, 2 - λ, 3 + λ)

Similarly, any point

Q

L_{2}

is defined by:

Q = (4 + μ, 5 + μ, 6 - μ)

The Vector Approach

The vector

P Q

is determined by the difference

Q - P

. Calculating this, we obtain:

P Q = (3 + μ - λ, 3 + μ + λ, 3 - μ - λ)

Because

P Q

represents the shortest distance, it must be perpendicular to the direction vectors of both lines, denoted as

b_{1}

and

b_{2}

. This geometric constraint requires that the dot products satisfy:

P Q \cdot b_{1} = 0 and P Q \cdot b_{2} = 0

Solving the System

By calculating these dot products, we transform the 3D geometric problem into a system of two linear equations:

μ + 3 λ = 3

3 μ + λ = - 3

Solving this system of equations yields the specific parameters:

λ = \frac{3}{2}, μ = - \frac{3}{2}

Final Calculation

Substituting these values back into our expressions for

P

and

Q

allows us to find the exact coordinates of the endpoints of the shortest path. The midpoint

M

is then calculated as the average of

P

and

Q

Following this rigorous process, we arrive at the final result. The culmination of our geometric journey is: 21

Common Traps & FAQs

Why can't we just use the formula for the shortest distance between two skew lines? The formula is

\frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}

That is a fantastic question! That formula is perfect for finding the *length* of the shortest distance, but our problem asks for the *midpoint* of the segment

P Q

. To find the midpoint, we need the specific coordinates of

P

and

Q

, which requires the parametric approach.

Why do we set the dot product to zero? Is it always true?

Yes, it is! In 3D geometry, the shortest distance between two skew lines is always along the common perpendicular. If a vector is perpendicular to a line, its dot product with the line's direction vector must be zero. Since

P Q

is the common perpendicular, it must be orthogonal to both lines.

Is there any other way to solve this without solving for

λ

and

μ

While you could potentially use vector projection methods, the parametric approach is the most robust and standard way to find the exact coordinates of the points

P

and

Q

. It turns a complex 3D geometry problem into a simple system of linear equations, which is much less prone to error.

Enter Numerical Value:

Visualized Solution

$V i s u a l i z in g t h e S k e w L in es L_{1} an d L_{2}$

$G e n er a l P o in t P o n L_{1}$

$G e n er a l P o in t Q o n L_{2}$

$D e f inin g t h e V ec t or P Q$

$T h e C o n d i t i o n f or S h or t es t D i s t an ce$

$A ppl y in g P Q \cdot b = 0$

$A ppl y in g P Q \cdot d = 0$

$S o l v in g f or λ an d μ$

$F in d in g C oor d ina t es o f P o in t P$

$F in d in g C oor d ina t es o f P o in t Q$

$C a l c u l a t in g t h e M i d p o in t M (α, β, γ)$

$F ina l C a l c u l a t i o n : 2 (α + β + γ)$

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Vector Approach

Solving the System

Final Calculation

Similar Questions

If the shortest distance between the lines $r_{1} = α \hat{i} + 2 \hat{j} + 2 \hat{k} + λ (\hat{i} - 2 \hat{j} + 2 \hat{k}), λ \in R, α > 0$ and $r_{2} = - 4 \hat{i} - \hat{k} + μ (3 \hat{i} - 2 \hat{j} - 2 \hat{k}), μ \in R$ is 9, then $α$ is equal to ___

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $330$ . Then the positive value of $5 α + β$ is

If the shortest distance between the line $r = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j})$ and $r = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\frac{2}{3}$ , then the integral value of $a$ is equal to

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is

The shortest distance between lines $L_{1}$ and $L_{2}$ , where $L_{1} : \frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 4}{2}$ and $L_{2}$ is the line passing through the points $A (- 4, 4, 3), B (- 1, 6, 3)$ and perpendicular to the line $\frac{x - 3}{- 2} = \frac{y}{3} = \frac{z - 1}{1}$ , is

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Skew Lines L1​ and L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } General Point P on L1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } General Point Q on L2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Defining the Vector PQ​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Condition for Shortest Distance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying PQ​⋅b=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying PQ​⋅d=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for λ and μ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Coordinates of Point P

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Coordinates of Point Q

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Midpoint M(α,β,γ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation: 2(α+β+γ)

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Vector Approach

Solving the System

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the lines 2x−1​=3y−2​=4z−3​ and 1x​=αy​=1z−5​ is 6​5​, then the sum of all possible values of α is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the shortest distance between the lines 3x−3​=−1y−α​=1z−3​ and −3x+3​=2y+7​=4z−β​ be 330​. Then the positive value of 5α+β is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the line r=(−i^+3k^)+λ(i^−aj^​) and r=(−j^​+2k^)+μ(i^−j^​+k^) is 32​​, then the integral value of a is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the shortest distance between the lines 2x−1​=3y−2​=λz−3​ and 1x−2​=4y−4​=5z−5​ is 3​1​, then the sum of all possible values of λ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between the lines 1x−5​=2y−2​=−3z−4​ and 1x+3​=4y+5​=−5z−1​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of all values of α, for which the shortest distance between the lines αx+1​=−1y−2​=−αz−4​ and αx​=2y−1​=2αz−1​ is 2​, is

$V i s u a l i z in g t h e S k e w L in es L_{1} an d L_{2}$

$G e n er a l P o in t P o n L_{1}$

$G e n er a l P o in t Q o n L_{2}$

$D e f inin g t h e V ec t or P Q$

$T h e C o n d i t i o n f or S h or t es t D i s t an ce$

$A ppl y in g P Q \cdot b = 0$

$A ppl y in g P Q \cdot d = 0$

$S o l v in g f or λ an d μ$

$F in d in g C oor d ina t es o f P o in t P$

$F in d in g C oor d ina t es o f P o in t Q$

$C a l c u l a t in g t h e M i d p o in t M (α, β, γ)$

$F ina l C a l c u l a t i o n : 2 (α + β + γ)$

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ is $\frac{5}{6}$ , then the sum of all possible values of $α$ is

Let the shortest distance between the lines $\frac{x - 3}{3} = \frac{y - α}{- 1} = \frac{z - 3}{1}$ and $\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - β}{4}$ be $330$ . Then the positive value of $5 α + β$ is

If the shortest distance between the line $r = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j})$ and $r = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$ is $\frac{2}{3}$ , then the integral value of $a$ is equal to

If the shortest distance between the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{λ}$ and $\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 5}{5}$ is $\frac{1}{3}$ , then the sum of all possible values of $λ$ is :

The shortest distance between the lines $\frac{x - 5}{1} = \frac{y - 2}{2} = \frac{z - 4}{- 3}$ and $\frac{x + 3}{1} = \frac{y + 5}{4} = \frac{z - 1}{- 5}$ is

The sum of all values of $α$ , for which the shortest distance between the lines $\frac{x + 1}{α} = \frac{y - 2}{- 1} = \frac{z - 4}{- α}$ and $\frac{x}{α} = \frac{y - 1}{2} = \frac{z - 1}{2 α}$ is $2$ , is