In Exercises 5–12, sketch each vector as a position vector and...

Question

In Exercises 5–12, sketch each vector as a position vector and find its magnitude. v = i - j

Accepted Answer

Welcome back. I am so glad you're here. We're asked to draw the following vector and consider the origin its initial point. Then calculate its magnitude. Our given vector is two I minus eight J. Then we are given a coordinate plane. We have a vertical Y axis, a horizontal X axis, they come together at the origin in the middle. The domain for our X values is from negative nine to positive nine. And the range for our Y values is from negative nine to positive nine. All right. So we look at our vector and we recognize that this is in the format of A I plus BJ where A is the coefficient of I here A is equal to two and B is the coefficient of J here B is equal to negative eight. And we recall from previous lessons that A is our horizontal component of our vector and that B is our vertical component of our vector. So what does that mean? Well, the origin is the initial point. So we draw on the origin, our initial point and then for the horizontal component, it's a positive two. That means along the X values we're going to the right two units. And then for the vertical component, we'll be going down eight because B is equal to negative eight. So from that positive two along the X axis, then we move down eight units along the Y axis. And so we can label that terminal point as two negative eight. Then to finish our vector from the initial point, we draw a line to the terminal point and draw an arrow pointing to the terminal point. So there we have our vector. Now we just need to calculate its magnitude. We recall from previous lessons that the magnitude of a vector when given in terms of I and JJ is equal to the square rut of A squared plus B squared. We know A and B. So we're taking the square rut of instead of A two squared plus instead of B negative eight squared and two squared is four plus negative eight squared is 64. So inside the radical, we have four plus 64 which is 68. So we have the square rut of 68. Well, 68 can be simplified. Well, the square rut of 68 can be simplified. That's the same as the square rut of four multiplied by 17. So that simplifies 22 square rut 17. So there is our vector's magnitude two square root 17. And there we have the graph of our vector. Well done. We'll catch you on the next one.

In Exercises 5–12, sketch each vector as a position vector and find its magnitude. v = i - j

Key Concepts

Position Vector

Vector Components

Magnitude of a Vector

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