Vector Addition and Parallelogram Rule

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In the following diagram, quadrilateral

$FBCE$ is a parallelogram. Let sides

$bar{FB}$ ,

$bar{FE}$ , and

$bar{EC}$ represent vectors

$vec{FB}$ ,

$vec{FE}$ , and

$vec{EC}$ . Also, let diagonal

$bar{FC}$ represent vector

$vec{FC}$ . Assume that points F, E, and D are collinear. Not pictured is line segment

$bar{CD}$ , which is perpendicular to

$bar{FD}$ .

Let

$ang BFE = theta$ .

According to the parallelogram rule,

$vec{FB} + vec{FE} = vec{FC}$ . The following questions will derive formulas for the magnitude and direction of

$vec{FC}$ , depending on the magnitudes and relative directions of vectors

$vec{FB}$ and

$vec{FE}$ .
Parallelogram ABCDEF v3

Grade 11 Vectors CCSS: HSN-VM.B.4, HSN-VM.B.4a

Considering parallelogram

$FBCE$ , 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).

$||vec{FE}|| + ED = FD$
$vec{FB} = vec{EC}$
$ang CED = theta$
$vec{FB} + vec{EC} = 2 vec{FE}$

Grade 11 Vectors CCSS: HSN-VM.B.4, HSN-VM.B.4a

Considering right triangle CDE, how is the magnitude of

$vec{EC}$ related to

$theta$ and the length of

$bar{ED}?$

$ED = ||vec{EC}|| cos theta$
$ED = ||vec{EC}|| sin theta$
$||vec{EC}|| = ED cos theta$
$tan theta = ||vec{EC}|| / (ED)$

Grade 11 Vectors CCSS: HSN-VM.B.4, HSN-VM.B.4a

Considering triangle CDE, how is the magnitude of

$vec{EC}$ related to the length of

$bar{CD}$ and

$theta ?$

$CD = ||vec{EC}|| cos theta$
$CD = ||vec{EC}|| sin theta$
$|| vec{EC}|| = CD cos theta$
$tan theta = ||vec{EC}|| / (CD)$

Grade 11 Vectors CCSS: HSN-VM.B.4, HSN-VM.B.4a

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

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

Taking the square root of both sides gives the formula for the magnitude of the sum of the vectors

$vec{FB}$ and

$vec{FE}$ :

$||vec{FC}|| = sqrt(||vec{FE}||^2 + ||vec{FB}||^2 + 2||vec{FE}|| \ ||vec{FB}||cos theta)$ .

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

Grade 11 Vectors CCSS: HSN-VM.B.4, HSN-VM.B.4a

Using in formation from previous questions, the value of

$ang 2$ can also be determined, as given below.

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

Why is the direction given in terms of the angle between

$vec{FC}$ and

$vec{FE}$ and not between

$vec{FC}$ and

$vec{FB}?$

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.

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Vector Addition and Parallelogram Rule

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