IB Mathematics: Analysis and Approaches Quiz: Vector Proofs

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Vector ProofsQuestion 1 of 20

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 $0 \le t \le 1$ ?

\vec{OP} = (1-t)\mathbf{a} + t\mu\mathbf{b}

\vec{OP} = t\mathbf{a} + (1-t)\mu\mathbf{a}

\vec{OP} = (1-t)\mathbf{a} + t\mathbf{b}

\vec{OP} = (1-t)\mu\mathbf{b} + t\mathbf{b}

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Vector Proofs

Practice Vector Proofs in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Vector Proofs, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$\vec{OP} = (1-t)\mathbf{a} + t\mu\mathbf{b}$ (correct answer)
$\vec{OP} = t\mathbf{a} + (1-t)\mu\mathbf{a}$
$\vec{OP} = (1-t)\mathbf{a} + t\mathbf{b}$
$\vec{OP} = (1-t)\mu\mathbf{b} + t\mathbf{b}$

Explanation: 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}

Question 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?

ABCD is a parallelogram.
ABCD is a trapezoid. (correct answer)
ABCD is a rhombus.
The area of $\triangle AMB$ is equal to the area of $\triangle CMD$ .

Explanation: 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

Question 3

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 $\vec{OP}$ ?

$\vec{OP} = \frac{2\mathbf{a}+\mathbf{b}}{3}$
$\vec{OP} = \frac{\mathbf{a}+2\mathbf{b}}{3}$ (correct answer)
$\vec{OP} = \mathbf{a} + \frac{2}{3}\mathbf{b}$
$\vec{OP} = \mathbf{a} + \frac{1}{3}(\mathbf{a}-\mathbf{b})$

Explanation: 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}

Question 4

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 $\vec{AN}$ ?

$\vec{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$ and $\vec{AN} = \frac{1}{2}\mathbf{a}-\mathbf{b}$
$\vec{OM} = \mathbf{b}-\mathbf{a}$ and $\vec{AN} = \frac{1}{2}\mathbf{b}-\mathbf{a}$
$\vec{OM} = \frac{1}{2}(\mathbf{a}+\mathbf{b})$ and $\vec{AN} = \frac{1}{2}\mathbf{b}-\mathbf{a}$ (correct answer)
$\vec{OM} = \mathbf{a}+\mathbf{b}$ and $\vec{AN} = \mathbf{a}-\frac{1}{2}\mathbf{b}$

Explanation: 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}

Question 5

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?

$\mathbf{h} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{b}$ and $\mathbf{h} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b}$
$\mathbf{h} \cdot (\mathbf{b}-\mathbf{a}) = 0$ and $(\mathbf{h}-\mathbf{a}) \cdot (\mathbf{h}-\mathbf{b}) = 0$
$\mathbf{h} \cdot \mathbf{a} = 0$ and $\mathbf{h} \cdot \mathbf{b} = 0$
$(\mathbf{h}-\mathbf{a}) \cdot \mathbf{b} = 0$ and $(\mathbf{h}-\mathbf{b}) \cdot \mathbf{a} = 0$ (correct answer)

Explanation: 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}

\mathbf{h}-\mathbf{a}

, and the vector

\vec{BC}

\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}

\mathbf{h}-\mathbf{b}

, and

\vec{AC}

\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.

Question 6

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?

$(\frac{\mathbf{a}+\mathbf{b}}{2}) \cdot (\mathbf{b}-\mathbf{a}) = 0$ (correct answer)
$\mathbf{a} \cdot \mathbf{b} = 0$
$\mathbf{a} = -\mathbf{b}$
$(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a} = 0$

Explanation: 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.

Question 7

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?

$\mathbf{b}-\mathbf{a} = k(\mathbf{c}-\mathbf{a})$ (correct answer)
$\mathbf{a} \cdot \mathbf{b} = k(\mathbf{b} \cdot \mathbf{c})$
$|\mathbf{b}-\mathbf{a}| = k|\mathbf{c}-\mathbf{b}|$
$\mathbf{c} = k(\mathbf{a}+\mathbf{b})$

Explanation: 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}

\mathbf{b}-\mathbf{a}

, and the vector

\vec{AC}

\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.

Question 8

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?

Show that $(\mathbf{c}-\mathbf{a}) \cdot \mathbf{d} = 0$
Show that $\mathbf{c} \cdot \mathbf{a} = t (\mathbf{a} \cdot \mathbf{d})$ for some scalar $t$
Show that $\mathbf{c}-\mathbf{a}$ is a scalar multiple of $\mathbf{d}$ (correct answer)
Show that $|\mathbf{c}-\mathbf{a}| = |t\mathbf{d}|$ for some scalar $t$

Explanation: 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}

Question 9

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 $\mathbf{c}$ ?

$|\mathbf{b}+\mathbf{c}|^2 + \frac{1}{2}|\mathbf{c}-\mathbf{b}|^2$
$|\mathbf{b}|^2 + |\mathbf{c}|^2$
$2\left(|\frac{\mathbf{b}+\mathbf{c}}{2}|^2 + |\frac{\mathbf{c}-\mathbf{b}}{2}|^2\right)$ (correct answer)
$(\mathbf{b}+\mathbf{c})\cdot(\mathbf{b}+\mathbf{c}) + (\mathbf{c}-\mathbf{b})\cdot(\mathbf{c}-\mathbf{b})$

Explanation: 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}

\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)

Question 10

$\mathbf{b}-\mathbf{a} = k(\mathbf{c}-\mathbf{a})$ (correct answer)
$\mathbf{a} \cdot \mathbf{b} = k(\mathbf{b} \cdot \mathbf{c})$
$|\mathbf{b}-\mathbf{a}| = k|\mathbf{c}-\mathbf{b}|$
$\mathbf{c} = k(\mathbf{a}+\mathbf{b})$

Explanation: 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}

\mathbf{b}-\mathbf{a}

, and the vector

\vec{AC}

\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.

