Honors Geometry Unit 14 Review: Vectors

Vectors give you an algebraic way to prove geometric facts. Treat each point as a position vector from the origin, and find the vector from $A$ to $B$ by subtracting, $\vec{AB} = B - A$.

Three tests cover most proofs. If $A$ and $B$ are distinct, points $A$, $B$, and $C$ are collinear exactly when $\vec{AC} = k\vec{AB}$ for some scalar $k$. Two distinct lines are parallel if their nonzero direction vectors are scalar multiples. Two nonzero segments are perpendicular if their direction vectors have a dot product of 0.

The midpoint of $\overline{AB}$ is the average of the position vectors, $M = \frac{A + B}{2}$. That gives a quick proof that a parallelogram's diagonals bisect each other, since both diagonals have the same midpoint.

Vectors also model forces. The resultant force is the sum of all the force vectors. If the sum is the zero vector, the object has no linear acceleration and is in translational equilibrium. An extended object also needs zero net torque to be in rotational equilibrium.

Worked example. Are $A(1, 2)$, $B(3, 5)$, and $C(7, 11)$ collinear? Compute $\vec{AB} = (3-1, 5-2) = (2, 3)$ and $\vec{AC} = (7-1, 11-2) = (6, 9)$. Since $(6, 9) = 3(2, 3)$, the vectors point the same way from the shared point $A$, so the three points lie on one line.

A vector has magnitude (length) and direction. A scalar is a plain number with size only. In component form, $\vec{v} = (a, b)$ means move $a$ horizontally and $b$ vertically. You may also see this written as $\langle a, b \rangle$ or $a\hat{i} + b\hat{j}$.

Magnitude comes from the Pythagorean theorem, $|\vec{v}| = \sqrt{a^2 + b^2}$. For a nonzero vector, the direction angle is measured counterclockwise from the positive x-axis. When $a \ne 0$, use $\tan\theta = \frac{b}{a}$. Inverse tangent returns an angle between $-90°$ and $90°$, so add $180°$ when $a < 0$, or add $360°$ when $a > 0$ and $b < 0$. When $a = 0$, the angle is $90°$ if $b > 0$ and $270°$ if $b < 0$. The zero vector has no direction angle.

To add vectors, add components, or draw them tip-to-tail and connect the first tail to the last head. Subtracting $\vec{b}$ means adding $-\vec{b}$. Multiplying by a scalar $c$ scales the length by $|c|$ and reverses direction when $c < 0$. A unit vector has length 1. For a nonzero vector $\vec{v}$, the unit vector in its direction is $\frac{\vec{v}}{|\vec{v}|}$.

Worked example. Find the magnitude, direction, and unit vector for $\vec{v} = (-3, 4)$. The magnitude is $\sqrt{9 + 16} = 5$. The calculator gives $\tan^{-1}\left(\frac{4}{-3}\right) \approx -53.1°$, but the vector points up and left into Quadrant II, so add $180°$ to get about $126.9°$. The unit vector is $\left(-\frac{3}{5}, \frac{4}{5}\right)$.

The dot product multiplies matching components and adds, $\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y$. The answer is a single number, not a vector. For nonzero vectors, it also equals $|\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between them. You can solve $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.

For nonzero vectors, the sign tells you the angle type. A positive dot product means an angle from 0° up to but not including 90°, zero means perpendicular (orthogonal), and negative means an angle greater than 90° through 180°.

A projection measures how much of $\vec{a}$ points along a nonzero vector $\vec{b}$, like the shadow $\vec{a}$ casts on the line through $\vec{b}$. The scalar projection is the signed length $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. The vector projection is the shadow itself, $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$. Subtracting the projection from $\vec{a}$ leaves the part perpendicular to $\vec{b}$.

Worked example. Let $\vec{a} = (2, 4)$ and $\vec{b} = (3, 1)$. The dot product is $6 + 4 = 10$. Since $|\vec{a}| = \sqrt{20}$ and $|\vec{b}| = \sqrt{10}$, $\cos\theta = \frac{10}{\sqrt{200}} = \frac{1}{\sqrt{2}}$, so $\theta = 45°$. The projection is $\frac{10}{10}(3, 1) = (3, 1)$. The perpendicular part is $(2, 4) - (3, 1) = (-1, 3)$, and $(-1, 3) \cdot (3, 1) = 0$ confirms it.