Class 12 — Mathematics (NCERT)

Three Dimensional Geometry

Class 12

✓Define and differentiate between direction cosines and direction ratios of a line in space.
✓Derive and apply the vector and Cartesian equations of a line in various forms.
✓Derive and apply the vector and Cartesian equations of a plane in various forms.
✓Calculate the angle between two lines, two planes, and a line and a plane.
✓Determine the shortest distance between two skew lines and two parallel lines.

Key concepts

Direction Cosines and Direction Ratios

If a directed line L passing through the origin makes angles α, β, γ with the positive directions of x, y, z-axes respectively, then cos α, cos β, cos γ are called the direction cosines of the line L. They are usually denoted by l, m, n. Any three numbers a, b, c which are proportional to the direction cosines l, m, n are called the direction ratios of the line. If a line passes through two points P(x1, y1, z1) and Q(x2, y2, z2), its direction ratios are (x2-x1), (y2-y1), (z2-z1).

l = cos α, m = cos β, n = cos γ; l² + m² + n² = 1; l = ±a/√(a²+b²+c²), m = ±b/√(a²+b²+c²), n = ±c/√(a²+b²+c²)

Equation of a Line in Space

A line in space can be uniquely determined if we know a point through which it passes and its direction. It can also be determined if it passes through two given points.

1. Through a point (x1, y1, z1) and parallel to vector with direction ratios (a, b, c):\n Vector form: r = a + λb\n Cartesian form: (x - x1)/a = (y - y1)/b = (z - z1)/c\n2. Through two points (x1, y1, z1) and (x2, y2, z2):\n Vector form: r = a + λ(b - a)\n Cartesian form: (x - x1)/(x2 - x1) = (y - y1)/(y2 - y1) = (z - z1)/(z2 - z1)

Angle between Two Lines

The angle between two lines is defined as the angle between their direction vectors.

Let lines be r = a1 + λb1 and r = a2 + μb2. Let θ be the angle between them.\nVector form: cos θ = |b1 . b2| / (|b1| |b2|)\nCartesian form: If direction ratios are (a1, b1, c1) and (a2, b2, c2), then cos θ = |a1a2 + b1b2 + c1c2| / (√(a1²+b1²+c1²) √(a2²+b2²+c2²))

Shortest Distance between Two Lines

The shortest distance between two lines is the length of the common perpendicular between them. This is relevant for skew lines (non-parallel and non-intersecting) and parallel lines.

1. Skew Lines: r = a1 + λb1 and r = a2 + μb2\n d = |(b1 x b2) . (a2 - a1)| / |b1 x b2|\n2. Parallel Lines: r = a1 + λb and r = a2 + μb\n d = |b x (a2 - a1)| / |b|

Equation of a Plane in Space

A plane in space can be uniquely determined by various conditions, such as a point on it and a normal vector to it, or three non-collinear points.

1. Normal Form (perpendicular distance 'd' from origin and unit normal vector n̂):\n Vector form: r . n̂ = d\n Cartesian form: lx + my + nz = d (where l, m, n are direction cosines of the normal)\n2. Through a point (x1, y1, z1) and perpendicular to vector n = Ai + Bj + Ck:\n Vector form: (r - a) . n = 0\n Cartesian form: A(x - x1) + B(y - y1) + C(z - z1) = 0\n3. Through three non-collinear points (x1, y1, z1), (x2, y2, z2), (x3, y3, z3):\n Cartesian form: | (x-x1) (y-y1) (z-z1) | = 0\n | (x2-x1) (y2-y1) (z2-z1) |\n | (x3-x1) (y3-y1) (z3-z1) |\n Vector form: (r - a) . [(b - a) x (c - a)] = 0\n4. Intercept Form (intercepts A, B, C on x, y, z axes):\n Cartesian form: x/A + y/B + z/C = 1\n5. Plane passing through the intersection of two planes P1: r . n1 = d1 and P2: r . n2 = d2:\n Vector form: (r . n1 - d1) + λ(r . n2 - d2) = 0\n Cartesian form: (A1x + B1y + C1z - D1) + λ(A2x + B2y + C2z - D2) = 0

Angle between Two Planes

The angle between two planes is defined as the angle between their normal vectors.

