Vector Proofs - IB Mathematics Analysis and Approaches

1. To prove using vectors that a given quadrilateral ABCD is not a parallelogram, which of the following conditions would be sufficient?
   Answer: $\vec{AB} \ne \vec{DC}$ and $\vec{AD} \ne \vec{BC}$ A quadrilateral is a parallelogram if and only if one pair of opposite sides are equal and parallel. In vector terms, this means $\vec{AB} = \vec{DC}$ OR $\vec{AD} = \vec{BC}$. To prove it is not a parallelogram, we must show that neither of these conditions holds. Therefore, showing that $\vec{AB} \ne \vec{DC}$ AND $\vec{AD} \ne \vec{BC}$ is a sufficient condition. Option A is about dot products and doesn't directly relate to the definition. Option B compares diagonals which isn't a defining property. Option C is true for any closed polygon.
2. In quadrilateral ABCD, the diagonals intersect at M. It is given that $\vec{AM} = k\vec{MC}$ and $\vec{BM} = k\vec{MD}$ for the same positive scalar $k$. Which geometric property of the quadrilateral can be proven from this information?
   Answer: ABCD is a trapezoid. Let the origin be M. Then $\vec{MA} = -k\vec{MC}$ and $\vec{MB} = -k\vec{MD}$. Let $\vec{MC} = \mathbf{c}$ and $\vec{MD} = \mathbf{d}$. Then $\vec{MA} = -k\mathbf{c}$ and $\vec{MB} = -k\mathbf{d}$. Consider the sides AB and DC. $\vec{AB} = \vec{MB} - \vec{MA} = -k\mathbf{d} - (-k\mathbf{c}) = k(\mathbf{c}-\mathbf{d})$. And $\vec{DC} = \vec{MC} - \vec{MD} = \mathbf{c}-\mathbf{d}$. Since $\vec{AB} = k\vec{DC}$, the vectors are parallel. Thus, the side AB is parallel to the side DC. This proves that ABCD is a trapezoid. It is only a parallelogram if $k=1$.
3. In a triangle OAB, with $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$, the altitude from O meets the line segment AB at point P. Which vector condition must be satisfied?
   Answer: $\mathbf{p} \cdot (\mathbf{b}-\mathbf{a}) = 0$ The altitude from O is the line segment OP, represented by the vector $\vec{OP} = \mathbf{p}$. This altitude is, by definition, perpendicular to the opposite side AB. The vector representing the side AB is $\vec{AB} = \vec{OB} - \vec{OA} = \mathbf{b}-\mathbf{a}$. For two vectors to be perpendicular, their scalar (dot) product must be zero. Therefore, $\mathbf{p} \cdot (\mathbf{b}-\mathbf{a}) = 0$.
4. The circumcenter of a triangle is the intersection of the perpendicular bisectors of its sides. Let the circumcenter of $\triangle ABC$ be at the origin P, with $\vec{PA} = \mathbf{a}$ and $\vec{PB} = \mathbf{b}$. Which vector equation represents the condition that P lies on the perpendicular bisector of the side AB?
   Answer: $(\frac{\mathbf{a}+\mathbf{b}}{2}) \cdot (\mathbf{b}-\mathbf{a}) = 0$ The perpendicular bisector of side AB passes through the midpoint M of AB and is perpendicular to the vector $\vec{AB}$. The position vector of M relative to P (the origin) is $\vec{PM} = \frac{\mathbf{a}+\mathbf{b}}{2}$. The vector representing the side AB is $\vec{AB} = \vec{PB} - \vec{PA} = \mathbf{b}-\mathbf{a}$. The line containing P and M must be perpendicular to AB. This means the vector $\vec{PM}$ must be perpendicular to $\vec{AB}$. Their dot product must be zero: $(\frac{\mathbf{a}+\mathbf{b}}{2}) \cdot (\mathbf{b}-\mathbf{a}) = 0$. This simplifies to $|\mathbf{b}|^2-|\mathbf{a}|^2=0$, which confirms P is equidistant from A and B.
5. Points A, B, and C have position vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$ respectively. A student wants to prove that the points are collinear. Which of the following vector equations, if true for some non-zero scalar $k$, would be sufficient for the proof?
   Answer: $\mathbf{b}-\mathbf{a} = k(\mathbf{c}-\mathbf{a})$ For points A, B, and C to be collinear, the vector from A to B must be parallel to the vector from A to C. The vector $\vec{AB}$ is $\mathbf{b}-\mathbf{a}$, and the vector $\vec{AC}$ is $\mathbf{c}-\mathbf{a}$. If $\vec{AB} = k\vec{AC}$ for some scalar $k$, the vectors are parallel. Since they share a common point A, the points A, B, and C must lie on the same line.
