01 · Explore
Direction Cosines and Direction Ratios
A line in three-dimensional space is characterized by the angles it makes with the coordinate axes. These angles determine the orientation of the line.
If a directed line L passing through the origin makes angles α, β, and γ with the positive x, y, and z-axes respectively, then cos α, cos β, and cos γ are called the direction cosines (d.c.'s) of the line, denoted by l, m, and n. A key property of these cosines is that the sum of their squares is always unity: l² + m² + n² = 1.
Direction ratios (d.r.'s) are any three numbers a, b, and c that are proportional to the direction cosines. This means a = λl, b = λm, and c = λn for some non-zero real number λ. Unlike direction cosines, which are unique for a directed line, a line has infinitely many sets of direction ratios.
To convert direction ratios (a, b, c) into direction cosines (l, m, n), we use the magnitude √(a² + b² + c²). The relations are l = ±a/√(a² + b² + c²), m = ±b/√(a² + b² + c²), and n = ±c/√(a² + b² + c²). The choice of sign depends on the directed orientation of the line.
| Quantity | Symbol | Constraint/Relation |
|---|---|---|
| Direction Angles | α, β, γ | Angles with x, y, z axes |
| Direction Cosines | l, m, n | l² + m² + n² = 1 |
| Direction Ratios | a, b, c | a/l = b/m = c/n = λ |
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Question
If a line has direction ratios 2, -1, -2, find its direction cosines.
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02 · Explore
Line Passing Through Two Points
The direction of a line is uniquely determined by any two distinct points lying on it.
Consider two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂). The direction ratios of the line passing through these points can be taken as the differences of their coordinates: (x₂ - x₁), (y₂ - y₁), and (z₂ - z₁).
The distance between the points, PQ, is given by the formula √((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²). The direction cosines are then obtained by dividing each direction ratio by this distance PQ.
For example, the direction cosines are l = (x₂ - x₁)/PQ, m = (y₂ - y₁)/PQ, and n = (z₂ - z₁)/PQ. This ensures that the sum of the squares of l, m, and n equals 1, satisfying the fundamental property of direction cosines.
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Question
What are the direction cosines of the y-axis?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 3, 4.
03 · Explore
Vector and Cartesian Equations of a Line
A line in space is defined by a point it passes through and its direction.
Vector Equation: Let a line pass through a point A with position vector 'a' and be parallel to a vector 'b'. The position vector 'r' of any arbitrary point P on the line is given by r = a + λb, where λ is a real parameter.
Cartesian Equation: If the line passes through (x₁, y₁, z₁) and has direction ratios a, b, c, the equation is (x - x₁)/a = (y - y₁)/b = (z - z₁)/c. If the line is defined specifically by direction cosines l, m, n, the equation is (x - x₁)/l = (y - y₁)/m = (z - z₁)/n.
To convert from vector form r = (x₁i + y₁j + z₁k) + λ(ai + bj + ck) to Cartesian form, we equate the components: x = x₁ + λa, y = y₁ + λb, z = z₁ + λc. Solving for λ in each gives the standard Cartesian ratio.
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Question
If a line has the Cartesian equation (x - 5)/3 = (y + 4)/7 = (z - 6)/2, find a point on the line and its direction ratios.
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04 · Explore
Angle Between Two Lines
The angle between two lines is determined by the angle between their direction vectors.
If two lines have direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂), the acute angle θ between them is given by cos θ = |(a₁a₂ + b₁b₂ + c₁c₂)| / [√(a₁² + b₁² + c₁²) · √(a₂² + b₂² + c₂²)].
In vector form, if the lines are r = a₁ + λb₁ and r = a₂ + μb₂, then cos θ = |b₁ · b₂| / (|b₁| |b₂|). This is the standard dot product formula applied to the direction vectors.
Two lines are perpendicular if their dot product is zero: a₁a₂ + b₁b₂ + c₁c₂ = 0. They are parallel if their direction ratios are proportional: a₁/a₂ = b₁/b₂ = c₁/c₂.
Calculating Angle Between Lines
L1: r = (3i+2j-4k) + λ(i+2j+2k); L2: r = (5i-2j) + μ(3i+2j+6k)
b₁ = i+2j+2k, b₂ = 3i+2j+6k. b₁·b₂ = (1)(3) + (2)(2) + (2)(6) = 3 + 4 + 12 = 19. |b₁| = √(1+4+4) = 3. |b₂| = √(9+4+36) = 7. cos θ = 19 / (3 * 7) = 19/21. θ = cos⁻¹(19/21).
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Question
What is the condition for two lines with direction cosines (l₁, m₁, n₁) and (l₂, m₂, n₂) to be perpendicular?
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05 · Explore
Shortest Distance Between Skew Lines
Skew lines are lines in space that are neither parallel nor intersecting. They lie in different planes.
The shortest distance between two skew lines is the length of the common perpendicular. For lines r = a₁ + λb₁ and r = a₂ + μb₂, the shortest distance 'd' is given by the magnitude of the projection of (a₂ - a₁) on the vector (b₁ × b₂).
The formula is d = |(b₁ × b₂) · (a₂ - a₁)| / |b₁ × b₂|. If the result is zero, the lines intersect.
In Cartesian form, the numerator is the absolute value of the determinant of a 3x3 matrix with rows (x₂-x₁, y₂-y₁, z₂-z₁), (a₁, b₁, c₁), and (a₂, b₂, c₂). The denominator is given by the formula √((b₁c₂ - b₂c₁)² + (c₁a₂ - c₂a₁)² + (a₁b₂ - a₂b₁)²).
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Question
What does it mean if the shortest distance between two skew lines is zero?
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06 · Explore
Distance Between Parallel Lines
Parallel lines have the same direction vector, so the skew-line formula cannot be used directly as the cross product of direction vectors would be zero.
Two lines r = a₁ + λb and r = a₂ + μb are parallel because they share the same direction vector 'b'. The distance between them is the perpendicular distance from any point on one line to the other.
The formula for the distance 'd' is d = |b × (a₂ - a₁)| / |b|. Here, (a₂ - a₁) is the vector joining a point on the first line to a point on the second line.
This formula uses the cross product to find the component of the joining vector that is perpendicular to the direction of the lines. The magnitude is then normalized by dividing by the magnitude of the direction vector 'b'.
Distance Between Parallel Lines
L1: r = (i+2j-4k) + λ(2i+3j+6k); L2: r = (3i+3j-5k) + μ(2i+3j+6k)
a₁ = i+2j-4k, a₂ = 3i+3j-5k, b = 2i+3j+6k. a₂ - a₁ = 2i + j - k. b × (a₂ - a₁) = det([i, j, k; 2, 3, 6; 2, 1, -1]) = i(-3-6) - j(-2-12) + k(2-6) = -9i + 14j - 4k. |b × (a₂ - a₁)| = √((-9)² + 14² + (-4)²) = √(81 + 196 + 16) = √293. |b| = √(2² + 3² + 6²) = √49 = 7. d = √293 / 7.
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Question
Why is the direction vector 'b' the same in both equations for parallel lines?
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Sign inNew here? Sign up freeNCERT reference: chapter PDF pages 11, 12, 13.