With The Diagram And The Above Data Answer These Questions: What Is The X Component Of Vector A?

Component Method of Vector Addition

Earlier in this lesson, nosotros learned that vectors oriented at right angles to one another can be added together using the Pythagorean theorem. For instance, 2 displacement vectors with magnitude and management of 11 km, North and 11 km, East can be added together to produce a resultant vector that is directed both due north and east. When the 2 vectors are added head-to-tail as shown beneath, the resultant is the hypotenuse of a right triangle. The sides of the right triangle have lengths of eleven km and 11 km. The resultant can be determined using the Pythagorean theorem; it has a magnitude of fifteen.6 km. The solution is shown beneath the diagram.

This Pythagorean approach is a useful approach for adding whatever two vectors that are directed at right angles to one another. A right triangle has two sides plus a hypotenuse; so the Pythagorean theorem is perfect for adding ii right angle vectors. But at that place are limits to the usefulness of the Pythagorean theorem in solving vector addition problems. For instance, the improver of 3 or iv vectors does non lead to the formation of a right triangle with two sides and a hypotenuse. So at first glance it may seem that information technology is impossible to use the Pythagorean theorem to determine the resultant for the addition of 3 or four vectors. Furthermore, the Pythagorean theorem works when the 2 added vectors are at right angles to one another - such as for calculation a north vector and an east vector. But what tin one do if the two vectors that are existence added are not at right angles to one some other? Is there a ways of using mathematics to reliably determine the resultant for such vector add-on situations? Or is the pupil of physics left to determining such resultants using a scaled vector diagram? Here on this folio, nosotros will larn how to approach more complex vector improver situations by combining the concept of vector components (discussed earlier) and the principles of vector resolution (discussed earlier) with the use of the Pythagorean theorem (discussed before).

Addition of Iii or More Right Bending Vectors

As our first case, consider the following vector addition problem:

Example 1:

A student drives his car 6.0 km, Due north before making a right hand turn and driving six.0 km to the East. Finally, the student makes a left manus plough and travels another 2.0 km to the north. What is the magnitude of the overall displacement of the pupil?

Like whatever problem in physics, a successful solution begins with the development of a mental pic of the situation. The construction of a diagram like that below oftentimes proves useful in the visualization process.

When these three vectors are added together in head-to-tail fashion, the resultant is a vector that extends from the tail of the first vector (half-dozen.0 km, North, shown in crimson) to the arrowhead of the third vector (2.0 km, North, shown in dark-green). The head-to-tail vector addition diagram is shown below.

Equally can exist seen in the diagram, the resultant vector (drawn in black) is not the hypotenuse of any right triangle - at to the lowest degree non of any immediately obvious right triangle. Only would information technology exist possible to strength this resultant vector to be the hypotenuse of a right triangle? The answer is Yep! To do so, the order in which the iii vectors are added must be changed. The vectors above were fatigued in the order in which they were driven. The student drove north, then eastward, then northward over again. Just if the three vectors are added in the order 6.0 km, Northward + 2.0 km, N + 6.0 km, Eastward, and so the diagram will look like this:

After rearranging the order in which the 3 vectors are added, the resultant vector is at present the hypotenuse of a correct triangle. The lengths of the perpendicular sides of the correct triangle are 8.0 m, North (vi.0 km + 2.0 km) and 6.0 km, East. The magnitude of the resultant vector (R) tin can be determined using the Pythagorean theorem.

