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Three Dimensional Geometry - Coplanar and skew lines, shortest distance between two lines

Grade 12CBSE

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

🔑Concepts

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Skew Lines: Skew lines are a pair of lines in three-dimensional space that are neither parallel nor intersecting. Visually, imagine two wires passing through a room at different heights and in different directions; they will never meet and are not oriented in the same way. They always lie in different, non-parallel planes.

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Coplanar Lines: Two lines are said to be coplanar if they lie in the same geometric plane. This occurs if the lines either intersect at a single point or are parallel to each other. Visually, these are lines that could be drawn together on a single flat sheet of paper.

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The Shortest Distance between Skew Lines: The shortest distance between two skew lines is the length of the unique line segment that is perpendicular to both lines simultaneously. This segment is known as the 'common perpendicular'. Visually, it represents the narrowest gap between the two lines in space.

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Shortest Distance between Parallel Lines: If two lines are parallel, they lie in the same plane and maintain a constant distance from each other. The shortest distance is the perpendicular distance from any arbitrary point on one line to the other line. Visually, this is similar to the distance between the two rails of a straight train track.

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Vector Representation of Lines: A line in 3D is defined by a point it passes through (position vector a⃗\vec{a}) and its direction (direction vector b⃗\vec{b}). The general equation is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where λ\lambda is a scalar. Visually, a⃗\vec{a} locates the line in space, and b⃗\vec{b} determines its orientation.

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Condition for Coplanarity: For two lines to be coplanar, the volume of the parallelepiped formed by the vector joining their reference points and their two direction vectors must be zero. Mathematically, this means the scalar triple product (a⃗2−a⃗1)⋅(b⃗1×b⃗2)(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) must equal zero.

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Intersecting Lines: Intersecting lines are a specific type of coplanar lines where the shortest distance between them is exactly zero. They share exactly one common point of coordinates (x,y,z)(x, y, z).

📐Formulae

Vector equation of a line: r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}

Cartesian equation of a line: x−x1l=y−y1m=z−z1n\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}, where (x1,y1,z1)(x_1, y_1, z_1) is a point on the line and l,m,nl, m, n are direction ratios.

Shortest distance between skew lines 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: d=∣(b⃗1×b⃗2)⋅(a⃗2−a⃗1)∣b⃗1×b⃗2∣∣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|

Shortest distance between parallel lines r⃗=a⃗1+λb⃗\vec{r} = \vec{a}_1 + \lambda \vec{b} and r⃗=a⃗2+μb⃗\vec{r} = \vec{a}_2 + \mu \vec{b}: d=∣b⃗×(a⃗2−a⃗1)∣b⃗∣∣d = \left| \frac{\vec{b} \times (\vec{a}_2 - \vec{a}_1)}{|\vec{b}|} \right|

Condition for coplanarity in Cartesian form: ∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣=0\begin{vmatrix} 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{vmatrix} = 0

Shortest distance for Cartesian skew lines: d=∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣(b1c2−b2c1)2+(c1a2−c2a1)2+(a1b2−a2b1)2d = \frac{\begin{vmatrix} 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{vmatrix}}{\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}}

💡Examples

Problem 1:

Find the shortest distance between the lines L1:r⃗=(i^+2j^+3k^)+λ(i^−3j^+2k^)L_1: \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k}) and L2:r⃗=(4i^+5j^+6k^)+μ(2i^+3j^+k^)L_2: \vec{r} = (4\hat{i} + 5\hat{j} + 6\hat{k}) + \mu(2\hat{i} + 3\hat{j} + \hat{k}).

Solution:

  1. Identify vectors: a⃗1=i^+2j^+3k^\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, b⃗1=i^−3j^+2k^\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}, a⃗2=4i^+5j^+6k^\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}, b⃗2=2i^+3j^+k^\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}.
  2. Calculate a⃗2−a⃗1=(4−1)i^+(5−2)j^+(6−3)k^=3i^+3j^+3k^\vec{a}_2 - \vec{a}_1 = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k} = 3\hat{i} + 3\hat{j} + 3\hat{k}.
  3. Find b⃗1×b⃗2=∣i^j^k^1−32231∣=i^(−3−6)−j^(1−4)+k^(3−(−6))=−9i^+3j^+9k^\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = \hat{i}(-3 - 6) - \hat{j}(1 - 4) + \hat{k}(3 - (-6)) = -9\hat{i} + 3\hat{j} + 9\hat{k}.
  4. Magnitude ∣b⃗1×b⃗2∣=(−9)2+32+92=81+9+81=171=319|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-9)^2 + 3^2 + 9^2} = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}.
  5. Scalar product (b⃗1×b⃗2)⋅(a⃗2−a⃗1)=(−9)(3)+(3)(3)+(9)(3)=−27+9+27=9(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-9)(3) + (3)(3) + (9)(3) = -27 + 9 + 27 = 9.
  6. Distance d=∣9319∣=319d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}} units.

Explanation:

We use the shortest distance formula for skew lines. We first determine the vector connecting points on both lines and the cross product of the direction vectors to find the common perpendicular direction. The projection of the connecting vector onto the common perpendicular gives the shortest distance.

Problem 2:

Show that the lines x−12=y−23=z−34\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and x−45=y−12=z\frac{x - 4}{5} = \frac{y - 1}{2} = z are coplanar.

Solution:

  1. Extract points and directions: P1(1,2,3)P_1(1, 2, 3), b⃗1=(2,3,4)\vec{b}_1 = (2, 3, 4) and P2(4,1,0)P_2(4, 1, 0), b⃗2=(5,2,1)\vec{b}_2 = (5, 2, 1).
  2. Calculate x2−x1=3,y2−y1=−1,z2−z1=−3x_2 - x_1 = 3, y_2 - y_1 = -1, z_2 - z_1 = -3.
  3. Evaluate the determinant: D=∣3−1−3234521∣D = \begin{vmatrix} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}.
  4. D=3(3(1)−2(4))−(−1)(2(1)−5(4))+(−3)(2(2)−5(3))D = 3(3(1) - 2(4)) - (-1)(2(1) - 5(4)) + (-3)(2(2) - 5(3)).
  5. D=3(3−8)+1(2−20)−3(4−15)=3(−5)+1(−18)−3(−11)=−15−18+33=0D = 3(3 - 8) + 1(2 - 20) - 3(4 - 15) = 3(-5) + 1(-18) - 3(-11) = -15 - 18 + 33 = 0.
  6. Since D=0D = 0, the lines are coplanar.

Explanation:

To check if two lines are coplanar in Cartesian form, we use the determinant condition. This determinant represents the scalar triple product of the vector joining the two points and the two direction vectors. If the volume of the resulting parallelepiped is zero, the vectors must lie in the same plane.