Vectors and geometry: properties of the scalar product, the Cauchy-Schwarz inequality and its equality case, the centroid and medians, concurrency of altitudes, perpendicular bisectors and the circumcentre, the angle bisector and the point where it meets the opposite side (new emphasis in the 2024 Extension 2 syllabus)

Syllabus dot point

“Prove and apply the Cauchy-Schwarz inequality $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|$ , and use vectors to describe and prove properties of medians, altitudes, perpendicular bisectors and angle bisectors of a triangle in two and three dimensions”

HSCMaths Extension 2Further Work with Vectors (MEX-V1)13 min readUpdated 7 Oct 2026

Quick answer

Cauchy-Schwarz: $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|$ , with equality for parallel vectors (prove it from $|\mathbf{a} - t\mathbf{b}|^2 \geq 0$ ). Medians meet at $\frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ in the ratio $2 : 1$ ; altitudes satisfy a zero scalar product with the opposite side and are concurrent; the perpendicular bisector of $BC$ is $\left(\mathbf{p} - \frac{1}{2}(\mathbf{b} + \mathbf{c})\right) \cdot (\mathbf{c} - \mathbf{b}) = 0$ , and the three meet at the circumcentre; the angle bisector at $A$ has direction $\hat{\mathbf{u}} + \hat{\mathbf{v}}$ and meets $BC$ in the ratio $AB : AC$ . New emphasis in the 2024 Extension 2 syllabus.

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What this dot point is asking
The answer
Practice questions

What this dot point is asking

NESA's Year 12 Further work with vectors in the Mathematics Extension 2 11-12 Syllabus (2024), first examined in the 2027 HSC, has a sub-heading Vectors and geometry. Under it students "Prove and use the Cauchy-Schwarz inequality for vectors", examine the properties of the scalar product (commutative, distributive, scalar multiples, unit vector results, orthogonality), "Define the medians, altitudes, perpendicular bisectors and angle bisectors of a triangle and recognise these definitions in vector proofs" and "Solve problems and prove geometric results in two dimensions and three dimensions using vectors". Our page on vector geometric proofs covers parallelograms, midpoints and rhombus diagonals; this page covers the Cauchy-Schwarz inequality and the four types of triangle line.

Note

Think of two sticks joined at a hinge. The scalar product measures how much one stick points along the other. If they point exactly the same way the "agreement" is as large as it can be, which is the product of their lengths. If the hinge opens up, the agreement shrinks, and at a right angle it is zero. So the agreement can never exceed the product of the lengths. That is the Cauchy-Schwarz inequality in a sentence. The rest of the page uses vectors to find four special lines in a triangle: lines to midpoints (medians), perpendicular lines from a vertex (altitudes), lines that split an angle in half (angle bisectors) and lines at right angles through the middle of a side (perpendicular bisectors).

The answer

Key fact

Properties: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$ , $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$ , $(k\mathbf{a}) \cdot \mathbf{b} = k(\mathbf{a} \cdot \mathbf{b})$ , $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$ , and $\mathbf{a} \cdot \mathbf{b} = 0$ for non-zero vectors means perpendicular.
Cauchy-Schwarz: $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|$ , with equality exactly when $\mathbf{a}$ and $\mathbf{b}$ are parallel.
Triangle $ABC$ with position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$

Median from $A$ goes to $\mathbf{m} = \frac{1}{2}(\mathbf{b} + \mathbf{c})$ . The medians meet at the centroid $\mathbf{g} = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ , which divides each median $2 : 1$ from the vertex.
Altitude from $A$ satisfies $(\mathbf{h} - \mathbf{a}) \cdot (\mathbf{c} - \mathbf{b}) = 0$ .
Perpendicular bisector of $BC$ is the set of points $\mathbf{p}$ with $\left(\mathbf{p} - \frac{1}{2}(\mathbf{b} + \mathbf{c})\right) \cdot (\mathbf{c} - \mathbf{b}) = 0$ , equivalently $|\mathbf{p} - \mathbf{b}| = |\mathbf{p} - \mathbf{c}|$ . The three perpendicular bisectors meet at the circumcentre, which is equidistant from $A$ , $B$ and $C$ .
Angle bisector at $A$ has direction $\hat{\mathbf{u}} + \hat{\mathbf{v}}$ (sum of unit vectors along $AB$ and $AC$ ), and meets $BC$ at the point $\mathbf{d} = \dfrac{|AC|\,\mathbf{b} + |AB|\,\mathbf{c}}{|AB| + |AC|}$ .