6. In $\triangle ABC$, the altitudes from A and B intersect at the orthocenter H. Let the origin be at vertex C, so $\vec{CA} = \mathbf{a}$ and $\vec{CB} = \mathbf{b}$. The position vector of H is $\mathbf{h}$. Which pair of vector equations most directly expresses the geometric conditions that define H?
   Answer: $(\mathbf{h}-\mathbf{a}) \cdot \mathbf{b} = 0$ and $(\mathbf{h}-\mathbf{b}) \cdot \mathbf{a} = 0$ With the origin at C, the position vectors are $\vec{C}=\mathbf{0}$, $\vec{A}=\mathbf{a}$, and $\vec{B}=\mathbf{b}$. The altitude from A is the line AH, which is perpendicular to the side BC. The vector $\vec{AH}$ is $\mathbf{h}-\mathbf{a}$, and the vector $\vec{BC}$ is $\vec{C}-\vec{B} = \mathbf{0}-\mathbf{b} = -\mathbf{b}$. Perpendicularity means $(\mathbf{h}-\mathbf{a}) \cdot (-\mathbf{b}) = 0$, which simplifies to $(\mathbf{h}-\mathbf{a}) \cdot \mathbf{b} = 0$. The altitude from B is BH, perpendicular to AC. The vector $\vec{BH}$ is $\mathbf{h}-\mathbf{b}$, and $\vec{AC}$ is $\vec{C}-\vec{A} = \mathbf{0}-\mathbf{a} = -\mathbf{a}$. Perpendicularity means $(\mathbf{h}-\mathbf{b}) \cdot (-\mathbf{a}) = 0$, which simplifies to $(\mathbf{h}-\mathbf{b}) \cdot \mathbf{a} = 0$. These two conditions in D directly represent the geometric definitions.
7. A rhombus is a parallelogram with equal-length sides. A square is a rhombus with perpendicular adjacent sides. For a rhombus defined by adjacent vectors $\mathbf{u}$ and $\mathbf{v}$ from a common vertex, which additional condition is sufficient to prove it is a square?
   Answer: $|\mathbf{u}+\mathbf{v}| = |\mathbf{u}-\mathbf{v}|$ When you encounter vector geometry problems involving parallelograms and squares, focus on the key geometric properties: equal side lengths for rhombi, and perpendicular sides for squares. Since we already have a rhombus (where $|\mathbf{u}| = |\mathbf{v}|$ by definition), we need an additional condition that forces the adjacent sides to be perpendicular, making it a square. Option C, $|\mathbf{u}+\mathbf{v}| = |\mathbf{u}-\mathbf{v}|$, is correct because it ensures perpendicularity. Let's see why: squaring both sides gives us $|\mathbf{u}+\mathbf{v}|^2 = |\mathbf{u}-\mathbf{v}|^2$. Expanding using the dot product formula: $(\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}+\mathbf{v}) = (\mathbf{u}-\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v})$. This simplifies to $|\mathbf{u}|^2 + 2\mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2 = |\mathbf{u}|^2 - 2\mathbf{u} \cdot \mathbf{v} + |\mathbf{v}|^2$. The terms cancel except for $4\mathbf{u} \cdot \mathbf{v} = 0$, which means $\mathbf{u} \cdot \mathbf{v} = 0$, confirming perpendicularity. Option A restates the rhombus condition we already have. Option B, $(\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = 0$, expands to $|\mathbf{u}|^2 - |\mathbf{v}|^2 = 0$, which again just confirms equal side lengths. Option D gives a specific angle relationship (60°) that doesn't guarantee perpendicularity. Remember: geometric vector problems often test whether you can translate between algebraic conditions and geometric properties. Practice expanding dot products to reveal hidden perpendicularity conditions.
8. In a tetrahedron OABC, let $\vec{OA} = \mathbf{a}$, $\vec{OB} = \mathbf{b}$, and $\vec{OC} = \mathbf{c}$. It is given that the pair of opposite edges OA and BC are perpendicular, and the pair of opposite edges OB and AC are perpendicular. What geometric property can be proven as a direct consequence of these two conditions?