Let planes be r . n1 = d1 and r . n2 = d2. Let θ be the angle between them.\nVector form: cos θ = |n1 . n2| / (|n1| |n2|)\nCartesian form: If normal direction ratios are (A1, B1, C1) and (A2, B2, C2), then cos θ = |A1A2 + B1B2 + C1C2| / (√(A1²+B1²+C1²) √(A2²+B2²+C2²))

Angle between a Line and a Plane

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

Let line be r = a + λb and plane be r . n = d. Let φ be the angle between them.\nVector form: sin φ = |b . n| / (|b| |n|)\nCartesian form: If line direction ratios are (a, b, c) and plane normal direction ratios are (A, B, C), then sin φ = |Aa + Bb + Cc| / (√(A²+B²+C²) √(a²+b²+c²))

Key facts to remember

1Direction cosines (l, m, n) of a line satisfy l² + m² + n² = 1.
2The vector equation of a line passing through a point with position vector 'a' and parallel to vector 'b' is r = a + λb.
3The Cartesian equation of a line passing through (x1, y1, z1) with direction ratios (a, b, c) is (x-x1)/a = (y-y1)/b = (z-z1)/c.
4The shortest distance between two skew lines r = a1 + λb1 and r = a2 + μb2 is d = |(b1 x b2) . (a2 - a1)| / |b1 x b2|.
5The vector equation of a plane at a perpendicular distance 'd' from the origin and having n̂ as the unit normal vector is r . n̂ = d.
6The Cartesian equation of a plane passing through (x1, y1, z1) and normal to a vector with direction ratios (A, B, C) is A(x-x1) + B(y-y1) + C(z-z1) = 0.
7The angle θ between two planes with normal vectors n1 and n2 is given by cos θ = |n1 . n2| / (|n1| |n2|).
8The angle φ between a line with direction vector b and a plane with normal vector n is given by sin φ = |b . n| / (|b| |n|).

Worked examples

Example 1

Find the shortest distance between the lines:\nL1: r = (i + 2j + k) + λ(i - j + k)\nL2: r = (2i - j - k) + μ(2i + j + 2k)

IThe given lines are of the form r = a1 + λb1 and r = a2 + μb2.

IIFrom L1: a1 = i + 2j + k, b1 = i - j + k

IIIFrom L2: a2 = 2i - j - k, b2 = 2i + j + 2k

IVFirst, calculate (a2 - a1):

Va2 - a1 = (2i - j - k) - (i + 2j + k) = (2-1)i + (-1-2)j + (-1-1)k = i - 3j - 2k

VINext, calculate the cross product b1 x b2:

VIIb1 x b2 = | i j k |\n | 1 -1 1 |\n | 2 1 2 |

VIII = i((-1)(2) - (1)(1)) - j((1)(2) - (1)(2)) + k((1)(1) - (-1)(2))

9 = i(-2 - 1) - j(2 - 2) + k(1 + 2)

10 = -3i - 0j + 3k = -3i + 3k

11Now, find the magnitude of b1 x b2:

12|b1 x b2| = √((-3)² + 0² + 3²) = √(9 + 9) = √18 = 3√2

13Calculate the dot product (b1 x b2) . (a2 - a1):

14(b1 x b2) . (a2 - a1) = (-3i + 3k) . (i - 3j - 2k)

15 = (-3)(1) + (0)(-3) + (3)(-2)

16 = -3 + 0 - 6 = -9

17Finally, use the formula for shortest distance d = |(b1 x b2) . (a2 - a1)| / |b1 x b2|:

18d = |-9| / (3√2) = 9 / (3√2) = 3/√2

19Rationalise the denominator: d = (3√2) / 2 units.

Answer

The shortest distance between the lines is (3√2)/2 units.

Remember to take the absolute value of the dot product in the numerator as distance is always non-negative.

Example 2

Find the equation of the plane passing through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 and passing through the point (2, 2, 1).

