In Class 12, we extend our knowledge of geometry to three dimensions using vectors. The key topics are direction cosines and direction ratios, equations of lines and planes in 3D, and distances between points, lines, and planes.
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Direction Cosines and Direction Ratios
- If a line makes angles alpha, beta, gamma with the positive x, y, z axes respectively, then:
- Direction cosines: l = cos(alpha), m = cos(beta), n = cos(gamma)
- Key property: l2 + m2 + n2 = 1
- Direction ratios (DRs): Any three numbers a, b, c proportional to l, m, n. DRs are not unique; direction cosines are unique (up to sign).
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Equation of a Line in 3D
Vector form: r = a + lambda × b
where a is the position vector of a point on the line, b is the direction vector, lambda is a scalar parameter.
Cartesian (symmetric) form:
(x - x1)/a = (y - y1)/b = (z - z1)/c
Line through two points A(x1,y1,z1) and B(x2,y2,z2):
(x - x1)/(x2 - x1) = (y - y1)/(y2 - y1) = (z - z1)/(z2 - z1)
Angle between two lines with direction ratios a1,b1,c1 and a2,b2,c2:
cos(theta) = |a1 a2 + b1 b2 + c1 c2| / (√(a12+b12+c12) × √(a22+b22+c22))
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Equation of a Plane
Vector form: r · n = d, where n is the normal vector to the plane.
Cartesian form: ax + by + cz = d (a, b, c are direction ratios of the normal)
Plane through three points: Use the determinant condition or find the normal via cross product.
Intercept form: x/a + y/b + z/c = 1 (where a, b, c are x-, y-, z-intercepts)
Distance from a point (x1,y1,z1) to plane ax + by + cz + d = 0:
Distance = |ax1 + by1 + cz1 + d| / √(a2 + b2 + c2)
Angle between two planes: cos(theta) = |n1 · n2| / (|n1| |n2|)
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Worked Examples
Find the direction cosines of the line joining A(2, 3, -4) and B(1, -2, 1).
Direction ratios: 1-2, -2-3, 1-(-4) = -1, -5, 5.
Magnitude = √(1 + 25 + 25) = √(51).
DCs = (-1/√(51), -5/√(51), 5/√(51)).
Write the equation of the line through (1, 2, 3) with direction (2, -1, 4).
Symmetric form: (x-1)/2 = (y-2)/(-1) = (z-3)/4
Find the angle between lines with DRs (1, 2, 3) and (3, 2, 1).
cos(theta) = |3 + 4 + 3|/(√(14) × √(14)) = 10/14 = 5/7.
theta = arccos(5/7).
Find the equation of the plane passing through (1, 2, -3) with normal vector n = i - 2j + 3k.
Equation: 1(x-1) + (-2)(y-2) + 3(z+3) = 0
=> x - 1 - 2y + 4 + 3z + 9 = 0
=> x - 2y + 3z + 12 = 0
Find the distance from point (3, -4, 1) to the plane 2x - 3y + z + 5 = 0.
Distance = |2(3) - 3(-4) + 1(1) + 5| / √(4 + 9 + 1) = |6 + 12 + 1 + 5| / √(14) = 24/√(14)
= 24/√(14) units
Find the angle between planes x + y + z = 1 and 2x - y + z = 2.
n1 = (1,1,1), n2 = (2,-1,1).
cos(theta) = |2 - 1 + 1| / (√(3) × √(6)) = 2 / 3sqrt(2) = √(2)/3
Find the equation of a plane with intercepts 2, 3, and 4 on the x, y, z axes.
Intercept form: x/2 + y/3 + z/4 = 1, or multiplying by 12: 6x + 4y + 3z = 12
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Common mistakes
- Direction cosines vs. direction ratios: DCs always satisfy l2 + m2 + n2 = 1; DRs need not.
- Perpendicular vs. parallel conditions: Lines are perpendicular if a1 a2 + b1 b2 + c1 c2 = 0; parallel if a1/a2 = b1/b2 = c1/c2.
- Sign in distance formula: Use absolute value in the numerator when calculating distance from a point to a plane.
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Summary
3D geometry uses direction cosines, direction ratios, and vector equations for lines and planes. The symmetric form of a line and the normal form of a plane are the standard representations. Key computations include angles between lines/planes and distances from points to planes.