Why Cauchy-Schwarz is true

Using angles, $|\mathbf{a} \cdot \mathbf{b}| = |\mathbf{a}||\mathbf{b}||\cos\theta| \leq |\mathbf{a}||\mathbf{b}|$ because $|\cos\theta| \leq 1$ . The angle-free proof (see the third practice question) starts from $|\mathbf{a} - t\mathbf{b}|^2 \geq 0$ and chooses $t = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}$ . The second proof is more useful because it works in any number of dimensions, and it is the same idea as the algebraic version you meet in proving inequalities.

Applying Cauchy-Schwarz

Write the given expression as a scalar product. For example, $x + 2y + 2z = (x, y, z) \cdot (1, 2, 2)$ , so $(x + 2y + 2z)^2 \leq 9(x^2 + y^2 + z^2)$ . The maximum of a linear expression on a sphere (or minimum of a sum of squares on a plane) then occurs where the two vectors are parallel, which tells you the point.

Medians and the centroid

The midpoint of $BC$ has position vector $\frac{1}{2}(\mathbf{b} + \mathbf{c})$ . The point $2/3$ of the way along the median from $A$ is

\mathbf{a} + \frac{2}{3}\left( \frac{\mathbf{b} + \mathbf{c}}{2} - \mathbf{a} \right) = \frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c}),

a formula that is symmetric in $A$ , $B$ and $C$ . Therefore the same point lies on all three medians, which proves they are concurrent.

Altitudes and concurrency

An altitude is a line through a vertex perpendicular to the opposite side, so the scalar product of the direction from the vertex with the side direction is zero. Placing the origin at one vertex simplifies the algebra, as in the proof in the practice questions.

Perpendicular bisectors

The perpendicular bisector of $BC$ passes through the midpoint $\frac{1}{2}(\mathbf{b} + \mathbf{c})$ and is perpendicular to $\overrightarrow{BC}$ , so a point $\mathbf{p}$ lies on it exactly when $\left(\mathbf{p} - \frac{1}{2}(\mathbf{b} + \mathbf{c})\right) \cdot (\mathbf{c} - \mathbf{b}) = 0$ . Expanding gives $\mathbf{p} \cdot (\mathbf{c} - \mathbf{b}) = \frac{1}{2}\left(|\mathbf{c}|^2 - |\mathbf{b}|^2\right)$ , which is the same as $|\mathbf{p} - \mathbf{b}| = |\mathbf{p} - \mathbf{c}|$ : every point on the line is equally far from $B$ and $C$ . If a point lies on the perpendicular bisectors of $AB$ and $BC$ it is equidistant from all three vertices, so it also lies on the perpendicular bisector of $AC$ . The three bisectors are therefore concurrent, at the circumcentre.

Angle bisectors

The sum of two unit vectors is the diagonal of a rhombus, which bisects the angle between them. So the bisector at $A$ has direction $\frac{\overrightarrow{AB}}{|AB|} + \frac{\overrightarrow{AC}}{|AC|}$ . The angle bisector theorem says it meets $BC$ at the point dividing $BC$ in the ratio $AB : AC$ , which gives the formula for $\mathbf{d}$ above.

How exam questions use this

"Use Cauchy-Schwarz to show ..." Choose two vectors whose scalar product is the expression, then square.
"Prove the medians (altitudes) are concurrent." Use position vectors, a clever origin and scalar products.
"Find the circumcentre." Write two perpendicular bisector conditions and solve them together.
"Find the point where the bisector meets the opposite side." Use the ratio $AB : AC$ .
"Show the vector $\mathbf{w}$ bisects the angle." Show it makes equal cosines with the two sides.

Worked examples

A minimum under a linear constraint

Find the minimum value of $x^2 + y^2 + z^2$ given that $x + 2y + 2z = 18$ , and the point where it occurs.

Step 1: apply Cauchy-Schwarz

With $\mathbf{a} = (x, y, z)$ and $\mathbf{b} = (1, 2, 2)$ , $(x + 2y + 2z)^2 \leq 9(x^2 + y^2 + z^2)$ , so $18^2 \leq 9(x^2 + y^2 + z^2)$ .

