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

Question

In Exercises 5–12, sketch each vector as a position vector and find its magnitude.v = -6i - 2j

Accepted Answer

Welcome back. I am so glad you're here. We're asked to draw the following vector and consider the origin as its initial point. Then calculate the magnitude. Our vector is equal to negative 13, I minus five J. And then we are given a graph. We have a vertical Y axis, a horizontal X axis that come together at the origin in the middle, the domain for our X axis is from negative 16 to 16. And the range for our Y axis is also from negative 16 to 16. All right. So we look at our vector and we see that that's in the format of a vector equals A I plus BJ where A is the coefficient of I and we can see here that A is equal to negative 13 and B is the coefficient of J. We can see that J or excuse me, B is equal to negative five. So how do we draw this? Well, we recall from previous lessons that A is the horizontal component of our vector. So that's going to be along the X axis and it's negative 13. So because our starting point, the initial point is the origin, we are going to go to the left 13 units for the value of our horizontal component. Our A and then we recall that B is the vertical component of our vector. So that tells us B is equal to negative five. We're going to go down five units from our negative 13 on the X value. And so that's going to put us at negative 13, negative five. We put a point there, we know our initial point was the origin, the terminal point is negative 13, negative five. And we draw a line connecting the two and then an arrow pointing to the terminal point for our vector. So that's how we graph that. Now, how about the magnitude we recall from previous lessons that the magnitude of a vector when it's given in terms of I and J is equal to the square root of A squared plus B squared. All right, we know our A and our B. So we're taking the square root of instead of A, that's negative 13 squared plus instead of B, that's negative five squared. And now all still underneath the radical negative 13 squared is 169 then plus negative five squared, that's 25. Now we can add 169 plus 25 that's 194. So we're taking the square rut of 194 and that one is as simplified as it gets. So that is our magnitude and there's our graph and our magnitude 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 = -6i - 2j

Key Concepts

Position Vectors

Vector Magnitude

Vector Components

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