R² = (8.0 km)² + (half-dozen.0 km)²
R² = 64.0 km²+ 36.0 km²
R² = 100.0 km²
R = SQRT (100.0 km2)
R = ten.0 km

(SQRT indicates square root)

In the first vector addition diagram above, the three vectors were added in the order in which they are driven. In the second vector addition diagram (immediately above), the club in which the vectors were added was switched effectually. The size of the resultant was not affected past this change in order. This illustrates an important betoken almost calculation vectors: the resultant is independent by the order in which they are added. Adding vectors A + B + C gives the same resultant equally adding vectors B + A + C or fifty-fifty C + B + A. As long as all 3 vectors are included with their specified magnitude and direction, the resultant will be the same. This property of vectors is the cardinal to the strategy used in the conclusion of the answer to the above example problem. To further illustrate the strategy, allow'southward consider the vector improver state of affairs described in Example 2 below.
Instance 2:

Mac and Tosh are doing the Vector Walk Lab. Starting at the door of their physics classroom, they walk 2.0 meters, south. They make a right hand turn and walk sixteen.0 meters, west. They plough right over again and walk 24.0 meters, north. They and then plough left and walk 36.0 meters, west. What is the magnitude of their overall displacement?

A graphical representation of the given problem will help visualize what is happening. The diagram below depicts such a representation.

When these iv vectors are added together in head-to-tail manner, the resultant is a vector that extends from the tail of the first vector (2.0 one thousand, South, shown in ruby-red) to the arrowhead of the quaternary vector (36.0 m, West, shown in light-green). The head-to-tail vector addition diagram is shown below.

The resultant vector (fatigued in black and labeled R) in the vector addition diagram to a higher place is not the hypotenuse of any immediately obvious right trangle. But past irresolute the order of improver of these four vectors, one can forcefulness this resultant vector to be the hypotenuse of a right triangle. For instance, past adding the vectors in the order of 2.0 one thousand, Southward + 24.0 thousand, Northward + 16.0 m, Westward + 36.0 one thousand. Westward, the resultant becomes the hypotenuse of a right triangle. This is shown in the vector improver diagram below.

With the vectors rearranged, the resultant is at present the hypotenuse of a correct triangle that has 2 perpendicular sides with lengths of 22.0 chiliad, N and 52.0 m, West. The 22.0 1000, Due north side is the result of 2.0 m, Due south and 24.0 thou, North added together. The 52.0 m, West side is the result of 16.0 m, West and 36.0 m, West added together. The magnitude of the resultant vector (R) can be determined using the Pythagorean theorem.

R² = (22.0 m)² + (52.0 thousand)²
R² = 484.0 1000^two + 2704.0 m²
R² = 3188.0 m²
R = SQRT (3188.0 m2^two)
R = 56.5 thousand

(SQRT indicates square root)

Equally tin be seen in these two examples, the resultant of the addition of three or more right bending vectors tin can be easily adamant using the Pythagorean theorem. Doing so involves the adding of the vectors in a different order.

SOH CAH TOA and the Direction of Vectors

The to a higher place discussion explains the method for determining the magnitude of the resultant for three or more perpendicular vectors. The topic of direction has been ignored in the discussion. Now we will turn our attention to the method for determining the direction of the resultant vector. As a quick review, recall that earlier in this lesson a convention for expressing the direction of a vector was introduced. The convention is known as the counter-clockwise from east convention, often abbreviated as the CCW convention. Using this convention, the direction of a vector is often expressed as a counter-clockwise angle of rotation of the vector about its tail from due E.

To begin our give-and-take, let's return to Example i above where we made an effort to add three vectors: 6.0 km, N + 6.0 km, E + 2.0 km, N. In the solution, the order of addition

of the iii vectors was rearranged and so that a right triangle was formed with the resultant beingness the hypotenuse of the triangle. The triangle is redrawn at the right. Discover that the bending in the lower left of the triangle has been labeled as theta (Θ). Theta (Θ) represents the angle that the vector makes with the northward centrality. Theta (Θ) can exist calculated using one of the three trigonometric functions introduced earlier in this lesson - sine, cosine or tangent. The mnemonic SOH CAH TOA is a helpful way of remembering which function to use. In this problem, nosotros wish to determine the angle measure of theta (Θ) and nosotros know the length of the side reverse theta (Θ) - 6.0 km - and the length of the side next the bending theta (Θ) - 8.0 km. The TOA of SOH CAH TOA indicates that the tangent of any angle is the ratio of the lengths of the side opposite to the side adjacent that angle. Thus, the tangent role volition be used to calculate the angle measure of theta (Θ). The piece of work is shown beneath.

