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Find the magnitude and direction of each vector. Find the unit vector in the direction of the given vector.$$\mathbf{v}=2 \mathbf{i}-4 \mathbf{j}$$

$=4.5,=296.6^{\circ},\left\langle\frac{\sqrt{5}}{5}, \frac{-2 \sqrt{5}}{5}\right\rangle \quad$

Precalculus

Algebra

Chapter 5

Applications of Trigonometry and Trigonometric Identities

Section 7

Vectors

Trigonometry

Introduction to Trigonometry

Missouri State University

Baylor University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

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Find a vector with the giv…

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Find a unit vector in the …

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Magnitude and Direction of…

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Find the vector $\mathbf{v…

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okay we are going to find the magnitude and direction of our vector and then we also are going to give our unit vector so real quickly um we can have our equations written here over at the side but remember to dis means the amount in the X. Direction where negative four is the amount in the Y direction. So it's really just finding a high pot knows. So we can take our two squared and our negative four squared. Which when I write in my calculator I'm just going to say four squared because since it's square, the negative won't count. And then um I end up with the square root of 20, well 20 is four times five, so take the square root of four and we can write that as two square root of five. Now our angle will be gotten from the inverse tangent of negative 4/2. Now notice our actual point by following its X and Y component puts us in quadrant four. So this is good because our answer actually places us also in quadrant four by giving us a negative angle. So you can use the negative angle. You could also explain it by just stating the angle is below the positive X axis. So giving a kind of a frame of reference or you could even add 360 to it and give it as the pure angle, you know following kind of the direction of the unit circle. Now as far as our unit vector. Um we take our original too and we divide by two square root of five. So the two is clean up but I still have a square root of five in the denominator same thing. Um with the negative four over to it cleans up to a negative two and I still have that square root of five in the denominator. Um The book answer probably rationalizes this. So you can also go one more step and multiply top and bottom by the square root of five.

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