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pps/blackboard/content/listContent jsp?course id- 19938 1&content.id; 4681594.1 Q Your Sets l Quiziet Pinckney AP Bology D corse APAmaica wecome, Manana-% NHS Vector Challenge Now that you have learned a litle about vectors, see if you can complete the vector challenge. Find vectors that satisfies each of the four challenges below 1. A vector whose y-component is more than twice its x-component. 2. A vector which when added to half its negative equals j 3. Three vectors that add up to zero. 4. Two vectors that are perpendicular to each other, but neither of which lines up directly with the x or y axis. To find these vectors in the upper right. The x- and y-component of each vector are specified by Rx and Ry at the top of the screen vectors, experiment with this vector addition simulation. To create a new vector, drag one from the bax of Create a document that shows vectors satistying each of the four challenges above. For each challenge, show the vectors and specify the components of each. You can hand write this document and scan/photograph to upload, or you can type the document and use screen captures of the simulation to show your vectors When you are finished, click on the Ch 6 Activity link below to upload and submit your work Ch 6 Activity Click the link to upload and submit your work DELL
Explanation / Answer
let A = Rx i + Ry j = <Rx , Ry>
1) y component is more than twice its x component
=> Ry > 2Rx
Now Rx be 1 => Ry > 2
=> let Ry be 3
Therefore vector A = i + 3j = <1 , 3>
2) A vector which when added to half its negative equals j
=> Rx i + Ry j + (1/2) (-Rx i + -Ry j) = j
==> (Rx/2) i + (Ry/2) j = 0 i + 1 j
by comparing Rx/2 = 0 => Rx = 0 and Ry/2 = 1 => Ry = 2
Therefore A = 0 i + 2 j = 2 j = < 0 , 2 j>
3) There vectors that add upto zero
let U = 1 i + 2 j = <1 , 2> , V = -3 i + 4 j = <-3 , 4> , W = 2 i - 6 j = <2 , -6>
A = U + V + W = (1 i + 2 j) + (-3 i + 4 j) + (2 i - 6 j) = 0 i + 0 j = 0
4) 2 vectors are said to be perpendicular when their dot product is zero
U = 2 i - 3 j = <2 , -3> , V = 3 i + 2 j = <3 , 2>
dot product = (2 i - 3 j ) . (3 i + 2 j) = 2(3) + (-3)(2) = 6 - 6 = 0
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