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

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.

Question Info

Vectors Question

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