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Three-Dimensional Geometry - Shortest Distance between Skew Lines

Grade 12ICSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Definition of Skew Lines: Skew lines are a pair of lines in three-dimensional space that are neither parallel to each other nor do they intersect. Visually, imagine two distinct planesβ€”like the floor and the ceiling of a room. If one line is drawn on the floor and another on the ceiling in a different direction, they will never touch and are not parallel; these are skew lines.

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Shortest Distance (SD) Property: The shortest distance between two skew lines is the length of the unique line segment that is perpendicular to both lines simultaneously. Geometrically, this segment acts as a 'bridge' at the point where the lines appear to cross when viewed from a perpendicular angle.

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Vector Representation: In 3D geometry, lines are represented as r⃗=a⃗1+λb⃗1\vec{r} = \vec{a}_1 + \lambda\vec{b}_1 and r⃗=a⃗2+μb⃗2\vec{r} = \vec{a}_2 + \mu\vec{b}_2. Here, a⃗1\vec{a}_1 and a⃗2\vec{a}_2 are position vectors of points on the lines, while b⃗1\vec{b}_1 and b⃗2\vec{b}_2 define the directions of the lines.

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Common Perpendicular Direction: The direction of the shortest distance segment is perpendicular to both direction vectors b⃗1\vec{b}_1 and b⃗2\vec{b}_2. This direction is obtained by the cross product b⃗1×b⃗2\vec{b}_1 \times \vec{b}_2. Visually, this resulting vector sticks out perpendicular to the 'virtual' planes containing each line.

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Shortest Distance as a Projection: The shortest distance can be understood as the magnitude of the projection of the vector joining any two points on the lines (usually aβƒ—2βˆ’aβƒ—1\vec{a}_2 - \vec{a}_1) onto the unit vector along the common perpendicular (bβƒ—1Γ—bβƒ—2)(\vec{b}_1 \times \vec{b}_2).

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Condition for Intersection: If the shortest distance between two lines is zero, the lines intersect and are said to be coplanar. This happens when the scalar triple product (aβƒ—2βˆ’aβƒ—1)β‹…(bβƒ—1Γ—bβƒ—2)(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) equals zero.

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Distance between Parallel Lines: If the direction vectors b⃗1\vec{b}_1 and b⃗2\vec{b}_2 are parallel (proportional), the lines are parallel rather than skew. In this case, the distance is constant and is calculated using the magnitude of the cross product of the direction vector with the difference of position vectors.

πŸ“Formulae

Vector Form (Skew Lines): d=∣(aβƒ—2βˆ’aβƒ—1)β‹…(bβƒ—1Γ—bβƒ—2)∣bβƒ—1Γ—bβƒ—2∣∣d = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|

Cartesian Form: d=∣∣x2βˆ’x1y2βˆ’y1z2βˆ’z1l1m1n1l2m2n2∣∣(m1n2βˆ’m2n1)2+(n1l2βˆ’n2l1)2+(l1m2βˆ’l2m1)2d = \frac{\left| \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \end{vmatrix} \right|}{\sqrt{(m_1n_2 - m_2n_1)^2 + (n_1l_2 - n_2l_1)^2 + (l_1m_2 - l_2m_1)^2}}

Distance between Parallel Lines: d=∣bβƒ—Γ—(aβƒ—2βˆ’aβƒ—1)∣∣bβƒ—βˆ£d = \frac{|\vec{b} \times (\vec{a}_2 - \vec{a}_1)|}{|\vec{b}|}

Vector Cross Product: bβƒ—1Γ—bβƒ—2=∣i^j^k^b1xb1yb1zb2xb2yb2z∣\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ b_{1x} & b_{1y} & b_{1z} \\ b_{2x} & b_{2y} & b_{2z} \end{vmatrix}

πŸ’‘Examples

Problem 1:

Find the shortest distance between the lines rβƒ—=(i^+2j^+k^)+Ξ»(i^βˆ’j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and rβƒ—=(2i^βˆ’j^βˆ’k^)+ΞΌ(2i^+j^+2k^)\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k}).

