Vectors and Scalars
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Other suggestions are listed in the solutions/explanations as applicable.
These are the solutions to the ACT past questions on the topics: Vectors and Scalars.
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Symbols and Formulas
(1.) A(x, y)
Point: A
(): parenthesis
x = x-coordinate
y = y-coordinate
(2.) A$\langle x, y \rangle$
Vector: A
$\langle \rangle$: angle brackets
x = horizontal component
y = vertical component
(3.) Given a vector: $\overrightarrow{AB}$
$
\overrightarrow{AB} = \vec{B} - \vec{A} \\[3ex]
first\;\;vector = \vec{A} = \langle A_x, A_y \rangle \\[3ex]
second\;\;vector = \vec{B} = \langle B_x, B_y \rangle \\[3ex]
\overrightarrow{AB} = \langle B_x - A_x, B_y - A_y \rangle \\[3ex]
\overrightarrow{AB} = -\overrightarrow{BA} \\[5ex]
$
(4.) Given a vector: $\overrightarrow{BA}$
$
\overrightarrow{BA} = \vec{A} - \vec{B} \\[3ex]
first\;\;vector = \vec{B} = \langle B_x, B_y \rangle \\[3ex]
second\;\;vector = \vec{A} = \langle A_x, A_y \rangle \\[3ex]
\overrightarrow{BA} = \langle A_x - B_x, A_y - B_y \rangle \\[3ex]
\overrightarrow{BA} = -\overrightarrow{AB} \\[5ex]
$
(5.) Given several vectors:
Notice the colors and the pattern
$
(a.)\;\; \overrightarrow{AB} = -\overrightarrow{BA} \\[3ex]
(b.)\;\; \overrightarrow{CD} = -\overrightarrow{DC} \\[3ex]
(c.)\;\; \overrightarrow{A\color{red}{K}} + \overrightarrow{\color{red}{K}C} = \overrightarrow{AC} \\[3ex]
(d.)\;\; \overrightarrow{U\color{red}{K}} + \overrightarrow{\color{red}{K}S} = \overrightarrow{US} \\[3ex]
(e.)\;\; \overrightarrow{M\color{red}{P}} + \overrightarrow{\color{red}{P}C} + \overrightarrow{\color{red}{C}D}
=
\overrightarrow{MD} \\[5ex]
$
(6.) Given two position vectors: $\overrightarrow{OE}$ and $\overrightarrow{OF}$
$
\overrightarrow{EF} = \overrightarrow{EO} + \overrightarrow{OF} \\[3ex]
\overrightarrow{EF} = -\overrightarrow{OE} + \overrightarrow{OF} \\[5ex]
$
(7.) Given vectors in component form OR unit vectors:
Say for a two-dimensional vector A:
$
\boldsymbol{A}\langle x, y \rangle = x\boldsymbol{i} + y\boldsymbol{j} \\[3ex]
$
Say for a three-dimensional vector C:
$
\boldsymbol{C}\langle x, y, z \rangle = x\boldsymbol{i} + y\boldsymbol{j} + z\boldsymbol{k} \\[3ex]
$
Sum and Difference of Vectors
Assume we have two two-dimensional vectors: C and D:
$
\boldsymbol{C}\langle c_1, c_2 \rangle = c_1\boldsymbol{i} + c_2\boldsymbol{j} \\[3ex]
\boldsymbol{D}\langle d_1, d_2 \rangle = d_1\boldsymbol{i} + d_2\boldsymbol{j} \\[5ex]
\underline{Sum} \\[3ex]
\vec{C} + \vec{D} \\[3ex]
= \langle c_1 + d_1, c_2 + d_2 \rangle \\[3ex]
= (c_1 + d_1)\boldsymbol{i} + (c_2 + d_2)\boldsymbol{j} \\[5ex]
\underline{Difference} \\[3ex]
\vec{C} - \vec{D} \\[3ex]
= \langle c_1 - d_1, c_2 - d_2 \rangle \\[3ex]
= (c_1 - d_1)\boldsymbol{i} + (c_2 - d_2)\boldsymbol{j} \\[5ex]
$
Assume we have two three-dimensional vectors: C and D:
$
\boldsymbol{C}\langle c_1, c_2, c_3 \rangle = c_1\boldsymbol{i} + c_2\boldsymbol{j} + c_3\boldsymbol{k} \\[3ex]
\boldsymbol{D}\langle d_1, d_2, d_3 \rangle = d_1\boldsymbol{i} + d_2\boldsymbol{j} + d_3\boldsymbol{j} \\[5ex]
\underline{Sum} \\[3ex]
\vec{C} + \vec{D} \\[3ex]
= \langle c_1 + d_1, c_2 + d_2, c_3 + d_3 \rangle \\[3ex]
= (c_1 + d_1)\boldsymbol{i} + (c_2 + d_2)\boldsymbol{j} + (c_3 + d_3)\boldsymbol{k} \\[5ex]
\underline{Difference} \\[3ex]
\vec{C} - \vec{D} \\[3ex]
= \langle c_1 - d_1, c_2 - d_2, c_3 - d_3 \rangle \\[3ex]
= (c_1 - d_1)\boldsymbol{i} + (c_2 - d_2)\boldsymbol{j} + (c_3 - d_3)\boldsymbol{k} \\[5ex]
$
(8.) A displacement vector say $\overrightarrow{OA}$ beginning from the origin, O and ending at the point
A is the position vector, $\overrightarrow{A}$
(9.) Section Formula
Given two points say A and B with position vectors: $\overrightarrow{A}$ and $\overrightarrow{B}$
respectively, if a point say C divides the line segment |AC| in the ratio: m:n, then the
position vector of C is given by:
$
\overrightarrow{C} = \dfrac{m\overrightarrow{B} + n\overrightarrow{A}}{m + n} \\[5ex]
$
(10.) Collinearity of Points using Vectors
Given two vectors say: $\overrightarrow{A}$ and $\overrightarrow{B}$ and a scalar say k:
$
Say: \overrightarrow{A} = \langle a_1, a_2 \rangle \\[5ex]
\overrightarrow{B} = \langle b_1, b_2 \rangle \\[5ex]
$
(a.) Scalar Multiple Method
The two vectors are collinear if one is a scalar multiple of the other.
