This MCQ module is based on: Shortest Distance between Two Lines
TOPIC 20 OF 25
Shortest Distance between Two Lines
🎓 Class 12
Mathematics
CBSE
Theory
Ch 11 — Three Dimensional Geometry
⏱ ~15 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Shortest Distance between Two Lines
Targeting Class 12 level in Coordinate Geometry, with Advanced difficulty.
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11.5 Shortest Distance Between Two Lines
Two lines in 3D can be: (i) intersecting (distance = 0), (ii) parallel (distance = constant), or (iii) skew? — neither parallel nor intersecting. For non-intersecting lines we find the shortest distance: the length of the segment perpendicular to both.
Fig 11.5: Shortest distance between skew lines is the length of common perpendicular PQ
11.5.1 Distance Between Two Skew Lines
Let the two lines be:
\(L_1: \vec r = \vec a_1 + \lambda \vec b_1\), \(L_2: \vec r = \vec a_2 + \mu \vec b_2\).
The unit vector perpendicular to both \(\vec b_1\) and \(\vec b_2\) is \(\hat n = \dfrac{\vec b_1 \times \vec b_2}{|\vec b_1 \times \vec b_2|}\). The shortest distance is the projection of \(\vec a_2 - \vec a_1\) on \(\hat n\):
Vector Form
\[\boxed{d = \left|\frac{(\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1)}{|\vec b_1 \times \vec b_2|}\right|}\]
Cartesian Form
If \(L_1:\dfrac{x-x_1}{a_1}=\dfrac{y-y_1}{b_1}=\dfrac{z-z_1}{c_1}\) and \(L_2:\dfrac{x-x_2}{a_2}=\dfrac{y-y_2}{b_2}=\dfrac{z-z_2}{c_2}\):
\[d = \frac{\left|\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1\\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\end{array}\right|}{\sqrt{(b_1 c_2 - b_2 c_1)^2 + (c_1 a_2 - c_2 a_1)^2 + (a_1 b_2 - a_2 b_1)^2}}\]
Condition for Intersecting Lines
Two lines are coplanar (intersecting or parallel) iff \((\vec b_1 \times \vec b_2)\cdot(\vec a_2 - \vec a_1) = 0\), i.e. the numerator of \(d\) is zero.
11.5.2 Distance Between Parallel Lines
If \(L_1: \vec r = \vec a_1 + \lambda \vec b\) and \(L_2: \vec r = \vec a_2 + \mu \vec b\) (same direction \(\vec b\)):
Vector Form
\[d = \left|\frac{\vec b \times (\vec a_2 - \vec a_1)}{|\vec b|}\right|\]
This is the magnitude of the component of \(\vec a_2 - \vec a_1\) perpendicular to \(\vec b\).
Example 9
Find the shortest distance between the lines \(\vec r = (\hat i + 2\hat j + \hat k) + \lambda(\hat i - \hat j + \hat k)\) and \(\vec r = (2\hat i - \hat j - \hat k) + \mu(2\hat i + \hat j + 2\hat k)\).
\(\vec a_1 = (1,2,1),\; \vec a_2=(2,-1,-1)\), so \(\vec a_2 - \vec a_1 = (1,-3,-2)\).
\(\vec b_1 = (1,-1,1),\; \vec b_2=(2,1,2)\).
\(\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i(-2-1) - \hat j(2-2) + \hat k(1+2) = (-3, 0, 3)\).
\(|\vec b_1 \times \vec b_2| = \sqrt{9+0+9} = 3\sqrt 2\).
Numerator: \((-3)(1) + 0(-3) + 3(-2) = -3 - 6 = -9\). Absolute = 9.
\(d = \dfrac{9}{3\sqrt 2} = \dfrac{3}{\sqrt 2} = \dfrac{3\sqrt 2}{2}\) units.
\(\vec b_1 = (1,-1,1),\; \vec b_2=(2,1,2)\).
\(\vec b_1 \times \vec b_2 = \begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&1\\2&1&2\end{vmatrix} = \hat i(-2-1) - \hat j(2-2) + \hat k(1+2) = (-3, 0, 3)\).
\(|\vec b_1 \times \vec b_2| = \sqrt{9+0+9} = 3\sqrt 2\).
Numerator: \((-3)(1) + 0(-3) + 3(-2) = -3 - 6 = -9\). Absolute = 9.
