3D unit vectors Linear Combinations

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Presentation transcript:

3D unit vectors Linear Combinations

|a| = (r2 + s2 + t2) Magnitude of a 3D Vector (General) z a y o x t s “the magnitude is the square root of the sum of the squares of the 3 components.” x [Pythagoras in 3D]

i is the unit vector in the x-direction Unit Vectors (1) i is the unit vector in the x-direction z j y k j is the unit vector in the y-direction i x k is the unit vector in the z-direction All vectors can be expressed as a linear combination of these 3 vectors

a = i - 2j - 6k Unit Vectors (2) z y a x All 3D Vectors can be expressed in this form

a = i - 2j - 6k |a| = (12 + (-2)2 + (-6)2) |a| = (1 + 4 + 36) = 41 3D Displacement Vectors x y z a = i - 2j - 6k |a| = (12 + (-2)2 + (-6)2) a |a| = (1 + 4 + 36) = 41 |a| = 6.4

Linear Combinations 3a + 6b 2a - b

+ = (6i + 4j) + (i - 5j) = 7i - j Linear Combination Example 64 1 7 -5 A car drives from A to B with displacement = 6i + 4j Then he drives from B to C with displacement = i - 5j 6i + 4j B A C i - 5j 1 -5 64 + = 7 -1 The resultant displacement is from A to C (6i + 4j) + (i - 5j) = 7i - j The magnitude of the displacement = (72 + (-1)2) = 50 = 7.1m (1 d.p.)

3D Vectors Follow all the same rules of 2D Vectors!

3D Displacement Vectors - adding x y z a + b a = i - 2j - 6k b = 3i + 4j + 11k a b a + b = (i - 2j - 6k) + (3i + 4j + 11k) a + b = (4i + 2j + 5k)

Given: a = 3i + 4j and b = i - 3j Vector Problem Find x and y, if xa + yb = 11i + 6j xa + yb = x(3i + 4j) + y(i - 3j) = 3xi + 4xj + yi - 3yj = (3x+y)i + (4x-3y)j = 11i + 6j Therefore: 3x + y = 11 (i parts) x3 9x + 3y = 33 13x = 39 + and: 4x - 3y = 6 (j parts) 4x - 3y = 6 3x + y = 11 Substitute 9 + y = 11 x = 3 y = 2