   Answer: The third pair of opposite edges, OC and AB, are also perpendicular. The given conditions translate to vector equations. OA $\perp$ BC means $\vec{OA} \cdot \vec{BC} = 0$, so $\mathbf{a} \cdot (\mathbf{c}-\mathbf{b}) = 0$, which gives $\mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{b}$. OB $\perp$ AC means $\vec{OB} \cdot \vec{AC} = 0$, so $\mathbf{b} \cdot (\mathbf{c}-\mathbf{a}) = 0$, which gives $\mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{a}$. Combining these results, we have $\mathbf{a} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{b}$ and $\mathbf{b} \cdot \mathbf{c} = \mathbf{a} \cdot \mathbf{b}$, which implies $\mathbf{a} \cdot \mathbf{c} = \mathbf{b} \cdot \mathbf{c}$. Rearranging this gives $\mathbf{b} \cdot \mathbf{c} - \mathbf{a} \cdot \mathbf{c} = 0$, which is $(\mathbf{b}-\mathbf{a}) \cdot \mathbf{c} = 0$. This is the dot product of $\vec{AB}$ and $\vec{OC}$, so it proves that the third pair of opposite edges are also perpendicular.
9. A proof shows that the medians of a triangle are concurrent. Let $\triangle OAB$ have medians from O to AB (meeting at M) and from A to OB (meeting at N). The intersection point P can be expressed as $\vec{OP} = k\vec{OM}$ and also as $\vec{OP} = \vec{OA} + l\vec{AN}$ for scalars $k, l$. Given $\vec{OA}=\mathbf{a}$ and $\vec{OB}=\mathbf{b}$, what is the correct substitution for $\vec{OM}$ and $\vec{AN}$?
   Answer: $\vec{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$ and $\vec{AN} = \frac{1}{2}\mathbf{b}-\mathbf{a}$ When working with medians and concurrent points in triangles, you need to carefully track how vectors relate to midpoints and the specific geometric relationships involved. Let's find $\vec{OM}$ first. Since M is the midpoint of AB, we have $\vec{AM} = \frac{1}{2}\vec{AB}$. To find $\vec{OM}$, we use $\vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} - \mathbf{a}) = \mathbf{a} + \frac{1}{2}\mathbf{b} - \frac{1}{2}\mathbf{a} = \frac{1}{2}(\mathbf{a} + \mathbf{b})$. For $\vec{AN}$, since N is the midpoint of OB, we have $\vec{ON} = \frac{1}{2}\vec{OB} = \frac{1}{2}\mathbf{b}$. Therefore, $\vec{AN} = \vec{AO} + \vec{ON} = -\mathbf{a} + \frac{1}{2}\mathbf{b} = \frac{1}{2}\mathbf{b} - \mathbf{a}$. This confirms answer C is correct: $\vec{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$ and $\vec{AN} = \frac{1}{2}\mathbf{b}-\mathbf{a}$. Answer A has the correct $\vec{OM}$ but gives $\vec{AN} = \frac{1}{2}\mathbf{a}-\mathbf{b}$, which incorrectly uses $\mathbf{a}$ instead of $\mathbf{b}$ in the first term. Answer B gives $\vec{OM} = \mathbf{b}-\mathbf{a}$, which is actually $\vec{AB}$ rather than the vector to the midpoint. Answer D provides $\vec{OM} = \mathbf{a}+\mathbf{b}$, missing the crucial factor of $\frac{1}{2}$ needed for midpoints. Remember: when finding vectors to midpoints, always include the $\frac{1}{2}$ factor, and carefully track the direction of your vectors using the triangle method $\vec{OC} = \vec{OA} + \vec{AC}$.
10. Let A and B be the endpoints of a diameter of a circle with centre at the origin O. Let their position vectors be $\mathbf{a}$ and $-\mathbf{a}$ respectively. P is any other point on the circle with position vector $\mathbf{p}$. A proof that the angle APB is a right angle is based on calculating $\vec{AP} \cdot \vec{BP}$. Which calculation correctly demonstrates this?
