Vector A has magnitude of 8 units and makes an angle of 45⁰ with the positive x-axis.vector B also has the same magnitude of 8 units and directed along the negative x-axis. find the magnitude and direction of A+B and the magnitude and direction of A-B ?
Apr 09, 2023To find the magnitude and direction of A+B, we can use the following formula: A + B = √(A² + B² + 2ABcosθ) where A = 8 units, B = -8 units (since it is directed along the negative x-axis), θ = 45⁰ (since A makes an angle of 45⁰ with the positive x-axis). Substituting these values, we get: A + B = √(8² + (-8)² + 2(8)(-8)cos45⁰) A + B = √(64 + 64 - 128/√2) A + B = √(128 - 64√2) ≈ 1.14 units To find the direction of A+B, we can use the following formula: tanθ = (Sinθ)/Cosθ where θ = angle between A+B and the positive x-axis. Substituting the values, we get: tanθ = (8Sin45⁰)/(-8 + 8Cos45⁰) tanθ = 1 θ = tan⁻¹(1) = 45⁰ Therefore, the magnitude of A+B is approximately 1.14 units and its direction makes an angle of 45⁰ with the positive x-axis. To find the magnitude and direction of A-B, we can use the same formula as above: A - B = √(A² + B² - 2ABcosθ) Substituting the values, we get: A - B = √(8² + (-8)² - 2(8)(-8)cos45⁰) A - B = √(128 + 64√2) ≈ 11.31 units To find the direction of A-B, we can use the same formula as above: tanθ = (Sinθ)/Cosθ where θ = angle between A-B and the positive x-axis. Substituting the values, we get: tanθ = (8Sin45⁰)/(-8 - 8Cos45⁰) tanθ = -1 θ = tan⁻¹(-1) = -45⁰ Therefore, the magnitude of A-B is approximately 11.31 units and its direction makes an angle of -45⁰ with the positive x-axis.
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Apr 10, 2023To find the magnitude and direction of A+B, we can use vector addition. We can break down vector A into its x and y components using trigonometry: Ax = 8 cos(45°) = 8/√2 Ay = 8 sin(45°) = 8/√2 Since vector B is directed along the negative x-axis, its x component is -8 and its y component is 0. Now we can add the x and y components of A and B to find the x and y components of A+B: (A+B)x = Ax + Bx = 8/√2 - 8 = -8(√2 - 1) (A+B)y = Ay + By = 8/√2 + 0 = 8(√2)/2 = 4√2 The magnitude of A+B is the square root of the sum of the squares of its x and y components: |A+B| = sqrt(A+B)x^2 + (A+B)y^2 = sqrt(-8(√2 - 1))^2 + (4√2)^2 = 8 To find the direction of A+B, we can use trigonometry to find the angle it makes with the positive x-axis: tanθ = (A+B)y / (A+B)x = (4√2) / (-8(√2 - 1)) = -1/3 θ = tan^-1(-1/3) = -18.43° (measured clockwise from the positive x-axis) Therefore, the magnitude of A+B is 8 units and it makes an angle of -18.43° with the positive x-axis. To find the magnitude and direction of A-B, we can use vector subtraction. We can subtract the x and y components of B from the x and y components of A: (A-B)x = Ax - Bx = 8/√2 + 8 = 8(√2 + 1) (A-B)y = Ay - By = 8/√2 + 0 = 8(√2)/2 = 4√2 The magnitude of A-B is the square root of the sum of the squares of its x and y components: |A-B| = sqrt(A-B)x^2 + (A-B)y^2 = sqrt(8(√2 + 1))^2 + (4√2)^2 = 16 To find the direction of A-B, we can use trigonometry to find the angle it makes with the positive x-axis: tanθ = (A-B)y / (A-B)x = (4√2) / (8(√2 + 1)) = 1/3 θ = tan^-1(1/3) = 18.43° (measured counterclockwise from the positive x-axis) Therefore, the magnitude of A-B is 16 units and it makes an angle of 18.43° with the positive x-axis.
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Apr 10, 2023