In Exercises 1–4, u and v have the same direction. In each exe...

Question

In Exercises 1–4, u and v have the same direction. In each exercise: Find ||u||.

Graph showing vectors a and b with coordinates and directions in a Cartesian plane.

Accepted Answer

Welcome back. I am so glad you're here. We are told for the given vectors A and B find the magnitude as A and B. Then we're given a graph. We have a vertical Y axis, A horizontal X axis. They come together at the origin in the middle and in the 1st and 2nd quadrants, we have two vectors vector A which begins at negative 17 15, we can label that our X sub one and Y sub one and it has its terminal point at 15 19 which we can label our X sub two Y sub two. And then vector B has its initial point at negative 27, which we can label our X sub one, Y sub one and has its terminal point at 30 11, which we can label our X sub two Y sub two. For the vector B. Our answer choices are answer choice A, the magnitude of A is the square rut of 1040 and the magnitude of B is the square rut of 1040. Answer choice B, the magnitude of A is 1040 and the magnitude of B is the square rat of 916 answer trace C the square rut or excuse me, the magnitude of A is the square root of 1040. And the magnitude of B is the square rut of 884. In answer choice D, the magnitude of A is the square rut of 1040 and the magnitude of B is 884. All right. So we recall from previous lessons, that magnitude is the length of a vector. And we can use the distance formula to determine the length of a vector. So for the distance formula, we recall from previous lessons that that is the square rut of. Now all underneath the radical, we have the quantity of X sub two minus X sub one squared plus the quantity of Y sub two minus Y sub one squared. Now, it's just a matter of plugging in where our X sub twos and Y sub twos are. So for the magnitude of our vector A we've got the square root of and now plugging in for our X sub twos and Y sub twos, et cetera. For our X sub two, we've got 15 minus our X sub one which is negative 17 that needs to be squared. Then plus the quantity of our Y sub two is 19 minus our Y sub one which is and that needs to be squared. Now simplifying in our parentheses, 15 minus negative 17 is plus 17 which is 32. So this is the square rat of which still needs to be squared plus. Now, on the other set of parentheses, 19 minus 15 is four and then four squared. Well, doing our squaring. Now 32 squared is plus four squared is 16 and adding 1024 plus 16 gives us 1040 which is still underneath the radical. Now 1040 is not a perfect square. We can simplify it. But looking at our answer choices, they left it at the square root of 1040. So we will too. Now, for the magnitude of vector B, we've got the square root of the quantity of our X sub two which is 30 minus our X sub one which is negative two that needs to be squared. And then plus now we have the quantity of our Y sub two which is 11 minus our Y sub one which is seven and that will need to be squared. So doing what's inside of the parentheses, 30 minus negative two is 30 plus two. So that will be 32 which needs to be squared. Then plus 11 minus seven is four and four. Sones still needs to be squared. Now, this is awfully familiar and we just did this 32 squared plus four squared that will give us still in the radical. So the square rot of 1040 again. So our answer choice is answer choice. A, both the vectors. Uh A and B have magnitudes of the square ret of 1040. Well done. We'll catch you on the next one.

In Exercises 1–4, u and v have the same direction. In each exercise: Find ||u||.

Key Concepts

Vector Magnitude (Norm)

Vector Direction and Scalar Multiples

Coordinate Geometry of Vectors

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