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

− 135 ° -135\degree − 135°

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

8. Vectors

Direction of a Vector

8. Vectors

Direction of a Vector: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding vectors involves calculating their direction and components. The direction is found using the inverse tangent of the y component divided by the x component, expressed as a positive angle from the x-axis. For example, if a vector has a magnitude of 10 and a direction of 53 degrees, the x and y components can be calculated using $x = magnitude cosine (angle)$ and $y = magnitude sine (angle)$ . This foundational knowledge is crucial for further studies in mathematics and science.

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 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:

θ = \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 (3, 4), we can calculate the direction as:

θ = \tan^{-1}\left(\frac{3}{4}\right) ≈ 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), we first calculate:

θ = \tan^{-1}\left(\frac{-1}{2}\right) ≈ -27^\circ

Since this angle is negative, we convert it to a positive angle by adding 360°:

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

In another example, for a vector with components (-3, -3), we find:

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

To determine the total direction, we recognize that this vector is in the third quadrant, so we add 180° to our angle:

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

In summary, calculating the direction of a vector involves using the inverse tangent function and adjusting the angle based on the quadrant in which the vector lies. This skill is essential for accurately representing vectors in various applications, from physics to engineering.

Problem

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

$45\degree$

$135\degree$

$-135\degree$

$315\degree$

Problem

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

$60\degree$

$0.030\degree$

$30\degree$

$0.010\degree$

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

Video duration:

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 essential tools in this calculation, following the mnemonic SOHCAHTOA, which helps remember the relationships between the sides of a right triangle and its angles.

To find the y component of a vector, you can use the formula:

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°) ≈ 10 × 0.8 = 8

Similarly, the x component can be determined using the cosine function:

x = magnitude × cos(angle)

Continuing with the same example, the x component would be:

x = 10 × cos(53°) ≈ 10 × 0.6 = 6

Thus, for this vector, the x component is 6 and the y component is 8.

In another scenario, if a vector has a magnitude of 5 and a direction of $ \frac{2\pi}{3} $ radians, the calculations would proceed as follows:

x = 5 × cos$\left(\frac{2\pi}{3}\right)$ = 5 × $-\frac{1}{2}$ = $-\frac{5}{2}$

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

In this case, the x component is negative, indicating 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 fundamental 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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We have more practice problems on Direction of a Vector

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 the angle the vector makes with the positive x-axis. If the vector has components (x, y), the direction is calculated as:

$θ = tan - 1 (\frac{y}{x})$

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

$θ = tan - 1 (\frac{4}{3}) \approx 53.13 °$

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 following formulas:

$x = magnitude \cdot cos (θ)$

$y = magnitude \cdot sin (θ)$

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

$x = 10 \cdot cos (53 °) \approx 6$

$y = 10 \cdot sin (53 °) \approx 8$

How do you express the direction of a vector as a positive angle from the x-axis?

To express the direction of a vector as a positive angle from the x-axis, you need to consider the quadrant in which the vector lies. If the calculated angle θ is negative, add 360° to it. For example, if θ = -27°, the positive angle is:

$360 ° - 27 ° = 333 °$

For vectors in other quadrants, adjust the angle accordingly by adding 180° or 270° as needed to ensure the angle is measured counterclockwise from the positive x-axis.

How do you find the components of a vector if given in radians?

To find the components of a vector given in radians, use the same formulas as for degrees but ensure the angle is in radians. For example, if a vector has a magnitude of 5 and a direction of 2π/3 radians, the components are:

$x = 5 \cdot cos (\frac{2}{π} / 3) = - \frac{5}{2}$

$y = 5 \cdot sin (\frac{2}{π} / 3) = \frac{5}{\sqrt}$

What is the significance of a negative component in a vector?

A negative component in a vector indicates the direction of the vector relative to the coordinate axes. For example, if the x component is negative, the vector points to the left of the y-axis. If the y component is negative, the vector points downward. This helps in determining the vector's position in the coordinate plane. For instance, a vector with components (-3, -3) lies in the third quadrant, indicating it points left and down from the origin.

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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.

Do you want more practice?

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 express the direction of a vector as a positive angle from the x-axis?

How do you find the components of a vector if given in radians?

What is the significance of a negative component in a vector?

Your Trigonometry 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.