Tangent(Θ) = Contrary/Next
Tangent(Θ) = 6.0/8.0
Tangent(Θ) = 0.75
Θ = tan^-1 (0.75)
Θ = 36.869 …°
Θ =37°

The problem is not over once the value of theta (Θ) has been calculated. This bending measure must now be used to country the direction. One means of doing so is to merely state that the management of the resultant is 37° east of due north. Alternatively, the counter-clockwise convention could be used. Since the angle that the resultant makes with e is the complement of the angle that it makes with north, we could express the direction as 53° CCW.

We will now consider Case ii as a 2d example of how to use SOH CAH TOA to determine the direction of a resultant. In Case 2, nosotros were trying to determine the magnitude of 2.0 grand. S + sixteen.0 thou, W + 24.0 m, N + 36.0 m, W. The solution involved re-ordering the addition so that the resultant was the hypotenuse of a right triangle with known sides. The right triangle is shown below. The resultant is fatigued in black. Once more, discover that the bending in the lower right of the triangle has been labeled as theta (Θ). Theta (Θ) represents the angle that the vector makes with the north axis.

Theta (Θ) can be calculated using the tangent office. In this problem, we wish to determine the bending measure of theta (Θ) and we know the length of the side contrary theta (Θ) - 52.0 m - and the length of the side adjacent the angle theta (Θ) - 22.0 m. The TOA of SOH CAH TOA indicates that the tangent of any bending is the ratio of the lengths of the side opposite to the side adjacent that angle. Thus, the tangent role will be used to calculate the angle measure of theta (Θ). The work is shown below.

Tangent(Θ) = Opposite/Next
Tangent(Θ) = 52.0/22.0
Tangent(Θ) = 2.3636 …
Θ = tan^-1 (2.3636 …)
Θ = 67.067 …°
Θ =67.1°

The problem is not over once the value of theta (Θ) has been calculated. This bending measure must at present be used to land the direction. One means of doing and so is to just state that the direction of the resultant is 67.1° west of due north. Alternatively, the counter-clockwise convention could be used. The n centrality is rotated xc° counter-clockwise from east and this vector is an boosted 67.1° counter-clockwise past n. Thus the CCW management is 157.i° CCW.

In summary, the direction of a vector can exist determined in the same way that it is always determined - by finding the angle of rotation counter-clockwise from due east. Since the resultant is the hypotenuse of a right triangle, this tin can be accomplished by first finding an angle that the resultant makes with one of the nearby axes of the triangle. In one case done, a trivial thinking is required in order to relate the bending to a direction.

Addition of Non-Perpendicular Vectors

Now nosotros will consider situations in which the two (or more) vectors that are being added are not at right angles to each other. The Pythagorean theorem is non applicable to such situations since information technology applies merely to correct triangles. Two not-perpendicular vectors will non form a right triangle. Yet it is possible to force two (or more) non-perpendicular vectors to exist transformed into other vectors that do form a correct triangle. The trick involves the concept of a vector component and the procedure of vector resolution.

A vector component describes the event of a vector in a given management. Whatsoever angled vector has two components; 1 is directed horizontally and the other is directed vertically. For instance, a northwest vector has a northward component and a west component. Together, the issue these two components are equal to the overall result of the angled vector. Every bit an case, consider a airplane that flies northwest from Chicago O'Hare airport towards the Canada border. The northwest deportation vector of the plane has 2 components - a due north component and a due west component. When added together, these two components are equal to the overall northwest displacement. This is shown in the diagram below.

The northwest vector has north and westward components that are represented as A₁₀ and A_y . It can be said that

A = A_x + A_y

So whenever nosotros think of a northwest vector, nosotros can think instead of two vectors - a due north and a w vector. The two components A_x + A_y can be substituted in for the unmarried vector A in the problem.

Now suppose that your task involves adding 2 non-perpendicular vectors together. Nosotros will call the vectors A and B. Vector A is a nasty angled vector that is neither horizontal nor vertical. And vector B is a nice, polite vector directed horizontally. The situation is shown beneath.

