Vectors Question

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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

$vec{FE}$ .

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; $cos2x = cos^2x - sin^2x$

Question Info

Vectors Question

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