Question 11

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 $\vec{PQ}$ ?

$\frac{1}{2}(\mathbf{c}-\mathbf{a})$ (correct answer)
$\frac{1}{2}(\mathbf{b}+\mathbf{d})$
$\mathbf{c}-\mathbf{a}$
$\frac{1}{2}(\mathbf{d}-\mathbf{b})$

Explanation: 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}

Question 12

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?

The third pair of opposite edges, OC and AB, are also perpendicular. (correct answer)
The tetrahedron is regular (all edges are equal in length).
The faces OAB and OAC have equal area.
The altitude from O to the base ABC bisects an angle of the base.

Explanation: 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.

Question 13

$(\frac{\mathbf{a}+\mathbf{b}}{2}) \cdot (\mathbf{b}-\mathbf{a}) = 0$ (correct answer)
$\mathbf{a} \cdot \mathbf{b} = 0$
$\mathbf{a} = -\mathbf{b}$
$(\mathbf{a}+\mathbf{b}) \cdot \mathbf{a} = 0$

Explanation: 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.

Question 14

$\vec{OP} = \frac{2\mathbf{a}+\mathbf{b}}{3}$
$\vec{OP} = \frac{\mathbf{a}+2\mathbf{b}}{3}$ (correct answer)
$\vec{OP} = \mathbf{a} + \frac{2}{3}\mathbf{b}$
$\vec{OP} = \mathbf{a} + \frac{1}{3}(\mathbf{a}-\mathbf{b})$

Explanation: 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}

Question 15

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?

$(\mathbf{a}+\mathbf{c}) \cdot (\mathbf{c}-\mathbf{a}) = |\mathbf{c}|^2 - |\mathbf{a}|^2 = 0$ since for a rhombus $|\mathbf{a}|=|\mathbf{c}|$ (correct answer)
$|\mathbf{a}+\mathbf{c}|^2 = |\mathbf{a}|^2 + 2\mathbf{a}\cdot\mathbf{c} + |\mathbf{c}|^2 = 2|\mathbf{a}|^2(1+\cos\theta)$
$(\mathbf{a}+\mathbf{c}) \cdot (\mathbf{c}-\mathbf{a}) = |\mathbf{c}|^2 - |\mathbf{a}|^2 = 0$ since for a rhombus $\mathbf{a} \cdot \mathbf{c}=0$
$\mathbf{a} \cdot \mathbf{c} = |\mathbf{a}||\mathbf{c}|\cos\theta = 0$ because the diagonals are perpendicular

Explanation: 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.

Question 16

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?

$\mathbf{g} = \mathbf{a} + \frac{1}{3}(\frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a})$
$\mathbf{g} = \mathbf{a} + \frac{2}{3}(\frac{\mathbf{b}+\mathbf{c}}{2} - \mathbf{a})$ (correct answer)
$\mathbf{g} = \frac{\mathbf{b}+\mathbf{c}}{2} + \frac{2}{3}(\mathbf{a} - \frac{\mathbf{b}+\mathbf{c}}{2})$
$\mathbf{g} = \frac{1}{2}(\mathbf{a} + \frac{\mathbf{b}+\mathbf{c}}{2})$

Explanation: 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}

\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})

Question 17

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?

$\vec{AP} \cdot \vec{BP} = (\mathbf{p}-\mathbf{a}) \cdot (\mathbf{p}+\mathbf{a}) = |\mathbf{p}|^2 - |\mathbf{a}|^2 = r^2 - r^2 = 0$ (correct answer)
$\vec{AP} \cdot \vec{BP} = (\mathbf{a}-\mathbf{p}) \cdot (-\mathbf{a}-\mathbf{p}) = |\mathbf{a}|^2 + |\mathbf{p}|^2 = 2r^2$
$\vec{AP} \cdot \vec{BP} = (\mathbf{p}-\mathbf{a}) \cdot (\mathbf{p}+\mathbf{a}) = |\mathbf{p}|^2 + 2\mathbf{p}\cdot\mathbf{a} - |\mathbf{a}|^2 = 2\mathbf{p}\cdot\mathbf{a}$
$\vec{AP} \cdot \vec{BP} = (\mathbf{p}-\mathbf{a}) \cdot (\mathbf{p}-\mathbf{a}) = |\mathbf{p}-\mathbf{a}|^2 = 0$

Explanation: 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.

Question 18

$\frac{1}{2}(\mathbf{c}-\mathbf{a})$ (correct answer)
$\frac{1}{2}(\mathbf{b}+\mathbf{d})$
$\mathbf{c}-\mathbf{a}$
$\frac{1}{2}(\mathbf{d}-\mathbf{b})$

Explanation: 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}

Question 19

The third pair of opposite edges, OC and AB, are also perpendicular. (correct answer)
The tetrahedron is regular (all edges are equal in length).
The faces OAB and OAC have equal area.
The altitude from O to the base ABC bisects an angle of the base.

Explanation: 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.

Question 20

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?

$|\mathbf{u}| = |\mathbf{v}|$
$(\mathbf{u}+\mathbf{v}) \cdot (\mathbf{u}-\mathbf{v}) = 0$
$|\mathbf{u}+\mathbf{v}| = |\mathbf{u}-\mathbf{v}|$ (correct answer)
$\mathbf{u} \cdot \mathbf{v} = \frac{1}{2}|\mathbf{u}|^2$

Explanation: 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.