(This concept also demonstrates the linear dependency of two vectors.)
OR
The two vectors are collinear if the ratio of their corresponding components are equal.
$
\overrightarrow{A} = k \cdot \overrightarrow{B} \\[5ex]
OR \\[3ex]
\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2} \\[6ex]
$
(b.) Method of Determinants
Two vectors are collinear if their determinant is zero.
$
\begin{vmatrix}
a_1 & a_2 \\[4ex]
b_1 & b_2
\end{vmatrix} = a_1b_2 - b_1a_2 = 0 \\[7ex]
$
(c.) Dot Product Angle Between the Vectors
If the dot product angle between the two vectors is 0° or 180°, then the two vectors are
collinear.
$
\overrightarrow{A} \cdot \overrightarrow{B} = |\overrightarrow{A}| |\overrightarrow{B}| \cos \theta \\[3ex]
where \\[3ex]
\theta = \text{dot product angle between the two vectors}
$
Theorems
(1.) Proportional Segments Theorem for Diagonals in a Quadrilateral
In a quadrilateral, the line segment joining two points that divide opposite sides proportionally or at specific
ratios from a common vertex is parallel to and proportional to the diagonal that does not connect to the vertex.
(2.)
Point C will be placed on $\overline{AB}$ so that $\overrightarrow{AC} = \dfrac{1}{4}\overrightarrow{AB}$.
What will be the coordinates of point C?
$ A.\;\; \left(\dfrac{1}{2}, \dfrac{5}{2}\right) \\[5ex] B.\;\; \left(1, \dfrac{5}{2}\right) \\[5ex] C.\;\; \left(2, \dfrac{3}{2}\right) \\[5ex] D.\;\; (3, 4) \\[3ex] E.\;\; \left(5, \dfrac{11}{2}\right) \\[5ex] $
$ A(-1, 1) \hspace{5em} B(7, 7) \\[3ex] x_1 = -1 \hspace{5em} x_2 = 7 \\[4ex] y_1 = 1 \hspace{6em} y_2 = 7 \\[4ex] Vector\; \overrightarrow{AB} \\[3ex] \Delta x = x_2 - x_1 \\[4ex] = 7 - (-1) \\[3ex] = 7 + 1 \\[3ex] = 8 \\[5ex] \Delta y = y_2 - y_1 \\[4ex] = 7 - 1 \\[3ex] = 6 \\[3ex] \overrightarrow{AB} = (\overrightarrow{AB}_x, \overrightarrow{AB}_y) = (\Delta x, \Delta y) = (8, 6) \\[5ex] \overrightarrow{AC} = \dfrac{1}{4}\overrightarrow{AB} \\[5ex] = \left(\dfrac{1}{4} * 8, \dfrac{1}{4} * 6\right) \\[5ex] = \left(2, \dfrac{3}{2}\right) \\[5ex] = \left(\overrightarrow{AC}_x, \overrightarrow{AC}_y\right) \\[5ex] \text{Coordinates of Point C} = \left(x_1 + \overrightarrow{AC}_x, y_1 + \overrightarrow{AC}_y\right) \\[5ex] = \left(-1 + 2, 1 + \dfrac{3}{2}\right) \\[5ex] = \left(1, \dfrac{2}{2} + \dfrac{3}{2}\right) \\[5ex] = \left(1, \dfrac{2 + 3}{2}\right) \\[5ex] = \left(1, \dfrac{5}{2}\right) $
