If a vector has magnitude ∣v⃗∣=13|v ⃗ |=13∣v⃗∣=13 and direction θ=157.38°θ=157.38°θ=157.38°, find the vector’s horizontal and vertical components.

v x = − 12 v_{x}=-12 v x = − 12 and v y = 5 v_{y}=5 v y = 5

If a vector has magnitude ∣ v ⃗ ∣ = 13 |v ⃗ |=13 ∣ v ⃗∣ = 13 and direction θ = 157.38 ° θ=157.38° θ = 157.38° , find the vector’s horizontal and vertical components.

Let's try solving this problem. So here we're told if a vector has a magnitude equal to 13, vector b, and it has a direction a 157.38 degrees, Find the vectors horizontal and vertical components. Okay. Now, to solve this problem, whenever I'm dealing with these types of problems where I'm trying to find a direction, I'm given the magnitude and direction and I need to find components, I like to draw a simple graph. Now this is not a step that you have to do to solve the problem, but I personally find this just helps me to see what I'm dealing with. So we have this x and y axis here for our graph. What I can see is our vector has a magnitude of 13, and the direction is 157.38 degrees. Now, each of these cusps here are 90 degree cuffs So we have 90, then 180, and then this just keep going around so this would be 270, and then 360. Now, if I go here this is a 90 degree cusp right over here would be a 180 degree turn, and I can see that this is between 90 and 180 degrees 157.38. So our vector is gonna be somewhere in this quadrant, and I'm not drawing it at a perfect angle, but this is just where the vector would be. And our vector has a magnitude of 13 right there, and then our angle is a 157.38 degrees. So what we're ultimately looking for here is the vector components which would be this component vx and then that component vy. Now to calculate both these components we can use an equation for For the x component you want to take the magnitude of your vector and multiply it by the cosine of the angle. And for the y component you want to take the magnitude of your vector and multiply it by the sine of the angle. Now starting with the x component here we can see that the magnitude of our vector is 13 and we can multiply that by the cosine of our direction here which is 157.38 degrees. Now what you can do is actually actually plug this whole thing into a calculator and you want to make sure you're in degree mode when you do this. If you calculate this, you should get an approximate value of negative 12. So that means our x component is negative 12. Now, next we need to find the y component, and to find the y component, well I can take our magnitude which is 13, and I can multiply that by the sine of our angle which is 157.38 degrees. Now plugging this into a calculator you should get an answer of positive 5 for the y component. So this is the x and y component for this vector, and this actually makes sense. It makes sense that we got a negative 12 because notice this vector is back here on the x axis of points in the negative direction, and it makes sense that our vector was positive 5 for the for the y component, because this vector points up in the positive y direction. So this is how you can find the components of vectors using these equations, so I hope you found this video helpful. Thanks for watching.

If a vector has magnitude ∣ v ⃗ ∣ = 13 |v ⃗ |=13 ∣ v ⃗∣ = 13 and direction θ = 157.38 ° θ=157.38° θ = 157.38° , find the vector’s horizontal and vertical components.

v x = − 12 v_{x}=-12 v x = − 12 and v y = 5 v_{y}=5 v y = 5

v x = 5 v_{x}=5 v x = 5 and v y = − 12 v_{y}=-12 v y = − 12

v x = 12 v_{x}=12 v x = 12 and v y = − 5 v_{y}=-5 v y = − 5

v x = − 12 v_{x}=-12 v x = − 12 and v y = − 5 v_{y}=-5 v y = − 5

8. Vectors

Direction of a Vector

Trigonometry

8. Vectors

Direction of a Vector

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Multiple Choice

If a vector has magnitude $|v ⃗ |=13$ and direction $θ=157.38°$ , find the vector’s horizontal and vertical components.

$v_{x}=5$ and $v_{y}=-12$

$v_{x}=-12$ and $v_{y}=5$

$v_{x}=12$ and $v_{y}=-5$

$v_{x}=-12$ and $v_{y}=-5$

Verified step by step guidance

To find the horizontal and vertical components of a vector given its magnitude and direction, we use the formulas: $ v_x = |\vec{v}| \cdot \cos(\theta) $ and $ v_y = |\vec{v}| \cdot \sin(\theta) $.

Substitute the given magnitude $ |\vec{v}| = 13 $ and direction $ \theta = 157.38^\circ $ into the formulas.

Calculate the horizontal component: $ v_x = 13 \cdot \cos(157.38^\circ) $.

Calculate the vertical component: $ v_y = 13 \cdot \sin(157.38^\circ) $.

Evaluate the trigonometric functions $ \cos(157.38^\circ) $ and $ \sin(157.38^\circ) $ to find the numerical values of $ v_x $ and $ v_y $.

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