    Answer: $\vec{AP} \cdot \vec{BP} = (\mathbf{p}-\mathbf{a}) \cdot (\mathbf{p}+\mathbf{a}) = |\mathbf{p}|^2 - |\mathbf{a}|^2 = r^2 - r^2 = 0$ The vectors from point P to A and B are $\vec{PA} = \mathbf{a}-\mathbf{p}$ and $\vec{PB} = -\mathbf{a}-\mathbf{p}$. Alternatively, we can use vectors originating from A and B towards P: $\vec{AP} = \mathbf{p}-\mathbf{a}$ and $\vec{BP} = \mathbf{p}-(-\mathbf{a}) = \mathbf{p}+\mathbf{a}$. The angle APB is the angle between these two vectors. Their dot product is $\vec{AP} \cdot \vec{BP} = (\mathbf{p}-\mathbf{a}) \cdot (\mathbf{p}+\mathbf{a}) = \mathbf{p}\cdot\mathbf{p} + \mathbf{p}\cdot\mathbf{a} - \mathbf{a}\cdot\mathbf{p} - \mathbf{a}\cdot\mathbf{a} = |\mathbf{p}|^2 - |\mathbf{a}|^2$. Since P and A are on the circle with centre O and radius $r$, their position vectors have magnitude $r$. Thus, $|\mathbf{p}| = |\mathbf{a}| = r$. The dot product is $r^2 - r^2 = 0$, proving the vectors are perpendicular.
11. The vector equation of a line passing through point A with position vector $\mathbf{a}$ and parallel to vector $\mathbf{d}$ is $\mathbf{r} = \mathbf{a} + t\mathbf{d}$. How can this be used to prove that the point C with position vector $\mathbf{c}$ lies on this line?
    Answer: Show that $\mathbf{c}-\mathbf{a}$ is a scalar multiple of $\mathbf{d}$ If point C lies on the line, its position vector $\mathbf{c}$ must satisfy the line's equation for some scalar value of $t$. So, we must have $\mathbf{c} = \mathbf{a} + t\mathbf{d}$ for some $t$. Rearranging this equation gives $\mathbf{c} - \mathbf{a} = t\mathbf{d}$. This shows that the vector from A to C, which is $\mathbf{c} - \mathbf{a}$, must be a scalar multiple of the direction vector $\mathbf{d}$. This is the condition for $\vec{AC}$ to be parallel to $\mathbf{d}$, which means C lies on the line passing through A with direction $\mathbf{d}$.
12. A parallelogram has adjacent sides represented by vectors $\mathbf{p}$ and $\mathbf{q}$. Which condition is sufficient to prove the parallelogram is a rectangle, but not necessarily a square?
    Answer: $|\mathbf{p}+\mathbf{q}| = |\mathbf{p}-\mathbf{q}|$ The vectors for the diagonals are $\mathbf{d}_1 = \mathbf{p}+\mathbf{q}$ and $\mathbf{d}_2 = \mathbf{p}-\mathbf{q}$. A parallelogram is a rectangle if its diagonals are equal in length, i.e., $|\mathbf{d}_1| = |\mathbf{d}_2|$. So, $|\mathbf{p}+\mathbf{q}| = |\mathbf{p}-\mathbf{q}|$. Squaring both sides gives $|\mathbf{p}|^2 + 2\mathbf{p}\cdot\mathbf{q} + |\mathbf{q}|^2 = |\mathbf{p}|^2 - 2\mathbf{p}\cdot\mathbf{q} + |\mathbf{q}|^2$, which simplifies to $4\mathbf{p}\cdot\mathbf{q} = 0$, or $\mathbf{p}\cdot\mathbf{q}=0$. This means the adjacent sides are perpendicular. This condition does not require $|\mathbf{p}|=|\mathbf{q}|$, so it defines a rectangle that is not necessarily a square.
13. In $\triangle OAB$, M is the midpoint of AB. It is given that the median OM is perpendicular to the side AB. Which of the following properties of $\triangle OAB$ can be proven from this condition?
    Answer: $\triangle OAB$ is isosceles with $OA = OB$. Let $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$. The vector for the median OM is $\vec{OM} = \frac{\mathbf{a}+\mathbf{b}}{2}$ since M is the midpoint of AB. The vector for the side AB is $\vec{AB} = \mathbf{b}-\mathbf{a}$. The condition that OM is perpendicular to AB means their dot product is zero: $\vec{OM} \cdot \vec{AB} = 0$. This gives $(\frac{\mathbf{a}+\mathbf{b}}{2}) \cdot (\mathbf{b}-\mathbf{a}) = 0$, which simplifies to $(\mathbf{b}+\mathbf{a}) \cdot (\mathbf{b}-\mathbf{a}) = 0$. Expanding this gives $|\mathbf{b}|^2 - |\mathbf{a}|^2 = 0$, which means $|\mathbf{b}|^2 = |\mathbf{a}|^2$. Taking the square root, we get $|\mathbf{b}| = |\mathbf{a}|$, which means the lengths of sides OB and OA are equal. Therefore, the triangle is isosceles.
