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

Grade 12ICSE

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

πŸ”‘Concepts

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A line in three-dimensional space is uniquely determined if it passes through a given point and has a given direction, or if it passes through two given points. Visually, think of a line as a path traced by a point whose position vector r⃗\vec{r} changes relative to a fixed origin, guided by a direction vector b⃗\vec{b}.

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Direction Cosines and Direction Ratios: If a line makes angles α,β,γ\alpha, \beta, \gamma with the positive directions of the x,y,and zx, y, \text{and } z axes respectively, then cos⁑α,cos⁑β,cos⁑γ\cos \alpha, \cos \beta, \cos \gamma are called direction cosines (represented as l,m,nl, m, n). Any numbers a,b,ca, b, c proportional to l,m,nl, m, n are direction ratios. Visually, these values define the 'tilt' of the line in 3D space.

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Vector Equation of a Line: The line passing through a point with position vector a⃗\vec{a} and parallel to a vector b⃗\vec{b} is given by r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b}, where λ\lambda is a scalar parameter. Geometrically, as λ\lambda varies, the tip of the vector r⃗\vec{r} moves along the infinite line.

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Cartesian Equation of a Line: If a line passes through the point (x1,y1,z1)(x_1, y_1, z_1) and has direction ratios a,b,ca, b, c, its equation is xβˆ’x1a=yβˆ’y1b=zβˆ’z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}. This form represents the line as the intersection of two planes in the XYZ coordinate system.

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Line passing through two points: For a line passing through two 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 (x2βˆ’x1),(y2βˆ’y1),(z2βˆ’z1)(x_2 - x_1), (y_2 - y_1), (z_2 - z_1). The vector equation is rβƒ—=aβƒ—+Ξ»(bβƒ—βˆ’aβƒ—)\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a}).

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Angle between two lines: The angle ΞΈ\theta between two lines is the angle between their direction vectors. If the direction vectors are b1βƒ—\vec{b_1} and b2βƒ—\vec{b_2}, then cos⁑θ=∣b1βƒ—β‹…b2βƒ—βˆ£βˆ£b1βƒ—βˆ£βˆ£b2βƒ—βˆ£\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}. Visually, if the lines are perpendicular, their dot product is zero; if parallel, one is a scalar multiple of the other.

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Shortest Distance between Skew Lines: Skew lines are lines that are neither parallel nor intersecting and lie in different planes. The shortest distance is the length of the line segment perpendicular to both lines. Visually, imagine two wires in a room at different heights running in different directions; the shortest distance is the 'gap' between them.

πŸ“Formulae

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

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

Vector equation (Two points): rβƒ—=aβƒ—+Ξ»(bβƒ—βˆ’aβƒ—)\vec{r} = \vec{a} + \lambda (\vec{b} - \vec{a})

Cartesian equation (Two points): xβˆ’x1x2βˆ’x1=yβˆ’y1y2βˆ’y1=zβˆ’z1z2βˆ’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=∣(b1βƒ—Γ—b2βƒ—)β‹…(a2βƒ—βˆ’a1βƒ—)∣b1βƒ—Γ—b2βƒ—βˆ£βˆ£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=∣bβƒ—Γ—(a2βƒ—βˆ’a1βƒ—)∣bβƒ—βˆ£βˆ£d = \left| \frac{\vec{b} \times (\vec{a_2} - \vec{a_1})}{|\vec{b}|} \right|

πŸ’‘Examples

Problem 1:

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

Solution:

  1. Identify the given point vector: aβƒ—=5i^+2j^βˆ’4k^\vec{a} = 5\hat{i} + 2\hat{j} - 4\hat{k}. \n2. Identify the direction vector: bβƒ—=3i^+2j^βˆ’8k^\vec{b} = 3\hat{i} + 2\hat{j} - 8\hat{k}. \n3. Plug into the vector equation formula rβƒ—=aβƒ—+Ξ»bβƒ—\vec{r} = \vec{a} + \lambda \vec{b}: \n rβƒ—=(5i^+2j^βˆ’4k^)+Ξ»(3i^+2j^βˆ’8k^)\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k}). \n4. For Cartesian equation, use (x1,y1,z1)=(5,2,βˆ’4)(x_1, y_1, z_1) = (5, 2, -4) and direction ratios (a,b,c)=(3,2,βˆ’8)(a, b, c) = (3, 2, -8): \n xβˆ’53=yβˆ’22=zβˆ’(βˆ’4)βˆ’8β‡’xβˆ’53=yβˆ’22=z+4βˆ’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=yβˆ’15=z+34\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4} and xβˆ’11=yβˆ’41=zβˆ’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). \n2. Identify direction ratios of second line: (a2,b2,c2)=(1,1,2)(a_2, b_2, c_2) = (1, 1, 2). \n3. Use the cosine formula: \n 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}} \n4. Calculate numerator: ∣3+5+8∣=16|3 + 5 + 8| = 16. \n5. Calculate denominator: 9+25+16β‹…1+1+4=50β‹…6=52β‹…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}. \n6. cos⁑θ=16103=853\cos \theta = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}. \n7. ΞΈ=cosβ‘βˆ’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.