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

If a vector has magnitude ∣ v ⃗ ∣ = 5 |v ⃗ |=5 ∣ v ⃗∣ = 5 and direction θ = 7 π 4 \theta=\frac{7\pi}{4} θ = 4 7 π , find the vector’s horizontal and vertical components.

v x = 3.54 v_{x}=3.54 v x = 3.54 and v y = − 3.54 v_{y}=-3.54 v y = − 3.54

v x = 4.98 v_{x}=4.98 v x = 4.98 and v y = 0.479 v_{y}=0.479 v y = 0.479

v x = 0.479 v_{x}=0.479 v x = 0.479 and v y = 4.98 v_{y}=4.98 v y = 4.98

v x = − 3.54 v_{x}=-3.54 v x = − 3.54 and v y = 3.54 v_{y}=3.54 v y = 3.54

14. Vectors

Direction of a Vector

14. Vectors

Direction of a Vector: Videos & Practice Problems

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Topic summary

To find the direction of a vector, use the inverse tangent of the y component divided by the x component, applying the SOHCAHTOA principle. For example, if vector v has components (3, 4), the direction is approximately 37 degrees. To find x and y components from magnitude and direction, use sine and cosine functions: $y = magnitude \times \sin (angle)$ and $x = magnitude \times \cos (angle)$ .

concept

Finding Direction of a Vector

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Finding Direction of a Vector Video Summary

Understanding the direction of a vector is crucial in vector analysis, as it indicates the angle the vector makes with the positive x-axis. To calculate this direction, we can utilize the properties of right triangles, specifically through the tangent function, which is defined as the ratio of the opposite side to the adjacent side. This relationship is encapsulated in the mnemonic SOHCAHTOA, where "SOH" stands for sine being opposite over hypotenuse, "CAH" for cosine being adjacent over hypotenuse, and "TOA" for tangent being opposite over adjacent.

To find the angle θ that represents the direction of a vector, we can use the formula:

$$\theta = \tan^{-1}\left(\frac{y}{x}\right)$$

where $y$ is the y-component and $x$ is the x-component of the vector. For example, if we have a vector $v$ with components (4, 3), we can calculate the direction as:

$$\theta = \tan^{-1}\left(\frac{3}{4}\right) \approx 37^\circ$$

However, when the vector lies in different quadrants, we must adjust our calculations to express the angle as a positive number from the positive x-axis. For instance, if we have a vector with components (2, -1), the calculation yields:

$$\theta = \tan^{-1}\left(\frac{-1}{2}\right) \approx -27^\circ$$

To convert this to a positive angle, we add 360 degrees, resulting in:

$$360^\circ - 27^\circ = 333^\circ$$

In another example, for a vector with components (-3, -3), the calculation gives:

$$\theta = \tan^{-1}\left(\frac{-3}{-3}\right) = 45^\circ$$

Since this vector is in the third quadrant, we add 180 degrees to the angle calculated, leading to:

$$180^\circ + 45^\circ = 225^\circ$$

By following this method, we can accurately determine the direction of vectors in any quadrant, ensuring we express the angle as a positive value from the positive x-axis. This skill is essential for effectively working with vectors in various applications.

Problem

Find the direction of the following vector: $u ⃗=⟨-10,10⟩$ .

45°

135°

−135°

315°

Problem

Find the direction of the following vector: $u⃗=\langle\frac{5\surd3}{3},5\rangle$ .

60°

0.030°

30°

0.010°

example

Finding Direction of a Vector Example 1

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Finding Direction of a Vector Example 1 Video Summary

In this problem, we are tasked with finding the vector c, given vectors a and b. The vectors are defined as follows: a = (2, 2), b = (2, -3), and c = 2a - 3b. To begin, we calculate 2a and 3b.

Calculating 2a involves multiplying each component of vector a by 2:

2a = 2 * (2, 2) = (4, 4).

Next, we calculate 3b by multiplying each component of vector b by 3:

3b = 3 * (2, -3) = (6, -9).

Now, we can find vector c by subtracting 3b from 2a:

c = (4, 4) - (6, -9) = (4 - 6, 4 - (-9)) = (-2, 13).

With vector c determined as (-2, 13), we can now calculate its magnitude using the formula:

|c| = √(x² + y²),

where x and y are the components of vector c. Plugging in the values:

|c| = √((-2)² + (13)²) = √(4 + 169) = √(173).

The magnitude of vector c is therefore √173.

Next, we determine the direction of vector c. The direction can be found by calculating the angle θ with respect to the x-axis using the tangent function:

tan(θ) = y/x.

Substituting the components of vector c:

tan(θ) = 13 / (-2).

To find θ, we take the inverse tangent:

θ = arctan(13 / -2).

This calculation yields an angle of approximately 81.3 degrees. However, since the x-component is negative and the y-component is positive, vector c lies in the second quadrant. To find the total angle from the positive x-axis, we calculate:

Total angle = 180° - 81.3° = 98.7°.

Thus, the direction of vector c is 98.7 degrees.

In summary, the magnitude of vector c is √173, and its direction is 98.7 degrees.

concept

Finding Components from Direction and Magnitude

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Finding Components from Direction and Magnitude Video Summary

Understanding how to find the x and y components of a vector from its magnitude and direction is a crucial skill in both mathematics and science. This process can be visualized using a right triangle, where the hypotenuse represents the vector's magnitude and the angle corresponds to its direction. The sine and cosine functions are key tools in this calculation, following the mnemonic SOHCAHTOA.

