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

### Analyzing the Setup Every line in 3D space is defined by a point it passes through and a direction it follows. From the given equations, we extract the position vectors: $\vec{a}_1 = -\hat{i} + 3\hat{k}$ and $\vec{a}_2 = -\hat{j} + 2\hat{k}$. The corresponding direction vectors are: $\vec{b}_1 = \hat{i} - a\hat{j}$ and $\vec{b}_2 = \hat{i} - \hat{j} + \hat{k}$. ### The Bridge Vector To find the distance, we first calculate the vector connecting the two lines: $\vec{a}_2 - \vec{a}_1 = (0 - (-1))\hat{i} + (-1 - 0)\hat{j} + (2 - 3)\hat{k} = \hat{i} - \hat{j} - \hat{k}$. This vector acts as a bridge between our two lines. ### The Common Perpendicular We now determine a direction perpendicular to both lines using the cross product $\vec{b}_1 \times \vec{b}_2$: $$ \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 1 & -1 & 1 \end{vmatrix} = -a\hat{i} - \hat{j} + (a-1)\hat{k} $$ The magnitude of this normal vector is: $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-a)^2 + (-1)^2 + (a-1)^2} = \sqrt{2a^2 - 2a + 2}$. ### The Synthesis The shortest distance $d$ between two skew lines is given by the formula: $$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$ The numerator is the absolute value of the dot product: $|(1)(-a) + (-1)(-1) + (-1)(a-1)| = |-a + 1 - a + 1| = |2 - 2a|$. Equating this to the given distance $\sqrt{\frac{2}{3}}$, we obtain: $$\frac{|2 - 2a|}{\sqrt{2a^2 - 2a + 2}} = \sqrt{\frac{2}{3}}$$ ### The Algebraic Resolution Squaring both sides of the equation, we have: $$\frac{4(1-a)^2}{2(a^2 - a + 1)} = \frac{2}{3} \Rightarrow \frac{(a-1)^2}{a^2 - a + 1} = \frac{1}{3}$$ Cross-multiplying yields: $3(a^2 - 2a + 1) = a^2 - a + 1$ $3a^2 - 6a + 3 = a^2 - a + 1$ $2a^2 - 5a + 2 = 0$ Factoring the quadratic equation: $(2a - 1)(a - 2) = 0$ This results in $a = \frac{1}{2}$ or $a = 2$. Since the problem requires an integral value for $a$, we reject $\frac{1}{2}$. The final value is **$a = 2$**.

JEE Main 2022 (24 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Enter Numerical Value:

Visualized Solution

Identify $a_{1}, b_{1}, a_{2}, b_{2}$

Line 1: $r = (- \hat{i} + 3 \hat{k}) + λ (\hat{i} - a \hat{j})$
$a_{1} = (- 1, 0, 3)$ , $b_{1} = (1, - a, 0)$
Line 2: $r = (- \hat{j} + 2 \hat{k}) + μ (\hat{i} - \hat{j} + \hat{k})$
$a_{2} = (0, - 1, 2)$ , $b_{2} = (1, - 1, 1)$

Calculate $a_{2} - a_{1}$

Connecting vector: $a_{2} - a_{1}$
$a_{2} - a_{1} = (0 - (- 1)) \hat{i} + (- 1 - 0) \hat{j} + (2 - 3) \hat{k}$
$a_{2} - a_{1} = (1, - 1, - 1)$

Find Cross Product $b_{1} \times b_{2}$

Common perpendicular direction: $b_{1} \times b_{2}$
$b_{1} \times b_{2} = \hat{i} 11 \hat{j} - a - 1 \hat{k} 01$
$b_{1} \times b_{2} = - a \hat{i} - \hat{j} + (a - 1) \hat{k}$

Magnitude of $b_{1} \times b_{2}$

Magnitude: $∣ b_{1} \times b_{2} ∣$
$= (- a)^{2} + (- 1)^{2} + (a - 1)^{2}$
$= a^{2} + 1 + a^{2} - 2 a + 1$
$= 2 a^{2} - 2 a + 2$