Step 2: the minimum

$x^2 + y^2 + z^2 \geq \frac{324}{9} = 36$ .

Step 3: the point

Equality needs $(x, y, z) = \lambda(1, 2, 2)$ . Then $\lambda + 4\lambda + 4\lambda = 18$ gives $\lambda = 2$ and the point is $(2, 4, 4)$ , where $4 + 16 + 16 = 36$ .

Marker's note: one mark for the setup, one for the minimum, one for the point and the check.

The altitude through a vertex in coordinates

In the triangle with $A(0, 0)$ , $B(4, 0)$ and $C(1, 3)$ , find the orthocentre (where the altitudes meet).

Step 1: altitude from $C$

It is perpendicular to $AB$ , which runs along the $x$ -axis, so it is the line $x = 1$ .

Step 2: altitude from $A$

It is perpendicular to $\overrightarrow{BC} = (-3, 3)$ . A point $(x, y)$ on it satisfies $(x, y) \cdot (-3, 3) = 0$ , so $y = x$ .

Step 3: intersection

At $x = 1$ , $y = 1$ , so $H = (1, 1)$ . Check the altitude from $B$ : $\overrightarrow{BH} \cdot \overrightarrow{AC} = (-3, 1) \cdot (1, 3) = 0$ .

Marker's note: one mark for each altitude, one for the orthocentre and one for the check with the third altitude.

The perpendicular bisectors and the circumcentre

In the same triangle, $A(0, 0)$ , $B(4, 0)$ and $C(1, 3)$ , find the circumcentre (where the perpendicular bisectors meet).

Step 1: bisector of $AB$

The midpoint of $AB$ is $(2, 0)$ and $\overrightarrow{AB} = (4, 0)$ , so $(x - 2, y) \cdot (4, 0) = 0$ gives $x = 2$ .

Step 2: bisector of $AC$

The midpoint of $AC$ is $\left(\frac{1}{2}, \frac{3}{2}\right)$ and $\overrightarrow{AC} = (1, 3)$ , so $\left(x - \frac{1}{2}\right) + 3\left(y - \frac{3}{2}\right) = 0$ , which simplifies to $x + 3y = 5$ .

Step 3: intersection and check

At $x = 2$ , $3y = 3$ so $y = 1$ and the circumcentre is $O = (2, 1)$ . Check: $|OA|^2 = 4 + 1 = 5$ , $|OB|^2 = 4 + 1 = 5$ and $|OC|^2 = 1 + 4 = 5$ , so $O$ is equidistant from all three vertices and lies on the third bisector as well.

Marker's note: one mark for each bisector, one for the circumcentre and one for the equal-distance check.

Common traps

Applying Cauchy-Schwarz without a clear choice of vectors: State what $\mathbf{a}$ and $\mathbf{b}$ are before using the inequality.
Forgetting the absolute value or the square: The inequality is $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|$ ; the scalar product itself can be negative, which is why squaring is often the cleanest form.
Claiming equality without checking parallelism: Equality needs the vectors parallel (and in the same direction for the maximum of a linear expression).
Using the median formula with the wrong denominator: The centroid is $\frac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})$ , not $\frac{1}{2}$ or $\frac{1}{4}$ .
Mixing up the bisector ratio: The bisector meets $BC$ so that $BD : DC = AB : AC$ . The larger side gets the larger share of $BC$ , so $\mathbf{d}$ weights $\mathbf{b}$ by $|AC|$ and $\mathbf{c}$ by $|AB|$ .

Exam technique

In a proof, say which vectors you take as position vectors and why. Choosing the origin at a vertex or at the centre of a figure saves a great deal of algebra. For Cauchy-Schwarz questions, write the expression as a scalar product on the first line. In a "hence" question, the previous inequality or proof is usually the engine for the next part.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

For

\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}

and

\mathbf{b} = 4\mathbf{i} - \mathbf{k}

, verify the Cauchy-Schwarz inequality

|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|

Show worked solution →

Scalar product: $\mathbf{a} \cdot \mathbf{b} = (1)(4) + (-2)(0) + (3)(-1) = 4 - 3 = 1$ .
Magnitudes: $|\mathbf{a}| = \sqrt{1 + 4 + 9} = \sqrt{14}$ and $|\mathbf{b}| = \sqrt{16 + 0 + 1} = \sqrt{17}$ , so $|\mathbf{a}||\mathbf{b}| = \sqrt{238} \approx 15.4$ .
Compare: $|1| \leq \sqrt{238}$ , so the inequality holds, and it is far from equality because $\mathbf{a}$ and $\mathbf{b}$ are nearly perpendicular.

