5. Find the magnitude of u given that 𝒖 ∙ 𝒖 = √45
6.a force is given by the vector F=(2,3) and moves an object from the point (1,3) to (5,9). The force and displacement are in the same direction. Find the work in foot-pounds.
7.determine if u=-2i+j and v=i+2j are orthogonal, parallel, or neither. Show all your work that supports your answer.
5. Since 𝒖 ∙ 𝒖 = ||𝒖||^2, we can take the square root of both sides to get ||𝒖|| = √(𝒖 ∙ 𝒖) = √45. Therefore, the magnitude of u is √45.
6. The distance between the two points can be found using the distance formula:
d = √[(5-1)^2 + (9-3)^2] = √52
Since the force and displacement are in the same direction, the work done is given by:
W = ||F|| d = √(2^2 + 3^2) √52 ≈ 35.72 foot-pounds
7. To determine if 𝒖 and 𝒗 are orthogonal, we can take their dot product:
𝒖 ∙ 𝒗 = (-2)(1) + (1)(2) = 0
Since the dot product is 0, we know that 𝒖 and 𝒗 are orthogonal, meaning they are perpendicular to each other.
if 𝒖 ∙ 𝒖 = √45 then |u| = ∜45
No, if 𝒖 ∙ 𝒖 = √45, then ||𝒖|| = √(𝒖 ∙ 𝒖) = √45. The square root of 45 is not a perfect fourth power, so the magnitude of 𝒖 cannot be expressed as the fourth root of 45.
what rot. If 𝒖 ∙ 𝒖 = |u|^2 = √45 then |u| = ∜45
If |u| = √45 then 𝒖 ∙ 𝒖 = 45
You are correct, my apologies for the confusion.
If 𝒖 ∙ 𝒖 = |u|^2 = √45, then |u| = √(𝒖 ∙ 𝒖) = √45, as you mentioned.
And if |u| = √45, then 𝒖 ∙ 𝒖 = |u|^2 = (√45)^2 = 45.
To find the magnitude of vector u given that u ∙ u = √45, we need to use the dot product formula and solve for the magnitude. Here's how you can do it:
1. Recall that the dot product of two vectors u and v is given by the formula: u ∙ v = |u| |v| cos(θ), where |u| represents the magnitude of vector u, |v| represents the magnitude of vector v, and θ is the angle between the two vectors.
2. In this case, we have u ∙ u = √45. Since the dot product of a vector with itself is equal to the square of its magnitude, we can rewrite the equation as: |u|^2 = √45.
3. Take the square root of both sides to solve for |u|: |u| = √(√45) = (45)^(1/4).
4. Simplify the expression: |u| ≈ 3.35.
Therefore, the magnitude of vector u is approximately 3.35.
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To find the work done in moving an object given a force vector and displacement vector, we can use the formula: work (W) = force ∙ displacement.
In this case, the given force vector is F = (2,3) and the object moves from point A(1,3) to point B(5,9) in the same direction as the force vector. Here's how you can calculate the work in foot-pounds:
1. Calculate the displacement vector (d) by subtracting the initial position vector (A) from the final position vector (B): d = B - A = (5, 9) - (1, 3) = (4, 6).
2. Calculate the dot product between the force vector (F) and the displacement vector (d): W = F ∙ d = (2, 3) ∙ (4, 6) = 2*4 + 3*6 = 8 + 18 = 26.
3. The result, 26, represents the work done in units of force multiplied by distance. If the given force is in pounds and the displacement is in feet, then the work is expressed in foot-pounds.
Therefore, the work done in moving the object is 26 foot-pounds.
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To determine if vectors u = -2i + j and v = i + 2j are orthogonal, parallel, or neither, we can use the dot product. Here's how you can check:
1. Calculate the dot product of the two vectors: u ∙ v = (-2i + j) ∙ (i + 2j) = (-2*1) + (1*2) = -2 + 2 = 0.
2. If the dot product of two vectors is zero, it means that they are orthogonal (perpendicular) to each other.
Therefore, vectors u = -2i + j and v = i + 2j are orthogonal.