course Phy 241
I still have a few questions about the university textbook problems 73, 79, and 90. Our last exchange concerning problem 73:1.73
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A graphic artist is creating a new logo for her company? Website. In the graphics program she is using, each pixel in an image file has coordinates (x,y), where the origin (0,0) is at the upper-left corner of the image, the positive x-axis points to the right, and the positive y axis points down. Distances are measured in pixels.
A) The artist draws a line from the pixel location (10,20) to the location (210, 200).
She wishes to draw a second line that starts at (10,20), is 250 pixels long, and is at angle of 30 degrees measured clockwise form the first line.
At which pixel location would this second line end? Give your answer to the nearest pixel.
B) The artist now draws an arrow that connects the lower-right end of the first line to the lower right end of the second line. Find the length and direction of this arrow. Draw a diagram showing all three lines.
For this problem I drew a point 10, 20 then a line extending to 210, 200. These are in the fourth quandrant although the y axis in the problem is said to be positive. 30 degrees is in between that line and the leftmost line starting at 10, 20 and extending 250 pixels. If we draw dotted lines on the top and right side of the first line we can form a right triangle. The two legs are 200 and 80 pixels so the angle between the topmost leg and the hypotenuse is 21.8 degrees.
If we take 21.8 and add it to 30 and then subtract that sum from 90 we get 38.2.
The two vectors do not necessarily form the legs of a right triangle.
A line 250 pixels long at angle 30 degrees clockwise from the first line would be at 8.2 degrees above horizontal. So it would go 250 pixels * cos(8.2) deg to the right, and 250 pixels * sin(8.2 deg) up.
You can then figure out how many pixels to the right or left, and how many up or down, you would have to travel to get from the end of one vector to the end of the other.
This represents the angle if we continue the dotted line from 10, 20 straight down and across to meet B. This is the angle between the first described dotted line and B. If we know this angle and the hypotenuse we can find the components of B. So Bx is represented by the sin of 38.2=Bx/250, or 154.6. The other leg is found by the cos function and comes out to be 196.5. Therefore if B starts at 10, 20 we can just add these to the line lengths and get a coordinate of 164.6 and 216.5. The answer in the book is 87, 258. Also how would you set up the B part of this problem, is it just vector addition?
My picture for the problem looks like the following:
(The picture is sent to your email address dsmith@vhcc.edu)
I can not see how there would be a line 8.2 degrees above the horizontal.
It must have been late at night when I responded. The second line is clockwise, not counterclockwise. I also accepted your 21.8 degree angle, which I beleive is wrong. 30 degrees counterclockwise from that 21.8 degree angle would be 8.2 degrees above horizontal.
Now, to get it right:
From 10, 20 to 210, 200 the displacements are 200 to the right, 180 down. The corresponding vector has magnitude around 270 and is directed at about 42 degrees down and to the right. You should work this out to get the precise angle and magnitude.
30 degrees clockwise would put you around 72 degrees below horizontal. A 250 pixel displacement in this direction would therefore go 250 cos(72 deg) pixels = 80 pixels or so to the right, and 250 sin(72 deg) pixels = 230 pixels or so down. This would put you roughly at the point (90, 250).
The vector from the point (210, 200) to (90, 250) has x and y components -120 and +50, 120 units to the left and 50 units down. What is the magnitude of this vector? What is its direction?
On 79 our exchange was as follows:
1.79
You are canoeing on a lake. Starting at your camp on the shore, you travel 240 m in the direction 32 degrees south of east to reach a store to purchase supplies. You know the distance because you have located both you camp and the store on a map of the lake. On the return trip you travel distance B in the direction 48 degrees north of west, distance C in the direction 62 degrees south of west, and then you are back at you are back at your camp. You measure the directions of travel with your compass, but you don? know the distances. Since you are curious to know the total distance you rowed, use vector methods to calculate the distances B and C.
The picture makes a triangle with majority of it in the fourth quandrant, including one of the long legs and degree 32 which extends down from the x axis to the neg y axis. The point in the quandtrant is joined at the origin to a smaller leg that extends to the origin 62 degrees south of west. The next leg extends from the opposite point of the fourth quandrant leg and meets the small leg. Since it is a full triangle does Ax+Bx+Cx=0 and the same for the y?? I feel like when I try to find components to add I have way too many variables.