14. In $\triangle OAB$, P is a point on AB such that AP:PB = 2:1. A vector proof is being constructed to find the position vector of P. Given $\vec{OA} = \mathbf{a}$ and $\vec{OB} = \mathbf{b}$, which is the correct expression for $\vec{OP}$?
    Answer: $\vec{OP} = \frac{\mathbf{a}+2\mathbf{b}}{3}$ Using the section formula for a point P dividing a line segment AB in the ratio m:n (AP:PB = m:n), the position vector of P is given by $\vec{OP} = \frac{n\vec{OA} + m\vec{OB}}{m+n}$. In this case, m=2 and n=1. Substituting the given vectors, we get $\vec{OP} = \frac{1\cdot\mathbf{a} + 2\cdot\mathbf{b}}{2+1} = \frac{\mathbf{a}+2\mathbf{b}}{3}$. Alternatively, one can use vector addition: $\vec{OP} = \vec{OA} + \vec{AP} = \mathbf{a} + \frac{2}{3}\vec{AB} = \mathbf{a} + \frac{2}{3}(\mathbf{b}-\mathbf{a}) = \mathbf{a} + \frac{2}{3}\mathbf{b} - \frac{2}{3}\mathbf{a} = \frac{1}{3}\mathbf{a} + \frac{2}{3}\mathbf{b} = \frac{\mathbf{a}+2\mathbf{b}}{3}$.
15. In triangle ABC, with vertices having position vectors $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{c}$, the medians intersect at the centroid G. It is a known property that the centroid divides the median from vertex A to the midpoint M of BC in a 2:1 ratio (AG:GM = 2:1). Which vector equation correctly represents the position vector $\mathbf{g}$ of the centroid based on this property?
    Answer: $\mathbf{g} = \mathbf{a} + \frac{2}{3}(\frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a})$ The position vector of M, the midpoint of BC, is $\mathbf{m} = \frac{\mathbf{b}+\mathbf{c}}{2}$. The vector for the median AM is $\vec{AM} = \mathbf{m} - \mathbf{a} = \frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a}$. Since G divides AM in the ratio 2:1, the vector $\vec{AG}$ is $\frac{2}{3}$ of the vector $\vec{AM}$. The position vector of G is given by $\mathbf{g} = \vec{OA} + \vec{AG} = \mathbf{a} + \frac{2}{3}\vec{AM} = \mathbf{a} + \frac{2}{3}(\frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a})$. This expression simplifies to the well-known formula $\mathbf{g} = \frac{1}{3}(\mathbf{a}+\mathbf{b}+\mathbf{c})$.
16. In triangle ABC, M is the midpoint of BC. A vector proof of Apollonius' theorem, which states $AB^2 + AC^2 = 2(AM^2 + BM^2)$, is being constructed. Let A be the origin. Then $\vec{AB} = \mathbf{b}$ and $\vec{AC} = \mathbf{c}$. What is the vector expression for $2(AM^2 + BM^2)$ in terms of $\mathbf{b}$ and $\mathbf{c}$?
    Answer: $2\left(|\frac{\mathbf{b}+\mathbf{c}}{2}|^2 + |\frac{\mathbf{c}-\mathbf{b}}{2}|^2\right)$ With A as the origin, $\vec{A} = \mathbf{0}$. We have $\vec{AB} = \mathbf{b}$ and $\vec{AC} = \mathbf{c}$. The midpoint M of BC has position vector $\vec{AM} = \mathbf{m} = \frac{\mathbf{b}+\mathbf{c}}{2}$. The vector $\vec{BM}$ is $\mathbf{m} - \mathbf{b} = \frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{b} = \frac{\mathbf{c}-\mathbf{b}}{2}$. The theorem relates the squares of the lengths, which are the squared magnitudes of the vectors. So, $AM^2 = |\mathbf{m}|^2 = |\frac{\mathbf{b}+\mathbf{c}}{2}|^2$ and $BM^2 = |\vec{BM}|^2 = |\frac{\mathbf{c}-\mathbf{b}}{2}|^2$. The expression $2(AM^2 + BM^2)$ becomes $2\left(|\frac{\mathbf{b}+\mathbf{c}}{2}|^2 + |\frac{\mathbf{c}-\mathbf{b}}{2}|^2\right)$.