Solution:

  1. Identify vectors: aβƒ—1=i^+2j^+k^\vec{a}_1 = \hat{i} + 2\hat{j} + \hat{k}, bβƒ—1=i^βˆ’j^+k^\vec{b}_1 = \hat{i} - \hat{j} + \hat{k}, aβƒ—2=2i^βˆ’j^βˆ’k^\vec{a}_2 = 2\hat{i} - \hat{j} - \hat{k}, bβƒ—2=2i^+j^+2k^\vec{b}_2 = 2\hat{i} + \hat{j} + 2\hat{k}.
  2. Calculate aβƒ—2βˆ’aβƒ—1=(2βˆ’1)i^+(βˆ’1βˆ’2)j^+(βˆ’1βˆ’1)k^=i^βˆ’3j^βˆ’2k^\vec{a}_2 - \vec{a}_1 = (2-1)\hat{i} + (-1-2)\hat{j} + (-1-1)\hat{k} = \hat{i} - 3\hat{j} - 2\hat{k}.
  3. Calculate bβƒ—1Γ—bβƒ—2=∣i^j^k^1βˆ’11212∣=i^(βˆ’2βˆ’1)βˆ’j^(2βˆ’2)+k^(1+2)=βˆ’3i^+0j^+3k^\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\hat{i} + 0\hat{j} + 3\hat{k}.
  4. Find magnitude ∣bβƒ—1Γ—bβƒ—2∣=(βˆ’3)2+02+32=18=32|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}.
  5. Calculate dot product: (aβƒ—2βˆ’aβƒ—1)β‹…(bβƒ—1Γ—bβƒ—2)=(1)(βˆ’3)+(βˆ’3)(0)+(βˆ’2)(3)=βˆ’3βˆ’6=βˆ’9(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-3) + (-3)(0) + (-2)(3) = -3 - 6 = -9.
  6. Apply formula: d=βˆ£βˆ’932∣=32=322d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} units.

Explanation:

We first extracted the position and direction vectors for both lines. We then calculated the cross product of the direction vectors to find the common perpendicular. Finally, we projected the vector connecting the two lines onto this common perpendicular to find the magnitude of the shortest distance.

Problem 2:

Find the shortest distance between the parallel lines rβƒ—=(i^+2j^βˆ’4k^)+Ξ»(2i^+3j^+6k^)\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k}) and rβƒ—=(3i^+3j^βˆ’5k^)+ΞΌ(2i^+3j^+6k^)\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k}).

Solution:

  1. Identify: aβƒ—1=i^+2j^βˆ’4k^\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}, aβƒ—2=3i^+3j^βˆ’5k^\vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}, and common direction bβƒ—=2i^+3j^+6k^\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}.
  2. Find aβƒ—2βˆ’aβƒ—1=2i^+j^βˆ’k^\vec{a}_2 - \vec{a}_1 = 2\hat{i} + \hat{j} - \hat{k}.
  3. Calculate bβƒ—Γ—(aβƒ—2βˆ’aβƒ—1)=∣i^j^k^23621βˆ’1∣=i^(βˆ’3βˆ’6)βˆ’j^(βˆ’2βˆ’12)+k^(2βˆ’6)=βˆ’9i^+14j^βˆ’4k^\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\hat{i} + 14\hat{j} - 4\hat{k}.
  4. Magnitude ∣bβƒ—Γ—(aβƒ—2βˆ’aβƒ—1)∣=(βˆ’9)2+142+(βˆ’4)2=81+196+16=293|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-9)^2 + 14^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}.
  5. Magnitude ∣bβƒ—βˆ£=22+32+62=4+9+36=7|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = 7.
  6. Distance d=2937d = \frac{\sqrt{293}}{7} units.

Explanation:

Since the direction vectors are identical, the lines are parallel. We used the specific parallel distance formula, which involves the cross product of the common direction vector with the vector connecting points on each line, divided by the magnitude of the direction vector.