IThe equation of a plane passing through the intersection of two planes P1 = 0 and P2 = 0 is given by P1 + λP2 = 0.

IIHere, P1: 3x - y + 2z - 4 = 0

IIIAnd P2: x + y + z - 2 = 0

IVSo, the equation of the required plane is (3x - y + 2z - 4) + λ(x + y + z - 2) = 0 --- (1)

VThe plane passes through the point (2, 2, 1). Substitute x=2, y=2, z=1 into equation (1):

VI(3(2) - 2 + 2(1) - 4) + λ(2 + 2 + 1 - 2) = 0

VII(6 - 2 + 2 - 4) + λ(3) = 0

VIII2 + 3λ = 0

93λ = -2

10λ = -2/3

11Substitute the value of λ back into equation (1):

12(3x - y + 2z - 4) - (2/3)(x + y + z - 2) = 0

13Multiply the entire equation by 3 to eliminate the fraction:

143(3x - y + 2z - 4) - 2(x + y + z - 2) = 0

159x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0

16Combine like terms:

17(9x - 2x) + (-3y - 2y) + (6z - 2z) + (-12 + 4) = 0

187x - 5y + 4z - 8 = 0

Answer

The equation of the required plane is 7x - 5y + 4z - 8 = 0.

This method is efficient for finding a plane through the intersection of two given planes and satisfying an additional condition.

Example 3

Find the angle between the line (x+1)/2 = y/3 = (z-3)/6 and the plane 10x + 2y - 11z = 3.

IThe given line is (x+1)/2 = y/3 = (z-3)/6. Its direction ratios are (a, b, c) = (2, 3, 6).

IISo, the direction vector of the line is b = 2i + 3j + 6k.

IIIThe given plane is 10x + 2y - 11z = 3. Its normal vector's direction ratios are (A, B, C) = (10, 2, -11).

IVSo, the normal vector to the plane is n = 10i + 2j - 11k.

VThe angle φ between a line and a plane is given by the formula sin φ = |b . n| / (|b| |n|).

VICalculate the dot product b . n:

VIIb . n = (2i + 3j + 6k) . (10i + 2j - 11k)

VIII = (2)(10) + (3)(2) + (6)(-11)

9 = 20 + 6 - 66 = -40

10Calculate the magnitude of b:

11|b| = √(2² + 3² + 6²) = √(4 + 9 + 36) = √49 = 7

12Calculate the magnitude of n:

13|n| = √(10² + 2² + (-11)²) = √(100 + 4 + 121) = √225 = 15

14Substitute these values into the formula for sin φ:

15sin φ = |-40| / (7 * 15)

16sin φ = 40 / 105

17Simplify the fraction:

18sin φ = 8 / 21

19Therefore, φ = sin⁻¹(8/21).

Answer

The angle between the line and the plane is sin⁻¹(8/21).

Be careful to use the sine formula for the angle between a line and a plane, not cosine. The cosine formula is for the angle between the line and the normal to the plane.

Common mistakes

✗Confusing direction cosines with direction ratios, or using direction ratios directly in formulas that require direction cosines without normalising.
✗Incorrectly applying the formula for the angle between a line and a plane (using cosine instead of sine, or vice-versa).
✗Errors in vector operations, particularly cross products and dot products, leading to incorrect signs or magnitudes.
✗Not correctly identifying the position vectors (a1, a2) and direction vectors (b1, b2) when calculating the shortest distance between lines.
✗Algebraic errors when expanding or simplifying Cartesian equations, especially when dealing with fractions or negative signs.

Exam tips

★Memorise all vector and Cartesian forms of equations and formulas. Practice converting between them to enhance understanding.
★Draw diagrams whenever possible to visualise the geometric situation, especially for problems involving lines, planes, and shortest distances.
★Pay close attention to vector operations (dot product, cross product) and scalar multiplication. Double-check your calculations.
★Always write down the relevant formula before substituting values. This helps in avoiding errors and ensures you gain partial marks even if a calculation mistake occurs.

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