Marker's note: one mark for the scalar product, one for the magnitudes and the comparison.

core2 marks

Show that for

\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}

and

\mathbf{b} = 2\mathbf{i} + 4\mathbf{j} + 4\mathbf{k}

the Cauchy-Schwarz inequality holds with equality, and explain what this tells you about the vectors.

Show worked solution →

Compute both sides. $\mathbf{a} \cdot \mathbf{b} = 2 + 8 + 8 = 18$ . Also $|\mathbf{a}| = 3$ and $|\mathbf{b}| = \sqrt{4 + 16 + 16} = 6$ , so $|\mathbf{a}||\mathbf{b}| = 18$ .

Both sides equal $18$ , so there is equality.

Interpretation. Equality occurs exactly when the vectors are parallel. Here $\mathbf{b} = 2\mathbf{a}$ , so the angle between them is $0^\circ$ and $\cos\theta = 1$ .

Marker's note: one mark for the equality, one for the link to parallel vectors.

core4 marks

Prove the Cauchy-Schwarz inequality

|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|

for vectors

\mathbf{a}

and

\mathbf{b}

with

\mathbf{b} \neq \mathbf{0}

, without using the angle between them, and state when equality occurs.

Show worked solution →

A square is never negative. For any real number $t$ , $|\mathbf{a} - t\mathbf{b}|^2 \geq 0$ . Expanding with $\mathbf{v} \cdot \mathbf{v} = |\mathbf{v}|^2$ ,

|\mathbf{a}|^2 - 2t(\mathbf{a} \cdot \mathbf{b}) + t^2|\mathbf{b}|^2 \geq 0.

Choose $t$ to make the left side small. Take $t = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}$ :

|\mathbf{a}|^2 - \frac{(\mathbf{a} \cdot \mathbf{b})^2}{|\mathbf{b}|^2} \geq 0.

Rearrange. Multiplying by $|\mathbf{b}|^2 > 0$ gives $(\mathbf{a} \cdot \mathbf{b})^2 \leq |\mathbf{a}|^2|\mathbf{b}|^2$ , and taking positive square roots gives $|\mathbf{a} \cdot \mathbf{b}| \leq |\mathbf{a}||\mathbf{b}|$ .

Equality. It holds exactly when $|\mathbf{a} - t\mathbf{b}|^2 = 0$ , that is $\mathbf{a} = t\mathbf{b}$ , so when the vectors are parallel. (If $\mathbf{b} = \mathbf{0}$ both sides are $0$ .)

Marker's note: one mark for $|\mathbf{a} - t\mathbf{b}|^2 \geq 0$ expanded, one for the choice of $t$ , one for the rearrangement, one for the equality case.

core4 marks

Use the Cauchy-Schwarz inequality with suitable vectors to show that

(x + 2y + 2z)^2 \leq 9\left( x^2 + y^2 + z^2 \right)

for all real

x

y

z

. Hence find the maximum value of

x + 2y + 2z

when

x^2 + y^2 + z^2 = 1

, and the values of

x

y

z

at which it occurs.

Show worked solution →

Choose the vectors. Let $\mathbf{a} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}$ . Then $\mathbf{a} \cdot \mathbf{b} = x + 2y + 2z$ and $|\mathbf{b}|^2 = 1 + 4 + 4 = 9$ .

Apply the inequality in its squared form: $(\mathbf{a} \cdot \mathbf{b})^2 \leq |\mathbf{a}|^2|\mathbf{b}|^2$ , that is

(x + 2y + 2z)^2 \leq 9\left( x^2 + y^2 + z^2 \right).

Maximum. If $x^2 + y^2 + z^2 = 1$ then $(x + 2y + 2z)^2 \leq 9$ , so $x + 2y + 2z \leq 3$ .