17. In $\triangle OAB$, let $\vec{OA}=\mathbf{a}$ and $\vec{OB}=\mathbf{b}$. Point D is on OB such that $\vec{OD}=\mu\vec{OB}$, and point P lies on the line segment AD. Which of the following is a valid vector representation of $\vec{OP}$ for some scalar $t$ where $0 \le t \le 1$?
    Answer: $\vec{OP} = (1-t)\mathbf{a} + t\mu\mathbf{b}$ The position vector of any point P on the line segment connecting points A and D can be expressed using the section formula: $\vec{OP} = (1-t)\vec{OA} + t\vec{OD}$ for a scalar $t$ in the interval $[0, 1]$. We are given $\vec{OA} = \mathbf{a}$ and $\vec{OD} = \mu\vec{OB} = \mu\mathbf{b}$. Substituting these into the formula gives $\vec{OP} = (1-t)\mathbf{a} + t(\mu\mathbf{b}) = (1-t)\mathbf{a} + t\mu\mathbf{b}$.
18. Let $\mathbf{u}$ and $\mathbf{v}$ be non-zero vectors. A student is proving a geometric property and arrives at the equation $(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = 0$. What is the geometric interpretation of this result?
    Answer: The vectors $\mathbf{u}$ and $\mathbf{v}$ have the same magnitude. Expanding the dot product gives: $(\mathbf{u} - \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot \mathbf{v} - \mathbf{v} \cdot \mathbf{u} - \mathbf{v} \cdot \mathbf{v} = |\mathbf{u}|^2 - |\mathbf{v}|^2$. Setting this equal to zero gives $|\mathbf{u}|^2 - |\mathbf{v}|^2 = 0$, which implies $|\mathbf{u}|^2 = |\mathbf{v}|^2$. Since magnitudes are non-negative, this means $|\mathbf{u}| = |\mathbf{v}|$. Geometrically, this means the vectors $\mathbf{u}$ and $\mathbf{v}$ have the same length (magnitude).
19. Let ABCD be an arbitrary quadrilateral. Let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA, respectively. A vector proof is used to show that PQRS is a parallelogram. The proof involves showing $\vec{PQ} = \vec{SR}$. Which of the following correctly expresses $\vec{PQ}$?
    Answer: $\frac{1}{2}(\mathbf{c}-\mathbf{a})$ Let the position vectors of the vertices be $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$. The position vectors of the midpoints are $\mathbf{p} = \frac{\mathbf{a}+\mathbf{b}}{2}$ and $\mathbf{q} = \frac{\mathbf{b}+\mathbf{c}}{2}$. The vector $\vec{PQ}$ is given by $\mathbf{q} - \mathbf{p}$. Substituting the expressions for $\mathbf{p}$ and $\mathbf{q}$ gives: $\vec{PQ} = \frac{\mathbf{b}+\mathbf{c}}{2} - \frac{\mathbf{a}+\mathbf{b}}{2} = \frac{\mathbf{b}+\mathbf{c}-\mathbf{a}-\mathbf{b}}{2} = \frac{\mathbf{c}-\mathbf{a}}{2}$. This is $\frac{1}{2}\vec{AC}$.
20. Let OABC be a parallelogram with $\vec{OA} = \mathbf{a}$ and $\vec{OC} = \mathbf{c}$. A vector proof is used to show that if the parallelogram is a rhombus, its diagonals are perpendicular. Which calculation correctly represents this proof?
    Answer: $(\mathbf{a}+\mathbf{c}) \cdot (\mathbf{c}-\mathbf{a}) = |\mathbf{c}|^2 - |\mathbf{a}|^2 = 0$ since for a rhombus $|\mathbf{a}|=|\mathbf{c}|$ The diagonals of the parallelogram are $\vec{OB} = \mathbf{a}+\mathbf{c}$ and $\vec{AC} = \mathbf{c}-\mathbf{a}$. To prove the diagonals are perpendicular, we must show their dot product is zero. The dot product is $(\mathbf{a}+\mathbf{c}) \cdot (\mathbf{c}-\mathbf{a}) = \mathbf{a}\cdot\mathbf{c} - \mathbf{a}\cdot\mathbf{a} + \mathbf{c}\cdot\mathbf{c} - \mathbf{c}\cdot\mathbf{a} = |\mathbf{c}|^2 - |\mathbf{a}|^2$. For a rhombus, the adjacent sides have equal length, so $|\mathbf{a}| = |\mathbf{c}|$. Therefore, $|\mathbf{c}|^2 - |\mathbf{a}|^2 = 0$, which completes the proof.

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