To find the y component of a vector, you can use the sine function. The relationship is expressed as:

y = magnitude × sin(angle)

For example, if a vector has a magnitude of 10 and a direction of 53 degrees, the y component can be calculated as:

y = 10 × sin(53°)

Using a calculator, sin(53°) is approximately 0.8, leading to:

y ≈ 10 × 0.8 = 8

Similarly, to find the x component, the cosine function is used:

x = magnitude × cos(angle)

Continuing with the same vector, the x component is calculated as:

x = 10 × cos(53°)

With cos(53°) approximately equal to 0.6, we find:

x ≈ 10 × 0.6 = 6

In another example, consider a vector with a magnitude of 5 and a direction of $ \frac{2\pi}{3} $ radians. The x component is calculated as:

x = 5 × cos\left(\frac{2\pi}{3}\right)

Since $ \cos\left(\frac{2\pi}{3}\right) $ is $ -\frac{1}{2} $, we have:

x = 5 × -\frac{1}{2} = -\frac{5}{2}

For the y component, we use the sine function:

y = 5 × sin\left(\frac{2\pi}{3}\right)

Here, $ sin\left(\frac{2\pi}{3}\right) $ is $ \frac{\sqrt{3}}{2} $, leading to:

y = 5 × \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2}

In this case, the negative x component indicates that the vector lies in the second quadrant of the Cartesian plane, where x values are negative and y values are positive. This understanding of vector components is essential for analyzing motion, forces, and other physical phenomena in future studies.

Problem

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 $v_{y}=-5$

Problem

If a vector has magnitude $|v ⃗ |=5$ and direction $\theta=\frac{7\pi}{4}$ , find the vector’s horizontal and vertical components.

$v_{x}=4.98$ and $v_{y}=0.479$

$v_{x}=0.479$ and $v_{y}=4.98$

$v_{x}=-3.54$ and $v_{y}=3.54$

$v_{x}=3.54$ and $v_{y}=-3.54$

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Direction of a Vector

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Here’s what students ask on this topic:

How do you find the direction of a vector given its components?

To find the direction of a vector given its components, you use the inverse tangent function. The direction θ is calculated using the formula:

$θ = tan ⁻ ¹ (\frac{y}{x})$

Here, y is the y-component and x is the x-component of the vector. For example, if a vector has components (3, 4), the direction is:

$θ = tan ⁻ ¹ (\frac{4}{3})$

Using a calculator, this gives approximately 53.13 degrees.

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What is the formula to find the x and y components of a vector given its magnitude and direction?

To find the x and y components of a vector given its magnitude and direction, you use the sine and cosine functions. The formulas are:

$x = magnitude \times cos (θ)$

$y = magnitude \times sin (θ)$

For example, if a vector has a magnitude of 10 and a direction of 53 degrees, the components are:

$x = 10 \times cos (53) \approx 6$

$y = 10 \times sin (53) \approx 8$

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How do you find the direction of a vector in the second quadrant?

To find the direction of a vector in the second quadrant, first calculate the angle θ using the inverse tangent function:

$θ = tan ⁻ ¹ (\frac{y}{x})$

Since the vector is in the second quadrant, the direction is:

$180 ° + θ$

For example, if the vector components are (-3, 4), then:

$θ = tan ⁻ ¹ (\frac{4}{-}) \approx 53.13 °$

Thus, the direction is:

$180 ° + 53.13 ° = 233.13 °$

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How do you convert the direction of a vector from degrees to radians?

To convert the direction of a vector from degrees to radians, use the conversion factor:

$radians = degrees \times \frac{π}{180}$

For example, if the direction is 60 degrees, the conversion to radians is:

$60 ° \times \frac{π}{180} = \frac{π}{3}$

So, 60 degrees is equivalent to π/3 radians.

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What is the significance of a negative angle when finding the direction of a vector?

A negative angle when finding the direction of a vector indicates that the angle is measured clockwise from the positive x-axis. To express this angle as a positive number, you can add 360 degrees to the negative angle. For example, if the calculated angle is -45 degrees, the positive equivalent is:

$360 ° + - 45 ° = 315 °$

This means the vector direction is 315 degrees counterclockwise from the positive x-axis.

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Direction of a Vector: Videos & Practice Problems

Finding Direction of a Vector

Finding Direction of a Vector Video Summary

Find the direction of the following vector: u⃗=⟨−10,10⟩u ⃗=⟨-10,10⟩u⃗=⟨−10,10⟩.

Find the direction of the following vector: u⃗=⟨5√33,5⟩u⃗=\langle\frac{5\surd3}{3},5\rangleu⃗=⟨35√3,5⟩.

Finding Direction of a Vector Example 1

Finding Direction of a Vector Example 1 Video Summary

Finding Components from Direction and Magnitude

Finding Components from Direction and Magnitude Video Summary

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.

If a vector has magnitude ∣v⃗∣=5|v ⃗ |=5∣v⃗∣=5 and direction θ=7π4\theta=\frac{7\pi}{4}θ=47π, find the vector’s horizontal and vertical components.

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Go over this topic definitions with flashcards

Here’s what students ask on this topic:

How do you find the direction of a vector given its components?

What is the formula to find the x and y components of a vector given its magnitude and direction?

How do you find the direction of a vector in the second quadrant?

How do you convert the direction of a vector from degrees to radians?

What is the significance of a negative angle when finding the direction of a vector?

Your Precalculus tutors

Find the direction of the following vector: $u ⃗=⟨-10,10⟩$ .

Find the direction of the following vector: $u⃗=\langle\frac{5\surd3}{3},5\rangle$ .

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

If a vector has magnitude $|v ⃗ |=5$ and direction $\theta=\frac{7\pi}{4}$ , find the vector’s horizontal and vertical components.