Dot Product for Numerator

Numerator: $∣ (a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) ∣$
$= ∣ (1) (- a) + (- 1) (- 1) + (- 1) (a - 1) ∣$
$= ∣ - a + 1 - a + 1∣$
$= ∣2 - 2 a ∣$

Shortest Distance Formula

Formula: $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$
Given $d = \frac{2}{3}$
$\frac{∣2 - 2 a ∣}{2 a ^{2} - 2 a + 2} = \frac{2}{3}$

Squaring and Simplifying

Squaring both sides:
$\frac{4 ( 1 - a ) ^{2}}{2 ( a ^{2} - a + 1 )} = \frac{2}{3}$
$\frac{2 ( a - 1 ) ^{2}}{a ^{2} - a + 1} = \frac{2}{3}$
$\frac{( a - 1 ) ^{2}}{a ^{2} - a + 1} = \frac{1}{3}$

Form the Quadratic Equation

Cross-multiplying:
$3 (a - 1)^{2} = a^{2} - a + 1$
$3 (a^{2} - 2 a + 1) = a^{2} - a + 1$
$3 a^{2} - 6 a + 3 = a^{2} - a + 1$
$2 a^{2} - 5 a + 2 = 0$

Solve for $a$

Factoring:
$2 a^{2} - 4 a - a + 2 = 0$
$2 a (a - 2) - 1 (a - 2) = 0$
$(2 a - 1) (a - 2) = 0$
Possible values: $a = \frac{1}{2}$ or $a = 2$

Final Answer

The question asks for the integral value of $a$ .
Since $a = \frac{1}{2}$ is not an integer, we select $a = 2$ .
Final Answer: $a = 2$

00:00 / 00:00

The Sigma Insight: Shortest Distance Between Two Skew Lines

Solution Diagram

Analyzing the Setup

Every line in 3D space is defined by a point it passes through and a direction it follows. From the given equations, we extract the position vectors:

a_{1} = - \hat{i} + 3 \hat{k}

and

a_{2} = - \hat{j} + 2 \hat{k}

The corresponding direction vectors are:

b_{1} = \hat{i} - a \hat{j}

and

b_{2} = \hat{i} - \hat{j} + \hat{k}

The Bridge Vector

To find the distance, we first calculate the vector connecting the two lines:

a_{2} - a_{1} = (0 - (- 1)) \hat{i} + (- 1 - 0) \hat{j} + (2 - 3) \hat{k} = \hat{i} - \hat{j} - \hat{k}

This vector acts as a bridge between our two lines.

The Common Perpendicular

We now determine a direction perpendicular to both lines using the cross product

b_{1} \times b_{2}

b_{1} \times b_{2} = \hat{i} 11 \hat{j} - a - 1 \hat{k} 01 = - a \hat{i} - \hat{j} + (a - 1) \hat{k}

The magnitude of this normal vector is:

∣ b_{1} \times b_{2} ∣ = (- a)^{2} + (- 1)^{2} + (a - 1)^{2} = 2 a^{2} - 2 a + 2

The Synthesis

The shortest distance

d

between two skew lines is given by the formula:

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

The numerator is the absolute value of the dot product:

∣ (1) (- a) + (- 1) (- 1) + (- 1) (a - 1) ∣ = ∣ - a + 1 - a + 1∣ = ∣2 - 2 a ∣

Equating this to the given distance

\frac{2}{3}

, we obtain:

\frac{∣2 - 2 a ∣}{2 a ^{2} - 2 a + 2} = \frac{2}{3}

The Algebraic Resolution

Squaring both sides of the equation, we have:

\frac{4 ( 1 - a ) ^{2}}{2 ( a ^{2} - a + 1 )} = \frac{2}{3} \Rightarrow \frac{( a - 1 ) ^{2}}{a ^{2} - a + 1} = \frac{1}{3}

Cross-multiplying yields:

3 (a^{2} - 2 a + 1) = a^{2} - a + 1

3 a^{2} - 6 a + 3 = a^{2} - a + 1

2 a^{2} - 5 a + 2 = 0

Factoring the quadratic equation:

(2 a - 1) (a - 2) = 0

This results in

a = \frac{1}{2}

a = 2

. Since the problem requires an integral value for

a

, we reject

\frac{1}{2}

The final value is $a = 2$ .

