Direction of a Vector: Videos & Practice Problems

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.

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.

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.