Geometry and Trigonometry - Vectors: Geometric representation (HL)

In a parallelogram $ABCD$, let $\vec{AB} = \mathbf{u}$ and $\vec{AD} = \mathbf{v}$. Express the diagonals $\vec{AC}$ and $\vec{BD}$ in terms of $\mathbf{u}$ and $\mathbf{v}$.

1. In a parallelogram, opposite sides are equal in magnitude and parallel, so $\vec{BC} = \vec{AD} = \mathbf{v}$ and $\vec{DC} = \vec{AB} = \mathbf{u}$.
2. Using the triangle law for $\vec{AC}$: $\vec{AC} = \vec{AB} + \vec{BC} = \mathbf{u} + \mathbf{v}$
3. Using the triangle law for $\vec{BD}$: $\vec{BD} = \vec{BA} + \vec{AD} = -\vec{AB} + \vec{AD} = -\mathbf{u} + \mathbf{v}$ Therefore, $\vec{AC} = \mathbf{u} + \mathbf{v}$ and $\vec{BD} = \mathbf{v} - \mathbf{u}$.

This demonstrates the geometric application of vector addition and the property that $\vec{BA} = -\vec{AB}$.

Given points $A$ and $B$ with position vectors $\mathbf{a}$ and $\mathbf{b}$ respectively, find the position vector of point $M$ such that $M$ is the midpoint of the line segment $AB$.

1. The vector $\vec{AB}$ is given by $\mathbf{b} - \mathbf{a}$.
2. Since $M$ is the midpoint, $\vec{AM} = \frac{1}{2}\vec{AB} = \frac{1}{2}(\mathbf{b} - \mathbf{a})$.
3. The position vector $\vec{OM}$ is found by: $\vec{OM} = \vec{OA} + \vec{AM}$ $\vec{OM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a})$ $\vec{OM} = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a}$ $\vec{OM} = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$

This example uses the relationship between displacement vectors and position vectors to derive the midpoint formula in vector form.

If $\vec{p} = 2\vec{x} - 3\vec{y}$ and $\vec{q} = \vec{x} + k\vec{y}$, find the value of the scalar $k$ such that $\vec{p}$ and $\vec{q}$ are parallel.

1. For two vectors to be parallel, one must be a scalar multiple of the other: $\vec{p} = m\vec{q}$ for some $m \in \mathbb{R}$.
2. Substitute the expressions: $2\vec{x} - 3\vec{y} = m(\vec{x} + k\vec{y})$ $2\vec{x} - 3\vec{y} = m\vec{x} + mk\vec{y}$
3. Compare the coefficients of $\vec{x}$ and $\vec{y}$: For $\vec{x}$: $2 = m$ For $\vec{y}$: $-3 = mk$
4. Substitute $m=2$ into the second equation: $-3 = 2k \implies k = -\frac{3}{2}$

Two vectors are parallel if their components (or their geometric representations in terms of base vectors $\vec{x}$ and $\vec{y}$) are proportional.

1. First, express $\vec{BC}$ in terms of $\mathbf{b}$ and $\mathbf{c}$: $\vec{BC} = \vec{AC} - \vec{AB} = \mathbf{b} - \mathbf{c}$
2. Since $D$ divides $BC$ in the ratio $2:1$, $\vec{BD} = \frac{2}{3} \vec{BC} = \frac{2}{3}(\mathbf{b} - \mathbf{c})$
3. Now, find $\vec{AD}$ using the triangle law: $\vec{AD} = \vec{AB} + \vec{BD} = \mathbf{c} + \frac{2}{3}(\mathbf{b} - \mathbf{c})$
4. Simplify the expression: $\vec{AD} = \mathbf{c} + \frac{2}{3}\mathbf{b} - \frac{2}{3}\mathbf{c} = \frac{1}{3}\mathbf{c} + \frac{2}{3}\mathbf{b}$

This problem uses the vector addition law and the ratio property of line segments. By expressing the segment $BC$ as a vector difference, we can find any point along that line by applying a scalar fraction.

1. In a regular hexagon, the vector from the center $M$ to a vertex is equal to the side vector. Let $M$ be the center. $\vec{OM} = \vec{AB} = \mathbf{b}$.
2. Due to symmetry and properties of regular hexagons, $\vec{BC}$ is parallel and equal to $\vec{OM} - \vec{OA} = \mathbf{b} - \mathbf{a}$ is incorrect; rather, in a regular hexagon, $\vec{BC} = \vec{AO} + \vec{AB} = \mathbf{b} - \mathbf{a}$.
3. Therefore, $\vec{OC} = \vec{OA} + \vec{AB} + \vec{BC} = \mathbf{a} + \mathbf{b} + (\mathbf{b} - \mathbf{a}) = 2\mathbf{b}$.

A regular hexagon can be divided into six equilateral triangles. Using the center of the hexagon as a reference point helps identify parallel and equal vectors.