Common Traps & FAQs

Wait, why do we use the cross product here? Can't we just find the distance between two arbitrary points on the lines?

That is a great question! If you pick two arbitrary points, the distance between them changes depending on which points you choose. The shortest distance is unique and occurs only when the connecting vector is perpendicular to both lines. The cross product

b_{1} \times b_{2}

gives us the exact direction of that common perpendicular, which is the only direction along which the shortest distance can be measured.

Why is the formula for shortest distance written with an absolute value in the numerator?

Distance is a physical quantity and must always be non-negative. The dot product

(a_{2} - a_{1}) \cdot (b_{1} \times b_{2})

can result in a negative value depending on the order of vectors or the direction of the cross product. Taking the absolute value ensures our final distance

d

is a positive scalar, which is physically meaningful.

Why do we reject the value

a = 1/2

at the end?

Always read the question carefully! The problem explicitly asks for the 'integral value' of

a

. While

a = 1/2

is a valid algebraic solution to the quadratic equation, it is not an integer. Therefore, we must discard it and accept only

a = 2

as the correct answer.

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Enter Numerical Value:

Visualized Solution

Identify $a_{1}, b_{1}, a_{2}, b_{2}$

Calculate $a_{2} - a_{1}$

Find Cross Product $b_{1} \times b_{2}$

Magnitude of $b_{1} \times b_{2}$

Dot Product for Numerator

Shortest Distance Formula

Squaring and Simplifying

Form the Quadratic Equation

Solve for $a$

Final Answer

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Bridge Vector

The Common Perpendicular

The Synthesis

The Algebraic Resolution

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 straight lines $3 (x - 1) = 6 (y - 2) = 2 (z - 1)$ and $4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R$ is $\frac{1}{38}$ , then the integral value of $λ$ 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

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 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

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

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

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.

Enter Numerical Value:

Visualized Solution

Identify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a1​,b1​,a2​,b2​

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​−a1​

Find Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​×b2​

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​×b2​

Dot Product for Numerator

Shortest Distance Formula

Squaring and Simplifying

Form the Quadratic Equation

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Answer

The Sigma Insight: Shortest Distance Between Two Skew Lines

Analyzing the Setup

The Bridge Vector

The Common Perpendicular

The Synthesis

The Algebraic Resolution

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 straight lines 3(x−1)=6(y−2)=2(z−1) and 4(x−2)=2(y−λ)=(z−3),λ∈R is 38​1​, then the integral value of λ 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​=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; } 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 :

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.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−λ​=1y−2​=1z−1​ and 1x−3​​=−2y−1​=1z−2​ is 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; } If the shortest distance between the line joining the points (1,2,3) and (2,3,4) and the line 2x−1​=−1y+1​=0z−2​ is α, then 28α2 is equal to ______.

Identify $a_{1}, b_{1}, a_{2}, b_{2}$

Calculate $a_{2} - a_{1}$

Find Cross Product $b_{1} \times b_{2}$

Magnitude of $b_{1} \times b_{2}$

Solve for $a$

If the shortest distance between the straight lines $3 (x - 1) = 6 (y - 2) = 2 (z - 1)$ and $4 (x - 2) = 2 (y - λ) = (z - 3), λ \in R$ is $\frac{1}{38}$ , then the integral value of $λ$ 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

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 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

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

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

If the shortest distance between the line joining the points $(1, 2, 3)$ and $(2, 3, 4)$ and the line $\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 2}{0}$ is $α$ , then $28 α^{2}$ is equal to ______.