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Answer all 6 pictures/questions !!

Answer all 6 pictures/questions !!-example-1
Answer all 6 pictures/questions !!-example-1
Answer all 6 pictures/questions !!-example-2
Answer all 6 pictures/questions !!-example-3
Answer all 6 pictures/questions !!-example-4
Answer all 6 pictures/questions !!-example-5

1 Answer

5 votes

Answer:

See below for answers and explanations

Explanation:

Problem 1

First, identify the three vectors in component form (remember to have the direction angle for each vector account for the quadrants they are in!):


u=\langle90\cos220^\circ,90\sin220^\circ\rangle\\v=\langle 60\cos110^\circ,60\sin110^\circ\rangle\\w=\langle 40\cos30^\circ,40\sin30^\circ\rangle

Secondly, we add them:


u+v+w=\langle90\cos220^\circ+60\cos110^\circ+40\cos30^\circ,90\sin220^\circ+60\sin110^\circ+40\sin30^\circ\rangle\approx\langle-54.824,18.531\rangle

Thirdly, find the magnitude of the new vector:


||u+v+w||=√((-54.824)^2+(18.531)^2)\\||u+v+w||\approx57.871

And finally, find the direction of the new vector:


\theta=tan^(-1)((18.531)/(-54.824))\\\theta\approx-18.675^\circ\approx-19^\circ

Because our new vector is in Quadrant II, our found direction angle is a reference angle, telling us that the actual direction of our vector is 19° clockwise from the negative x-axis, which would be 180°-19°=161°

Thus, B is the best answer

Problem 2

To find the component form of a vector, we simply subtract the initial point from the terminal point. In this case, our initial point is at (7,-1) and our terminal point is at (1,6). We subtract the horizontal and vertical components separately:


\langle1-7,6-(-1)\rangle=\langle-6,7\rangle

Thus, A is the correct answer

Problem 3

Again, subtract the initial point from the terminal point to find the vector, and then apply scalar multiplication:


-(1)/(2)v=-(1)/(2) \langle-8-8,10-4\rangle=-(1)/(2)\langle-16,6\rangle=\langle8,-3\rangle

Thus, C is the correct answer

Problem 4 (top one in 4th image)

Remember that scalar multiplication only affects the magnitude of the vector and not the direction. Thus,
-3u=-3\bigr[20\langle\cos30^\circ,\sin30^\circ\rangle\bigr]=-60\langle\cos30^\circ,\sin30^\circ\rangle, which means our magnitude is -60 and our direction remains 30°.

Thus, B is the correct answer

Problem 5 (fake #6, bottom one in 4th image)

Basically, whatever is in front of the i represents the horizontal component, and whatever is in front of the j represents the vertical component, so:


u=-3i+8j=\langle-3,8\rangle

Thus, B is the correct answer

Problem 6 (real one, 5th image)

Recall:

  • Two vectors are said to be orthogonal if their dot product is 0 and are perpendicular to each other (form a 90° angle)
  • Two vectors are said to be parallel if the angle between the vectors is 0° or 180° i.e. cosθ=-1

Knowing these facts, let us compute the dot product:


u\cdot v=(-4*28)+(7*-49)=(-112)+(-343)=-455

Since the dot product of the two vectors is not 0, then the vectors are not orthogonal, so we can eliminate choices A and B

To check if the vectors are parallel, find the angle between them:


\displaystyle\theta=cos^(-1)\biggr((u\cdot v)/(||u||\:||v||)\biggr)\\\\\theta=cos^(-1)\biggr((-455)/(√((-4)^2+(7)^2)√((28)^2+(-49)^2))\biggr)\\\\\theta=cos^(-1)\biggr((-455)/(√(16+49)√(784+2401))\biggr)\\\\\theta=cos^(-1)\biggr((-455)/(√(65)√(3185))\biggr)\\\\\theta=cos^(-1)\biggr((-455)/(√(207025))\biggr)\\\\\theta=cos^(-1)\biggr((-455)/(455)\biggr)\\\\\theta=cos^(-1)(-1)\\\\\theta=180^\circ

So, since the angle between the two vectors is 180°, this implies that cosθ=-1, and the two vectors must be parallel.

Thus, D is the correct answer

User Nagaraja JB
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