Of course nasty vector A has 2 components - A_x and A_y . These two components together are equal to vector A. That is, A = A_ten + A_y .

And since this is truthful, it makes since to say that A + B = A_x + A_y + B.

And so the problem of A + B has been transformed into a trouble in which all vectors are at right angles to each other. Nasty has been replaced past nice and that should make any physics student happy. With all vectors existence at correct angles to one another, their addition leads to a resultant that is at the hypotenuse of a correct triangle. The Pythagorean theorem can then exist used to make up one's mind the magnitude of the resultant.

To see how this process works with an actual vector add-on problem, consider Instance 3 shown below.

Example 3:

Max plays middle linebacker for S's football team. During one play in last Friday nighttime'due south game against New Greer Academy, he made the post-obit movements subsequently the ball was snapped on third downward. First, he back-pedaled in the southern direction for two.vi meters. He so shuffled to his left (west) for a distance of ii.2 meters. Finally, he made a half-turn and ran downfield a distance of 4.eight meters in a direction of 240° counter-clockwise from east (30° W of Southward) earlier finally knocking the wind out of New Greer's wide receiver. Determine the magnitude and management of Max'due south overall displacement.

As is the usual instance, the solution begins with a diagram of the vectors being added.

To assist in the discussion, the three vectors accept been labeled as vectors A, B, and C. The resultant is the vector sum of these three vectors; a caput-to-tail vector addition diagram reveals that the resultant is directed southwest. Of the three vectors existence added, vector C is clearly the nasty vector. Its direction is neither due south nor due due west. The solution involves resolving this vector into its components.

The process of vector resolution was discussed earlier in this lesson. The process involves using the magnitude and the sine and cosine functions to determine the 10- and y-components of the vector. Vector C makes a thirty° angle with the southern direction. Past sketching a right triangle with horizontal and vertical legs and C as the hypotenuse, information technology becomes possible to determine the components of vector C. This is shown in the diagram below. The side adjacent this 30° bending in the triangle is the vertical side; the vertical side represents the vertical (s) component of C - C_y. And then to make up one's mind C_y, the cosine office is used. The side opposite the thirty° angle is the horizontal side; the horizontal side represents the horizontal (westward) component of C - C_x. The values of C_x and C_y tin be adamant past using SOH CAH TOA. The cosine role is used to determine the southward component since the due south component is adjacent to the 30° angle. The sine office is used to determine the west component since the westward component is the side opposite to the 30° bending. The work is shown below.

Now our vector addition trouble has been transformed from the addition of two nice vectors and i nasty vector into the addition of four nice vectors.

With all vectors oriented along are customary north-south and eastward-west axes, they can be added head-to-tail in any order to produce a correct triangle whose the hypotenuse is the resultant. Such a diagram is shown below.

The triangle's perpendicular sides have lengths of 4.vi meters and 6.756 meters. The length of the horizontal side (4.6 m) was adamant past adding the values of B (2.2 k) and C_x (2.4 m). The length of the vertical side (6.756… thou) was determined by adding the values of A (2.6 one thousand) and C_y (4.156… m). The resultant's magnitude (R) can now be determined using the Pythagorean theorem.

Rⁱⁱ = (6.756… m)² + (4.vi grand)²
R² = 45.655… m^two + 21.16 k²
R² = 66.815… m²
R = SQRT(66.815… m² )
R = viii.174 … m
R = ~8.2 1000

The direction of the resultant can be determined by finding the angle that the resultant makes with either the n-southward or the eastward-west vector. The diagram at the right shows the bending theta (Θ) marked inside the vector addition triangle. This angle theta is the bending that the resultant makes with west. Its value can be determined using the tangent function. The tangent function (as in TOA) relates the angle value to the ratio of the lengths of the reverse side to the adjacent side. That is,

tangent(Θ) = (half-dozen.756… m)/(iv.6 m) = one.46889…

Using the inverse tangent function, the angle theta (Θ) tin can be adamant. On most calculators, this involves using the 2nd-Tangent buttons.

