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Chapter 1: Problem 21

Express in cylindrical components: \((a)\) the vector from \(C(3,2,-7)\) to \(D(-1,-4,2) ;(b)\) a unit vector at \(D\) directed toward \(C ;(c)\) a unit vector at \(D\) directed toward the origin.

Short Answer

Expert verified
Question: Find the following based on the given solution: (a) The vector from point C to point D in cylindrical components (b) A unit vector at point D directed toward point C in cylindrical components (c) A unit vector at point D directed toward the origin in cylindrical components Answer: (a) Vector CD in cylindrical components: CD = (2√13, arctan(3/2), 9) (b) Unit vector from D to C in cylindrical components: Unit_Vector_DC = (1, arctan(3/2), -9/sqrt(97)) (c) Unit vector from D to the origin in cylindrical components: Unit_Vector_DO = (1, arctan(4), 2/3)

Step by step solution

01

(a) Find Vector CD in Rectangular Coordinates

To find the vector from point C to point D, we subtract the coordinates of point C from the coordinates of point D. CD = D - C = (-1, -4, 2) - (3, 2, -7) The rectangular components of CD are: CD_x = -1 - 3 = -4 CD_y = -4 - 2 = -6 CD_z = 2 - (-7) = 9 So the rectangular vector CD is: CD = (-4, -6, 9).
02

(a) Convert Rectangular Vector CD to Cylindrical Components

Now let's convert the rectangular coordinates of vector CD to cylindrical components. We can express the cylindrical coordinates as ρ = sqrt(CD_x^2 + CD_y^2), φ = arctan(CD_y / CD_x), and z = CD_z Computing the values: ρ = sqrt((-4)^2 + (-6)^2) = sqrt(52) = 2√13 φ = arctan(-6/-4) = arctan(3/2) (Keep in mind the signs when selecting the right quadrant for φ) z = 9 So the vector CD in cylindrical coordinates is: CD = (2√13, arctan(3/2), 9)
03

(b) Determine Unit Vector DC in Rectangular Components

First, we find the vector DC by reversing the direction of vector CD. DC = - CD = (4, 6, -9) Next, we need to find the magnitude of this vector: |DC| = sqrt(4^2 + 6^2 + (-9)^2) = sqrt(97) Now, divide the vector DC by its magnitude to get the unit vector in rectangular components: Unit_Vector_DC = DC/|DC| = (4/sqrt(97), 6/sqrt(97), -9/sqrt(97))
04

(b) Convert Unit Vector DC to Cylindrical Components

Now let's convert the rectangular coordinates of the unit vector DC to cylindrical components. Similar to part (a), we have ρ = 1 (since it's a unit vector), φ = arctan(6/4), z = -9/sqrt(97) So the unit vector from D to C in cylindrical components is: Unit_Vector_DC = (1, arctan(3/2), -9/sqrt(97))
05

(c) Determine Unit Vector DO in Rectangular Components

D is the point (-1, -4, 2). Finding the unit vector from D to the origin, we just divide the coordinates of D by its magnitude: |D| = sqrt((-1)^2 + (-4)^2 + (2)^2) = 9 Unit_Vector_DO = D/|D| = (-1/3, -4/3, 2/3)
06

(c) Convert Unit Vector DO to Cylindrical Components

Finally, we convert the rectangular coordinates of the unit vector DO to cylindrical components. Again, with the same approach as previous parts, we have ρ = 1 (since it's a unit vector), φ = arctan(-4/-1), z = 2/3 So the unit vector from D to the origin in cylindrical components is: Unit_Vector_DO = (1, arctan(4), 2/3)

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Most popular questions from this chapter

Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a flight along the \(x\) axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume \(\mathbf{a}_{x}\) to be directed to the east and \(\mathbf{a}_{y}\) to the north. The wind velocity at the operating altitude is assumed to be: $$ \mathbf{v}(x, y)=\frac{\left(0.01 x^{2}-0.08 x+0.66\right) \mathbf{a}_{x}-(0.05 x-0.4) \mathbf{a}_{y}}{1+0.5 y^{2}} $$ Determine the location and magnitude of \((a)\) the maximum tailwind encountered; \((b)\) repeat for headwind; \((c)\) repeat for crosswind; \((d)\) Would more favorable tailwinds be available at some other latitude? If so, where?

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Three vectors extending from the origin are given as \(\mathbf{r}_{1}=(7,3,-2)\), \(\mathbf{r}_{2}=(-2,7,-3)\), and \(\mathbf{r}_{3}=(0,2,3)\). Find \((a)\) a unit vector perpendicular to both \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(b)\) a unit vector perpendicular to the vectors \(\mathbf{r}_{1}-\mathbf{r}_{2}\) and \(\mathbf{r}_{2}-\mathbf{r}_{3} ;\) (c) the area of the triangle defined by \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2} ;(d)\) the area of the triangle defined by the heads of \(\mathbf{r}_{1}, \mathbf{r}_{2}\), and \(\mathbf{r}_{3}\).
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