Where it occurs. Equality needs $\mathbf{a}$ parallel to $\mathbf{b}$ , in the same direction, so $\mathbf{a} = \lambda\mathbf{b}$ with $\lambda > 0$ and $|\mathbf{a}| = 1$ . Then $\lambda = \frac{1}{3}$ , giving $(x, y, z) = \left( \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right)$ , and indeed $\frac{1}{3} + \frac{4}{3} + \frac{4}{3} = 3$ .

Marker's note: one mark for the vectors, one for the squared inequality, one for the maximum $3$ , one for the point of equality.

exam3 marks

The triangle

ABC

has vertices

A(1, 2, 3)

B(5, -2, 1)

and

C(3, 6, 5)

. Let

M

be the midpoint of

BC

. Find

M

and show that the point

G

with position vector

\tfrac{1}{3}(\mathbf{a} + \mathbf{b} + \mathbf{c})

lies on the median

AM

, dividing it in the ratio

2 : 1

Show worked solution →

Midpoint: $\mathbf{m} = \frac{1}{2}(\mathbf{b} + \mathbf{c}) = \frac{1}{2}(8, 4, 6) = (4, 2, 3)$ , so $M = (4, 2, 3)$ .
The point $G$: $\mathbf{a} + \mathbf{b} + \mathbf{c} = (9, 6, 9)$ , so $\mathbf{g} = (3, 2, 3)$ .
Show it lies on $AM$ at the two-thirds mark: $\overrightarrow{AM} = (3, 0, 0)$ and $\overrightarrow{AG} = (2, 0, 0) = \frac{2}{3}\overrightarrow{AM}$ . So $G$ lies on $AM$ with $AG : GM = 2 : 1$ .

In general, $\mathbf{a} + \frac{2}{3}(\mathbf{m} - \mathbf{a}) = \frac{1}{3}\mathbf{a} + \frac{1}{3}(\mathbf{b} + \mathbf{c})$ , and this expression is symmetric in $A$ , $B$ and $C$ , so $G$ lies on all three medians.

Marker's note: one mark for $M$ , one for $G$ , one for the $2 : 1$ conclusion.

exam4 marks

Prove using vectors that the three altitudes of a triangle

ABC

are concurrent.

Show worked solution →

Choose the origin at $C$ so that $\overrightarrow{CA} = \mathbf{a}$ and $\overrightarrow{CB} = \mathbf{b}$ . Then $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ .

Let $H$ be the intersection of the altitudes from $A$ and $B$ , with position vector $\mathbf{h}$ .

The altitude from $A$ is perpendicular to $CB$ : $(\mathbf{h} - \mathbf{a}) \cdot \mathbf{b} = 0$ , so $\mathbf{h} \cdot \mathbf{b} = \mathbf{a} \cdot \mathbf{b}$ .

The altitude from $B$ is perpendicular to $CA$ : $(\mathbf{h} - \mathbf{b}) \cdot \mathbf{a} = 0$ , so $\mathbf{h} \cdot \mathbf{a} = \mathbf{a} \cdot \mathbf{b}$ .

Subtract. $\mathbf{h} \cdot \mathbf{a} - \mathbf{h} \cdot \mathbf{b} = 0$ , so $\mathbf{h} \cdot (\mathbf{a} - \mathbf{b}) = 0$ . This says $\overrightarrow{CH} \perp \overrightarrow{AB}$ .

Conclude. $H$ lies on the line through $C$ perpendicular to $AB$ , which is the third altitude. The three altitudes meet at $H$ .

Marker's note: one mark for the choice of origin and the two altitude conditions, one for each of the two equations, one for the subtraction and conclusion.

exam4 marks

The vectors

\mathbf{u} = 2\mathbf{i} - \mathbf{j} + 2\mathbf{k}

and

\mathbf{v} = -4\mathbf{i} + 3\mathbf{k}

are the directions of two sides of a triangle meeting at a vertex. Find a vector along the bisector of the angle between them with integer components, and show that it makes equal angles with

\mathbf{u}

and

\mathbf{v}

Show worked solution →

Use unit vectors. $|\mathbf{u}| = 3$ and $|\mathbf{v}| = 5$ , so $\hat{\mathbf{u}} = \frac{1}{3}(2, -1, 2)$ and $\hat{\mathbf{v}} = \frac{1}{5}(-4, 0, 3)$ .

