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Three Dimensional Geometry - Cartesian equation and vector equation of a line

Grade 12CBSE

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

🔑Concepts

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Direction Ratios and Direction Cosines: These values define the orientation of a line in 3D space. Direction ratios (a,b,c)(a, b, c) are any three numbers proportional to the direction cosines (l,m,n)(l, m, n), which represent the cosines of the angles α,β,γ\alpha, \beta, \gamma the line makes with the x,y,zx, y, z axes. Visually, if you imagine a line passing through the origin, its direction can be visualized as a vector pointing toward the point (a,b,c)(a, b, c) from (0,0,0)(0, 0, 0).

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Vector Equation of a Line: A line is uniquely determined by a point AA with position vector a⃗\vec{a} and a direction vector b⃗\vec{b}. The equation is r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where λ\lambda is a scalar. Visually, this is like starting at point AA and moving any distance along a path parallel to vector b⃗\vec{b} in both positive and negative directions.

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Cartesian Equation of a Line: Also known as the 'Symmetrical 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 written as x−x1a=y−y1b=z−z1c\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}. Visually, this expresses the line as the intersection of two planes and relates the coordinates of any point on the line to its fixed point and slope components.

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Line Passing Through Two Points: If a line passes through two points A(a⃗)A(\vec{a}) and B(b⃗)B(\vec{b}), its direction vector is given by b⃗−a⃗\vec{b} - \vec{a}. The vector equation becomes r⃗=a⃗+λ(b⃗−a⃗)\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a}). Visually, the line acts as an infinite extension of the line segment connecting points AA and BB.

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Angle Between Two Lines: The angle θ\theta between two lines depends only on their direction vectors b1⃗\vec{b_1} and b2⃗\vec{b_2}. Visually, even if two lines in 3D are skew (do not intersect), we determine the angle between them by shifting them parallel to themselves until they meet at a common point.

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Shortest Distance Between Skew Lines: Skew lines are lines in 3D space that are neither parallel nor intersecting. They lie in different planes. Visually, imagine one line on the floor and another on the ceiling pointing in a different direction; the shortest distance is the length of the common perpendicular line segment connecting them.

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Shortest Distance Between Parallel Lines: Since parallel lines have the same direction, the distance between them is constant. Visually, this distance is the perpendicular length from any point on one line to the other line, similar to the distance between two parallel rails on a track.

📐Formulae

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

Cartesian equation of a line: x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

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

Cartesian equation (two-point form): 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=∣b1⃗⋅b2⃗∣b1⃗∣∣b2⃗∣∣\cos \theta = \left| \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \right|

Cartesian Angle: 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 = \frac{|\vec{b} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b}|}

💡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. Substitute 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. Identify coordinates (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).
  5. Substitute into the Cartesian formula: x−53=y−22=z+4−8\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}

Explanation:

This problem demonstrates the direct conversion from given geometric parameters (a point and a direction) to both standard equation forms in 3D geometry.

Problem 2:

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

Solution:

  1. Extract points and directions: a1⃗=(1,1,0)\vec{a_1} = (1, 1, 0), b1⃗=(2,−1,1)\vec{b_1} = (2, -1, 1); a2⃗=(2,1,−1)\vec{a_2} = (2, 1, -1), b2⃗=(3,−5,2)\vec{b_2} = (3, -5, 2).
  2. Find a2⃗−a1⃗=(2−1)i^+(1−1)j^+(−1−0)k^=i^−k^\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (1-1)\hat{j} + (-1-0)\hat{k} = \hat{i} - \hat{k}.
  3. Calculate cross product b1⃗×b2⃗\vec{b_1} \times \vec{b_2}: b1⃗×b2⃗=∣i^j^k^2−113−52∣=i^(−2+5)−j^(4−3)+k^(−10+3)=3i^−j^−7k^\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix} = \hat{i}(-2 + 5) - \hat{j}(4 - 3) + \hat{k}(-10 + 3) = 3\hat{i} - \hat{j} - 7\hat{k}
  4. Calculate magnitude: ∣b1⃗×b2⃗∣=32+(−1)2+(−7)2=9+1+49=59|\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + (-1)^2 + (-7)^2} = \sqrt{9 + 1 + 49} = \sqrt{59}.
  5. Calculate dot product: (b1⃗×b2⃗)⋅(a2⃗−a1⃗)=(3)(1)+(−1)(0)+(−7)(−1)=3+7=10(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (3)(1) + (-1)(0) + (-7)(-1) = 3 + 7 = 10.
  6. Apply formula: d=∣1059∣d = \left| \frac{10}{\sqrt{59}} \right|.

Explanation:

To find the distance between skew lines, we determine the vector that is perpendicular to both lines (using cross product) and project the vector joining points on both lines onto this common perpendicular.