\(d = \dfrac{9}{3\sqrt 2} = \dfrac{3}{\sqrt 2} = \dfrac{3\sqrt 2}{2}\) units.
Example 10
Find the shortest distance between the lines \(\dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1}\) and \(\dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}\).
\((x_1,y_1,z_1)=(-1,-1,-1), (x_2,y_2,z_2)=(3,5,7)\). d.r.s: (7,-6,1), (1,-2,1).
Numerator (determinant): \(\begin{vmatrix}4&6&8\\7&-6&1\\1&-2&1\end{vmatrix}\) = \(4(-6+2) - 6(7-1) + 8(-14+6)\) = \(4(-4) - 6(6) + 8(-8) = -16 - 36 - 64 = -116\). |·|=116.
Denom: \((b_1 c_2 - b_2 c_1)=-6-(-2)=-4\), \((c_1 a_2 - c_2 a_1)=1-7=-6\), \((a_1 b_2 - a_2 b_1)=-14+6=-8\). \(\sqrt{16+36+64}=\sqrt{116}\).
\(d = 116/\sqrt{116} = \sqrt{116} = 2\sqrt{29}\) units.
Numerator (determinant): \(\begin{vmatrix}4&6&8\\7&-6&1\\1&-2&1\end{vmatrix}\) = \(4(-6+2) - 6(7-1) + 8(-14+6)\) = \(4(-4) - 6(6) + 8(-8) = -16 - 36 - 64 = -116\). |·|=116.
Denom: \((b_1 c_2 - b_2 c_1)=-6-(-2)=-4\), \((c_1 a_2 - c_2 a_1)=1-7=-6\), \((a_1 b_2 - a_2 b_1)=-14+6=-8\). \(\sqrt{16+36+64}=\sqrt{116}\).
\(d = 116/\sqrt{116} = \sqrt{116} = 2\sqrt{29}\) units.
Example 11 (Parallel)
Find the distance between parallel lines \(\vec r = \hat i + 2\hat j - 4\hat k + \lambda(2\hat i + 3\hat j + 6\hat k)\) and \(\vec r = 3\hat i + 3\hat j - 5\hat k + \mu(2\hat i + 3\hat j + 6\hat k)\).
\(\vec a_2 - \vec a_1 = (2,1,-1)\), \(\vec b=(2,3,6)\), \(|\vec b|=7\).
\(\vec b \times (\vec a_2 - \vec a_1) = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&6\\2&1&-1\end{vmatrix} = \hat i(-3-6) - \hat j(-2-12) + \hat k(2-6) = (-9, 14, -4)\).
Magnitude = \(\sqrt{81+196+16}=\sqrt{293}\).
\(d = \sqrt{293}/7\) units.
\(\vec b \times (\vec a_2 - \vec a_1) = \begin{vmatrix}\hat i&\hat j&\hat k\\2&3&6\\2&1&-1\end{vmatrix} = \hat i(-3-6) - \hat j(-2-12) + \hat k(2-6) = (-9, 14, -4)\).
Magnitude = \(\sqrt{81+196+16}=\sqrt{293}\).
\(d = \sqrt{293}/7\) units.
Activity: Common Perpendicular with Straws
L4 AnalyseMaterials: Two straws or pencils, a third thin stick, ruler.
- Fix two pencils in skew configuration (e.g., one horizontal, one tilted through space not meeting it).
- Try to find a third thin stick that touches both pencils and is perpendicular to each.
- Measure its length — this is the shortest distance.
- Now note that any other segment joining the two lines is longer (drop a parallel and verify).
The common perpendicular is unique. Its direction is along \(\vec b_1 \times \vec b_2\), which is perpendicular to both direction vectors simultaneously — a unique line (up to sign).
Miscellaneous Exercise (Selected)
Q1. Find the shortest distance between lines whose vector equations are \(\vec r = (1-t)\hat i + (t-2)\hat j + (3-2t)\hat k\) and \(\vec r = (s+1)\hat i + (2s-1)\hat j - (2s+1)\hat k\).
