Vector a has magnitude of 60 at an angle of 500 o from the x...

Vector A has magnitude of 6.0 at an angle of 50.0 o from the x-axis. Vector B has magnitude of 11.0 at an angle of -20.0 o from the x-axis. Let C=A+B. Let D=A-B. Let V=2*A-B. What is the x-component of vector C? Tries 0/12 What is the y-component of vector C? Tries 0/12 What is the x-component of vector D? Tries 0/12 What is the y-component of vector D? Tries 0/12 What is the x-component of vector V? Tries 0/12 What is the y-component of vector V?

Transcript

00:01 In this problem, we have been given the two vectors.

00:04 Vector a and vector b, both have magnitude of nine units, and vector a is placed at an angle of 20 degree with positive x direction, and vector b is aligned at an angle of 50 degree with respect to positive x direction as indicated here.

00:22 And it's also given to us that vector a plus vector b is equal to vector c.

00:27 And along with this we have also been given that vector d is a vector minus b vector and finally vector v that is given as two times of a vector minus b vector so here we can first get these vectors in terms of components so vector a will have two components one along x direction and another along y direction so x component will be nine costs 20 degrees and y component will be 9 sine 20 degrees.

01:00 And now we can get these values so we can get vector a as 9 cost 20 degree along icap which is 8 .5 i cap and along y -direction it will be 9 sign 20 degree which is coming out to be 3 .1 j -cap.

01:19 And similarly here we have vector b that will also have two components along the x direction its component will be 9 cost 50 degree and along wide direction its component will be 9 sine 50 degree so putting the values we're going to get b vector as 5 .8 i cap minus 6 .9 j cap and now when we add a and b vector we're going to get vector c as 14 .3 i cap minus 3 .8 j cap and subtracting vector b from vector a we're going to get here 2 .7 icap plus 10 jcap and here once again we multiply vector a with two and from this we subtract vector b so we're going to get vector v as 11 .2 i cap plus 13 .1 jcap and now from these informations we have to figure out the x component of vector c so we can observe that x component of vector c that's cx is 14 .3 units...

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