Θ = tan^-1 (ane.46889…) = 55.7536… °
Θ = ~56°

This 56° angle is the angle betwixt the resultant vector (drawn in black to a higher place) and the due west direction. This makes the direction 56° south of westward. The direction of the resultant based on the counter-clockwise from east convention (CCW) tin can be adamant by adding 180° to the 56°. So the CCW direction is 236°.

Case 4 provides one concluding example of how to combine vector resolution with vector addition in order to add three or more not-perpendicular vectors. Considering this example includes 3 especially nasty vectors, a table will be used to organize the information about he magnitude and direction of the components. The utilize of a tabular array is a wise idea when issues become complicated.

Example iv:

Cameron Per (his friends call him Cam) and Baxter Nature are on a hike. Starting from home base, they make the following movements.
A: 2.65 km, 140° CCW
B: 4.77 km, 252° CCW
C: iii.18 km, 332° CCW
Determine the magnitude and direction of their overall displacement.

The visual representation of the situation is shown below.

To determine the resultant, the three individual vectors are resolved into horizontal and vertical components. The angle information about each vector is used to grade a right triangle in which the vector is the hypotenuse and the perpendicular sides are oriented along the eastward-west and north-southward axes. This is shown in the diagram below.

Trigonometric functions - sine, cosine and tangent - are and so used to make up one's mind the magnitude of the horizontal and vertical component of each vector. The work is shown and organized in the table below.

Vector	East-West Component	Northward-South Component
A ii.65 km 140° CCW	(ii.65 km)•cos(forty°) = 2.030… km, West	(ii.65 km)•sin(40°) = 1.703… km, North
B 4.77 km 252° CCW	(4.77 km)•sin(xviii°) = 1.474… km, West	(iv.77 km)•cos(18°) = 4.536… km, South
C 3.18 km 332° CCW	(3.18 km)•cos(28°) = 2.808… km, Eastward	(iii.xviii km)•sin(28°) = i.493… km, South
Sum of A + B + C	0.696 km, West	4.326 km, Due south

The last row of the in a higher place table represents the sum of all the Due east-Due west components and the sum of all the Northward-South components. The resultant consists of these two components. The resultant is determined past adding together these two the components to grade a right triangle that has a hypotenuse that is equal to the resultant. This typically involves adding all the horizontal components to determine the total length of the horizontal side of the correct triangle … and calculation all the vertical components to determine the total length of the vertical side of the correct triangle. This is done in the table to a higher place by uncomplicated adding another row to the table for the vector sum of all the components. In adding the e-westward components of all the private vectors, one must consider that an e component and a westward component would add together as a positive and a negative. Some students prefer to recollect of this as subtraction every bit opposed to improver. In actuality, it really is improver - the improver of vectors with opposite direction. Similarly, a northward and a south component would too add together together every bit a positive and a negative. In one case the bottom row is accurately adamant, the magnitude of the resultant tin can be adamant using Pythagorean theorem.

R² = (0.696 km)ⁱⁱ + (4.326 km)ⁱⁱ
R² = 0.484 km² + 18.714 km^two
R^two = 19.199 kmⁱⁱ
R = SQRT(19.199 km²)
R = ~four.38 km

The management of the resulting displacement tin can exist determined by constructing the final triangle from the components of the resultant. The components of the resultant are just the sum the east-west and north-south components. Once done, SOH CAH TOA is used to make up one's mind the angle that the resultant makes with a nearby axis. The diagram is shown at the correct. The angle labeled as theta (Θ) is the angle between the resultant vector and the west axis. This angle can exist calculated as follows:

Tangent(Θ) = opposite/side by side
Tangent(Θ) = (4.326 km)/(0.696 km)
Tangent(Θ) = half-dozen.216
Θ = tan^-1(half-dozen.216)
Θ = lxxx.9°

This bending mensurate represents the angle of rotation of the vector south of west. It would be worded as 80.9° south of west. Since west is 180° counterclockwise from east, the direction could also be expressed in the counterclockwise (CCW) from east convention every bit 260.9°.

