krit.club logo

Three-Dimensional Geometry - Equation of a Line (Vector and Cartesian)

Grade 12ICSE

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

🔑Concepts

•

The equation of a line in 3D space is uniquely determined if it passes through a given point AA with position vector a⃗\vec{a} and is parallel to a given direction vector b⃗\vec{b}. This is represented in vector form as r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda\vec{b}.

A line L passing through point A and parallel to direction vector b.
•

In Cartesian form, the equation of a line passing through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c is x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}. These ratios are the components of the direction vector b⃗=ai^+bj^+ck^\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}.

3D coordinate axes showing X, Y, and Z directions.
•

A line can also be defined by two distinct points A(x1,y1,z1)A(x_1, y_1, z_1) and B(x2,y2,z2)B(x_2, y_2, z_2). The direction ratios are found by the difference of their coordinates: (x2−x1,y2−y1,z2−z1)(x_2 - x_1, y_2 - y_1, z_2 - z_1).

A line segment defined by two points A and B.
•

The shortest distance between two skew lines is the length of the common perpendicular to both lines. If the distance is zero, the lines intersect.

Two skew lines in different planes with a perpendicular line segment representing the shortest distance.

📐Formulae

Vector equation (Point and Direction): r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}

Cartesian equation (Point and Direction): x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

Vector equation (Two points): r⃗=a⃗+λ(b⃗−a⃗)\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})

Cartesian equation (Two points): x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}

Angle between lines: cos⁡θ=∣a1a2+b1b2+c1c2a12+b12+c12a22+b22+c22∣\cos \theta = \left| \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \right|

Shortest distance (Skew lines): d=∣(b1⃗×b2⃗)⋅(a2⃗−a1⃗)∣b1⃗×b2⃗∣∣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 (Parallel lines): d=∣b⃗×(a2⃗−a1⃗)∣b⃗∣∣d = \left| \frac{\vec{b} \times (\vec{a_2} - \vec{a_1})}{|\vec{b}|} \right|

💡Examples

Problem 1:

Find the vector and Cartesian equations of the line passing through the point (5,2,−4)(5, 2, -4) and which is parallel to the vector 3i^+2j^−8k^3\hat{i} + 2\hat{j} - 8\hat{k}.

Solution:

  1. Identify the given point vector: a⃗=5i^+2j^−4k^\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}.
  2. Identify the direction vector: b⃗=3i^+2j^−8k^\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}.
  3. Plug into the vector equation formula r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}: r⃗=(5i^+2j^−4k^)+λ(3i^+2j^−8k^)\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}).
  4. For Cartesian equation, use (x1,y1,z1)=(5,2,−4)(x_1, y_1, z_1) = (5, 2, -4) and direction ratios (a,b,c)=(3,2,−8)(a, b, c) = (3, 2, -8): x−53=y−22=z−(−4)−8⇒x−53=y−22=z+4−8\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z - (-4)}{-8} \Rightarrow \frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}.

Explanation:

This problem demonstrates the direct conversion from physical coordinates and a direction vector into both standard forms of a 3D line.

Problem 2:

Find the angle between the pair of lines given by: x+33=y−15=z+34\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4} and x−11=y−41=z−52\frac{x-1}{1} = \frac{y-4}{1} = \frac{z-5}{2}.

Solution:

  1. Identify direction ratios of first line: (a1,b1,c1)=(3,5,4)(a_1, b_1, c_1) = (3, 5, 4).
  2. Identify direction ratios of second line: (a2,b2,c2)=(1,1,2)(a_2, b_2, c_2) = (1, 1, 2).
  3. Use the cosine formula: cos⁡θ=∣(3)(1)+(5)(1)+(4)(2)∣32+52+4212+12+22\cos \theta = \frac{|(3)(1) + (5)(1) + (4)(2)|}{\sqrt{3^2 + 5^2 + 4^2} \sqrt{1^2 + 1^2 + 2^2}}
  4. Calculate numerator: ∣3+5+8∣=16|3 + 5 + 8| = 16.
  5. Calculate denominator: 9+25+16⋅1+1+4=50⋅6=52⋅6=103\sqrt{9 + 25 + 16} \cdot \sqrt{1 + 1 + 4} = \sqrt{50} \cdot \sqrt{6} = 5\sqrt{2} \cdot \sqrt{6} = 10\sqrt{3}.
  6. cos⁡θ=16103=853\cos \theta = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}.
  7. θ=cos⁡−1(853)\theta = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right).

Explanation:

The angle between lines depends solely on their direction ratios. We extract the denominators from the Cartesian form and apply the dot product formula.

Problem 3:

Find the Cartesian equation of the line passing through the points A(1,−1,3)A(1, -1, 3) and B(2,3,1.1)B(2, 3, 1.1).

Graph of a line passing through points A and B on a 2D projection plane.

Solution:

  1. Identify coordinates: (x1,y1,z1)=(1,−1,3)(x_1, y_1, z_1) = (1, -1, 3) and (x2,y2,z2)=(2,3,1.1)(x_2, y_2, z_2) = (2, 3, 1.1).
  2. Calculate direction ratios: a=x2−x1=2−1=1a = x_2 - x_1 = 2 - 1 = 1 b=y2−y1=3−(−1)=4b = y_2 - y_1 = 3 - (-1) = 4 c=z2−z1=1.1−3=−1.9c = z_2 - z_1 = 1.1 - 3 = -1.9
  3. Apply the formula: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}
  4. Result: x−11=y+14=z−3−1.9\frac{x - 1}{1} = \frac{y + 1}{4} = \frac{z - 3}{-1.9}

Explanation:

To find the equation through two points, we first find the direction vector by subtracting the coordinates of the first point from the second, then substitute into the point-direction Cartesian form.

Problem 4:

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

Diagram showing the shortest distance d between two non-intersecting, non-parallel lines.

Solution:

  1. Identify vectors: a1⃗=(1,2,3)\vec{a_1} = (1, 2, 3), b1⃗=(1,−3,2)\vec{b_1} = (1, -3, 2), a2⃗=(4,5,6)\vec{a_2} = (4, 5, 6), b2⃗=(2,3,1)\vec{b_2} = (2, 3, 1).
  2. Compute a2⃗−a1⃗=(3,3,3)\vec{a_2} - \vec{a_1} = (3, 3, 3).
  3. Compute b1⃗×b2⃗=∣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 ∣b1⃗×b2⃗∣=(−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. Dot product (b1⃗×b2⃗)⋅(a2⃗−a1⃗)=(−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:

Skew lines are lines that are not parallel and do not intersect. The shortest distance is the projection of the vector joining two points (one on each line) onto the vector perpendicular to both lines.