Vector Addition and the Parallelogram Rule (Grades 11-12)

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Vector Addition and the Parallelogram Rule

When using the parallelogram rule to add vectors, the sum of the vectors can be either diagonal of the parallelogram formed.

True
False

Add the vectors using the parallelogram method. [math] \ \ \ vec{u} = <<4,-6>>, \ \ \ vec{v} = <<1,-2>>[/math]
Coordinate Plane - 10x10 - Blank

Add the vectors using the parallelogram method. [math] \ \ \ \ vec{u} = <<-5,3>>, \ \ \ vec{v} = <<9,2>>[/math]
Coordinate Plane - 10x10 - Blank

In the following diagram, quadrilateral [math]FBCE[/math] is a parallelogram. Let sides [math]bar{FB}[/math], [math]bar{FE}[/math], and [math]bar{EC}[/math] represent vectors [math]vec{FB}[/math], [math]vec{FE}[/math], and [math]vec{EC}[/math]. Also, let diagonal [math]bar{FC}[/math] represent vector [math]vec{FC}[/math]. Assume that points F, E, and D are collinear. Not pictured is line segment [math]bar{CD}[/math], which is perpendicular to [math]bar{FD}[/math].

Let [math]ang BFE = theta[/math].

According to the parallelogram rule, [math]vec{FB} + vec{FE} = vec{FC}[/math]. The following questions will derive formulas for the magnitude and direction of [math]vec{FC}[/math], depending on the magnitudes and relative directions of vectors [math]vec{FB}[/math] and [math]vec{FE}[/math].
Parallelogram ABCDEF v3

Considering parallelogram [math]FBCE[/math], and that point D is collinear with points F and E, which of the following is/are true? (There may be more than one correct answer).

[math]||vec{FE}|| + ED = FD[/math]
[math]vec{FB} = vec{EC}[/math]
[math]ang CED = theta[/math]
[math]vec{FB} + vec{EC} = 2 vec{FE}[/math]

Considering right triangle CDE, how is the magnitude of [math]vec{EC}[/math] related to [math]theta[/math] and the length of [math]bar{ED}?[/math]

[math] ED = ||vec{EC}|| cos theta[/math]
[math]ED = ||vec{EC}|| sin theta[/math]
[math]||vec{EC}|| = ED cos theta[/math]
[math]tan theta = ||vec{EC}|| / (ED) [/math]

Considering triangle CDE, how is the magnitude of [math]vec{EC}[/math] related to the length of [math]bar{CD}[/math] and [math]theta ?[/math]

[math]CD = ||vec{EC}|| cos theta[/math]
[math]CD = ||vec{EC}|| sin theta[/math]
[math] || vec{EC}|| = CD cos theta [/math]
[math] tan theta = ||vec{EC}|| / (CD)[/math]

In the following work, two reasons are not given (in steps 1 and 5). What are these reasons?

[math] \|\|vec{FC}\|\|^2 [/math]	[math] = [/math]	[math] FD^2 + CD^2[/math]	[math]"Step 1 "[/math]
[math] [/math]	[math] = [/math]	[math] (\|\|vec{FE}\|\| + ED)^2 + CD^2[/math]	[math] "Info from previous questions"[/math]
[math] [/math]	[math] = [/math]	[math] \|\|vec{FE}\|\|^2 + 2\|\|vec{FE}\|\| ED + ED^2 + CD^2 [/math]	[math] "Expanding the square"[/math]
[math] [/math]	[math] = [/math]	[math] \|\|vec{FE}\|\|^2 + 2\|\|vec{FE}\|\| \ \|\|vec{EC}\|\|cos theta + \|\|vec{EC}\|\|^2 cos^2 theta + \|\|vec{EC}\|\|^2sin^2theta [/math]	[math]"Info from previous questions"[/math]
[math] [/math]	[math] = [/math]	[math] \|\|vec{FE}\|\|^2 + 2\|\|vec{FE}\|\| \ \|\|vec{EC}\|\|cos theta + \|\|vec{EC}\|\|^2 [/math]	[math]"Step 5"[/math]
[math] [/math]	[math] = [/math]	[math] \|\|vec{FE}\|\|^2 + \|\|vec{FB}\|\|^2 + 2\|\|vec{FE}\|\| \ \|\|vec{FB}\|\|cos theta [/math]	[math]"Info from previous questions"[/math]

Taking the square root of both sides gives the formula for the magnitude of the sum of the vectors [math]vec{FB}[/math] and [math]vec{FE}[/math]:

[math] ||vec{FC}|| = sqrt(||vec{FE}||^2 + ||vec{FB}||^2 + 2||vec{FE}|| \ ||vec{FB}||cos theta)[/math].

Pythagorean Theorem; [math]sin^2x + cos^2x = 1[/math]
Pythagorean Theorem; Law of Sines
Law of Vector Addition; Law of Cosines
Parallelogram Law; [math]cos2x = cos^2x - sin^2x[/math]

Using in formation from previous questions, the value of [math]ang 2[/math] can also be determined, as given below.

[math]ang 2 = tan^{-1} ((||vec{FB}|| sin theta) / (||vec{FE}|| + ||vec{FB}|| cos theta)) [/math]

Why is the direction given in terms of the angle between [math]vec{FC}[/math] and [math]vec{FE}[/math] and not between [math]vec{FC} [/math] and [math]vec{FB}? [/math]

It has to be given relative to the horizontal vector.
It has to be given in relation to a vector that is parallel to one of the axis.
It doesn't matter, since the diagonals of a parallelogram bisect their respective angles.
It doesn't matter, but it is simpler to derive the equation this way given the work already done with finding the magnitude of the vector.

Given two vectors, with magnitudes 8 and 4, which have a common initial point and an angle of 37° between them, what is the resulting vector if they are added together, given as a magnitude and direction? Let the calculated angle be relative to the vector of magnitude 4. Hint: use the parallelogram rule.

Magnitude 11.5, angle 12.1°.
Magnitude 11.5, angle 24.9°.
Magnitude 5.4, angle 35.9°.
Magnitude 5.4, angle 17.1°.

For vectors [math]vec{a}, vec{b}[/math], if they are placed such that their initial points are coincident, they have an angle of 60° between them. Also, [math]||vec{a}|| = 5[/math]. If these vectors are added together, the resulting vector, [math]vec{S}[/math] has a magnitude of 15 and an angle of 43.4°, relative to [math]vec{a}[/math]. What is the magnitude of [math]vec{b} ?[/math] Round answer to the nearest unit.