So the consequence of our assay is that the overall deportation is 4.38 km with a direction of 260.9° (CCW).

The questions that take been addressed on this page are:

How tin can three or more than perpendicular vectors be added together to decide the resultant?
How tin 2 or more non-perpendicular vectors exist added together to determine the resultant?

For both questions, we have plant that whatsoever 2 or iii or more vectors can be transformed or rearranged so that they add together to form a right triangle with the hypotenuse being the resultant. Once the right triangle is formed, Pythagorean theorem and SOH CAH TOA tin can exist used to calculate the resultant.

We Would Like to Suggest ...

Sometimes information technology isn't enough to only read about it. You take to interact with it! And that'southward exactly what y'all exercise when you use i of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of our Name That Vector Interactive, our Vector Addition Interactive, or our Vector Guessing Game Interactive. All iii Interactives can exist constitute in the Physics Interactive department of our website and provide an interactive feel with the skill of adding vectors.

Experiment with the widget beneath and then endeavor the problems in the Cheque Your Understanding department to test your skill at calculation vectors using components.

Practice!

The widget below computes the sum of 3 vectors if the x- and y-components are known. Use the widget to practice and check a problem.

Bank check Your Understanding

Consider the diagram below. Ix unique, labeled locations are provided on a filigree. Each foursquare on the grid represents a 20-meter ten 20-meter area. Rightward on the grid is in the due east management and up on the grid in the northward direction. Utilise the filigree in answering the next few questions.

one. Suppose that a person starts at position A and walks to position E and then to position G. Fill in the table below to betoken the e-west and the due north-south components of the individual legs of the walk and the components of the resulting displacement. Make the measurements off the filigree. Finally, utilize the Pythagorean theorem and SOH CAH TOA to make up one's mind the magnitude and the direction of the resulting displacement.

Vector	East-West Component	North-South Component
A to East
Due east to One thousand
Resultant A to G

Magnitude of Resultant: _________________________
Direction of Resultant: _________________________

ii. Using the same filigree, echo the measurements for a walk from location C to location B to location F. Make the measurements off the grid and utilize the Pythagorean theorem and SOH CAH TOA to make up one's mind the magnitude and the direction of the resulting displacement.

Vector	East-Westward Component	North-Due south Component
C to B
B to F
Resultant C to F

Magnitude of Resultant: _________________________
Direction of Resultant: _________________________

3. Finally, use the same grid to repeat the measurements for a walk from location I to location B to location G to location H. Make the measurements off the grid and use the Pythagorean theorem and SOH CAH TOA to determine the magnitude and the management of the resulting deportation.

Vector	E-West Component	N-Due south Component
I to B
B to G
G to H
Resultant I to H

Magnitude of Resultant: _________________________
Management of Resultant: _________________________

4. During her recent trip to the grocery store, Claire de Iles walked 28 m to the stop of an aisle. She then made a correct paw turn and walked 12 m down the end aisle. Finally, she made another right hand turn and walked 12 grand in the contrary direction as her original direction. Determine the magnitude of Claire's resultant displacement. (The actual direction - east, west, northward, south are not the focus.)

v. In the final game of last year's regular flavor, South was playing New Greer Academy for the Conference Championship. In the last play of the game, star quarterback Avery took a snap from scrimmage and scooted backwards (northwards) eight.0 yards. He then ran sideways (westward) out of the pocket for 12.0 yards before finally throwing a 34.0 yard pass directly downfield (southward) to Kendall for the game-winning touchdown. Determine the magnitude and direction of the ball's deportation.

6. Mia Ander exits the front end door of her dwelling and walks along the path shown in the diagram at the right (not to scale). The walk consists of 4 legs with the following magnitudes:

A = 46 k
B = 142 m
C = 78 g
D = 89 m

Determine the magnitude and management of Mia's resultant displacement. Consider using a tabular array to organize your calculations.