You are absolutely correct. The only two unknowns are the magnitudes of B and C, and your conditions on x and y give you two equations. You can solve two equations for two unknowns.
Here is a solution submitted today by another student:
I used the equations Ax = A cos (theta) and Ay = A sin (theta) to calculate the components of each vector.
Vector A has a magnitude of 240 meters, a direction of -32 degrees, an x component of 203.53 meters and a y component of -127.18 meters
Vector B has a magnitude of B meters, a direction of 132 degrees (180 degrees - 48 degrees), an x component of -.6691 * B meters, and a y component of .7431 * B meters.
Vector C has a magnitude of C meters, a direction of 242 degrees (180 degrees + 62 degrees), an x component of -.4695 * C meters, and a y component of -.8829 * C meters.
Since the starting and ending points of the trip are the same, I wrote down this relationship between the components: Ax + Bx + Cx = 0 and Ay + By + Cy = 0. Then I solved each of them for the C component to get:
Ax + Bx = -Cx and Ay + By = -Cy.
Into the first equation, I substituted the components I calculated earlier:
203.53 - .6691 * B = .3746 * C
Then solved for C:
(203.35 - .6691 * B) / .3746 = C
I substitued this value for C into the other equation:
.7431 * B - 127.18 = (.9272 / .3746) *(203.35 - .6691 * B)
The right-hand side of the equation is simplified:
.7431 * B - 127.18 = 503.77 - 1.6562* B
Similar terms are grouped together:
.7431 * B + 1.6562 * B = 503.77 + 127.18
And simplified:
2.3993 * B = 630.95
Then solved for B = 263 meters.
The components of vector B are calculated:
Bx = 263 cos (132) = -176.0 meters and By = 263 sin (132) = 195.45 meters.
The x component of B is substituted into the equation Ax + Bx = -Cx to find the magnitude of C:
203.53 - 176.0 = .3746 * C , which is solved for C = 73.55
The components of vector C are calculated:
Cx = 73.55 cos (248) = -27.55 meters and Cy = 73.55 sin (248) = -68.19 meters.
I understand this answer until"" Ax + Bx = -Cx and Ay + By = -Cy.
Into the first equation, I substituted the components I calculated earlier:
203.53 - .6691 * B = .3746 * C
Where does the .3746 C come from.
That should be .4695, obtained from the correct value of cos(242 deg). Cx = C cos(242 deg) = - .4695 C.
The correct value of sin(242 deg) should also be used in obtaining the expression for Cy.
This equation represents Ax+Bx=-Cx. -Cx should be .469*C found by taking the cos of 242.
Our exchange for 90 was as follows:
Obtain a unit vector perpendicular to the two vectors given in Problem 1.85: A=-2.00i +3.00j+4.00k B=3.00i+1.00j-3.00k
Would a unit vector perpendicular to the vectors be the same numbers just the signs changed? Its hard to visualize a perpendicular vector in a 3 dimensional situation.
It is indeed difficult to visualize this, but it's worth the effort. Calculating the unit vector is a good start.
Remember that the cross product, or vector product, of two vectors is perpendicular to both. If you find the vector product, then divide the resulting vector by its magnitude, you will have the unit vector. Then sketch the three vectors on a 3-dimensional coordinate system, and try to visualize them from the sketch.
First I found the vector product using the equation Cx=AyBz-AzBy Cy=AzBx-AxBz Cz=AxBy-AyBx
Cx=(3)(-3)-(4)(1)=-13
Cy=(4)(3)-(-2)(-3)=6
Cz=(-2)(1)-(3)(3)=-9
To find the magnitude square all of these numbers add them together and take the square root of that sum to obtain 16.9. Do you divide the -13,6, and-9 by the 16.9? -0.769i+.355j-.533k
To get a unit vector in the direction of a given vector, you divide the given vector by its magnitude. If you divide a vector of magnitude 16.9 by 16.9, you end up with a vector of length 1, i.e., a unit vector.
You can check to be sure you have a unit vector by calculating the magnitude of the vector you obtained. The magnitude is obtained either using the pythagorean theorem or by taking the square root of the dot product of the vector with itself.
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