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Vectors (2)

Vectors (2). 3D unit vectors Linear Combinations. Magnitude of a 3D Vector (General). t. s. r. z. a. y. | a | = ( r 2 + s 2 + t 2 ). o. “the magnitude is the square root of the sum of the squares of the 3 components.”. [Pythagoras in 3D]. x. Unit Vectors (1).

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Vectors (2)

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  1. Vectors (2) 3D unit vectors Linear Combinations

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

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

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

  5. 3D Displacement Vectors z y a = i - 2j - 6k x |a| = (12 +(-2)2 + (-6)2) a |a| = (1+4+ 36) = 41 |a| = 6.4

  6. Linear Combinations 3a + 6b 2a - b

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

  8. 3D Vectors Follow all the same rules of 2D Vectors!

  9. 3D Displacement Vectors - adding z y a + b a = i - 2j - 6k x b = 3i + 4j + 11k a b a + b = (i - 2j - 6k) + (3i + 4j + 11k) a + b = (4i + 2j + 5k)

  10. Vector Problem + 13x = 39 Given: a = 3i + 4j andb =i - 3j 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 (iparts) x3 9x + 3y = 33 4x - 3y = 6 and: 4x - 3y = 6 (jparts) 3x + y = 11 Substitute x = 3 9 + y = 11 y = 2

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