Rewrite: \(L_1:\vec r=(1,-2,3)+t(-1,1,-2)\), \(L_2:\vec r=(1,-1,-1)+s(1,2,-2)\). \(\vec a_2-\vec a_1=(0,1,-4)\). \(\vec b_1\times\vec b_2=(1·(-2)-(-2)·2,\;(-2)·1-(-1)·(-2),\;(-1)·2-1·1)=(2,-4,-3)\). \(|·|=\sqrt{4+16+9}=\sqrt{29}\). Numerator=\(2·0+(-4)·1+(-3)·(-4)=8\). \(d=8/\sqrt{29}\).
Q2. Find the shortest distance between \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\) and \(\dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}\).
Points (1,2,3), (2,4,5) → diff (1,2,2). d.r.s (2,3,4), (3,4,5). Cross = \((15-16,\;12-10,\;8-9)=(-1,2,-1)\). Magnitude=\(\sqrt 6\). Num=\(-1+4-2=1\). \(d=1/\sqrt 6\).
Competency-Based Questions
Scenario: Two aircraft flying straight-line paths have trajectories \(L_1:\vec r=(2,-1,3)+\lambda(1,2,1)\) and \(L_2:\vec r=(0,3,-1)+\mu(2,1,-1)\) (km).
Q1. Calculate the shortest distance between the two flight paths.
L3 Apply\(\vec a_2-\vec a_1=(-2,4,-4)\). Cross of (1,2,1) and (2,1,-1) = (2·(-1)-1·1, 1·2-1·(-1), 1·1-2·2)=(-3,3,-3). Magnitude=\(3\sqrt 3\). Num=|(-3)(-2)+3·4+(-3)(-4)|=|6+12+12|=30. \(d=30/3\sqrt3=10/\sqrt3\approx5.77\) km.
Q2. Analyse whether the paths are skew, parallel, or intersecting.
L4 Analysed.r.s (1,2,1) not proportional to (2,1,-1), so not parallel. Since \(d > 0\), they don't intersect — so they are skew.
Q3. Evaluate: Air traffic safety requires at least 5 km separation. Are these paths safe?
L5 Evaluate\(d \approx 5.77\) km > 5 km, so minimum-distance criterion is satisfied. Safe, marginally. A more conservative rule (10 km) would flag this pair.
Q4. Create a new flight path for a third aircraft that is parallel to \(L_1\) yet at exactly 7 km from it.
L6 CreateKeep direction (1,2,1). Pick a starting point \((x_0,y_0,z_0)\) with perpendicular distance 7 from \(L_1\). E.g., displace (2,-1,3) along direction perpendicular to (1,2,1), say (2,-1,0)/\(\sqrt 5\). New start = (2,-1,3)+7·(2,-1,0)/\(\sqrt 5\). Resulting line: \(\vec r = (2+14/\sqrt 5, -1-7/\sqrt 5, 3)+t(1,2,1)\). Many valid answers.
Assertion–Reason Questions
A: If \((\vec b_1 \times \vec b_2)\cdot(\vec a_2-\vec a_1)=0\), the two lines intersect.
R: The numerator of the shortest-distance formula vanishes iff the lines are coplanar.
R: The numerator of the shortest-distance formula vanishes iff the lines are coplanar.
(d) A is false — coplanar lines could also be parallel (non-intersecting). R is true.
A: Parallel lines in 3D always have the same direction ratios up to a scalar.
R: For parallel lines, \(\vec b_1\times\vec b_2 = \vec 0\).
R: For parallel lines, \(\vec b_1\times\vec b_2 = \vec 0\).
(a) Cross product zero ⇔ vectors parallel ⇔ d.r.s proportional. R explains A.
A: The shortest distance between two intersecting lines is zero.
R: Intersecting lines share a common point.
R: Intersecting lines share a common point.
(a) At the common point, distance = 0. R explains A directly.
Frequently Asked Questions — Three Dimensional Geometry
What is Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool in NCERT Class 12 Mathematics?
Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool is a key concept covered in NCERT Class 12 Mathematics, Chapter 11: Three Dimensional Geometry. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool step by step?
To solve problems on Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 11: Three Dimensional Geometry?
The essential formulas of Chapter 11 (Three Dimensional Geometry) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool important for the Class 12 board exam?
Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool?
Common mistakes in Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool?
End-of-chapter NCERT exercises for Part 3 — Shortest Distance Between Skew & Parallel Lines | Class 12 Maths Ch 11 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 11, and solve at least one previous-year board paper to consolidate your understanding.
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