The sum of two unit vectors bisects the angle between them (they form a rhombus whose diagonal bisects the angle):

\hat{\mathbf{u}} + \hat{\mathbf{v}} = \left( \frac{2}{3} - \frac{4}{5}, -\frac{1}{3}, \frac{2}{3} + \frac{3}{5} \right) = \left( -\frac{2}{15}, -\frac{1}{3}, \frac{19}{15} \right).

Scale by $15$ : $\mathbf{w} = (-2, -5, 19)$ .

Check the angles. $\mathbf{w} \cdot \mathbf{u} = -4 + 5 + 38 = 39$ and $|\mathbf{w}| = \sqrt{4 + 25 + 361} = \sqrt{390}$ , so $\cos\theta_1 = \frac{39}{3\sqrt{390}} = \frac{13}{\sqrt{390}}$ . Also $\mathbf{w} \cdot \mathbf{v} = 8 + 57 = 65$ , so $\cos\theta_2 = \frac{65}{5\sqrt{390}} = \frac{13}{\sqrt{390}}$ . The cosines are equal, and the common angle is about $48.8^\circ$ .

Marker's note: one mark for the unit vectors, one for the sum, one for the integer vector, one for the matching cosines.

exam4 marks

A triangle has vertices

A(0, 0, 0)

B(3, 0, 4)

and

C(0, 12, 0)

. The bisector of angle

BAC

meets

BC

D

. Using the fact that

D

divides

BC

in the ratio

AB : AC

, find

D

and verify that

AD

bisects angle

BAC

Show worked solution →

Lengths. $AB = \sqrt{9 + 0 + 16} = 5$ and $AC = 12$ . So $BD : DC = 5 : 12$ .

Section formula. $\mathbf{d} = \dfrac{12\mathbf{b} + 5\mathbf{c}}{17} = \dfrac{12(3, 0, 4) + 5(0, 12, 0)}{17} = \dfrac{(36, 60, 48)}{17}$ .

So $D = \left( \frac{36}{17}, \frac{60}{17}, \frac{48}{17} \right)$ .

Verify. First, $\mathbf{b} \cdot \mathbf{c} = 0$ , so angle $BAC = 90^\circ$ and each half should be $45^\circ$ . Then $\mathbf{d} \cdot \mathbf{b} = \frac{108 + 192}{17} = \frac{300}{17}$ and $|\mathbf{d}| = \frac{\sqrt{36^2 + 60^2 + 48^2}}{17} = \frac{12\sqrt{50}}{17} = \frac{60\sqrt{2}}{17}$ , since $36^2 + 60^2 + 48^2 = 1296 + 3600 + 2304 = 7200$ .

\cos\angle BAD = \frac{300/17}{5 \times 60\sqrt{2}/17} = \frac{1}{\sqrt{2}},

so $\angle BAD = 45^\circ$ . Similarly $\mathbf{d} \cdot \mathbf{c} = \frac{720}{17}$ and $\cos\angle DAC = \frac{720/17}{12 \times 60\sqrt{2}/17} = \frac{1}{\sqrt{2}}$ , so $\angle DAC = 45^\circ$ .

Marker's note: one mark for the lengths and ratio, one for $\mathbf{d}$ , one for each angle check.

Practise this

Flashcards Maths Extension 2 deck

Sources & how we know this

Mathematics Extension 2 11-12 Syllabus (2024): Year 12 Further work with vectors — NESA (2024)
Mathematics Extension 2 11-12 Syllabus (2024): course overview and implementation — NESA (2024)
Mathematics Extension 2 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 2 HSC exam papers — NESA

What this dot point is asking

The answer

Why Cauchy-Schwarz is true

Applying Cauchy-Schwarz

Medians and the centroid

Altitudes and concurrency

Perpendicular bisectors

Angle bisectors

How exam questions use this

A minimum under a linear constraint

Step 1: apply Cauchy-Schwarz

Step 2: the minimum

Step 3: the point

The altitude through a vertex in coordinates

Step 1: altitude from CCC

Step 2: altitude from AAA

Step 3: intersection

The perpendicular bisectors and the circumcentre

Step 1: bisector of ABABAB

Step 2: bisector of ACACAC

Step 3: intersection and check

Practice questions

Practise this

Sources & how we know this

Step 1: altitude from $C$

Step 2: altitude from $A$

Step 1: bisector of $AB$

Step 2: bisector of $AC$