Collinear points and orthogonal vectors in 2D

Unit 10 Collinear points and orthogonal vectors in 2D

Key unit competence: Apply properties of collinearity and orthogonality in solving problems involving vectors.

Key ideas

• Points $A$, $B$, $C$ are collinear if they lie on one line. This happens when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel: $\overrightarrow{AC}=k\overrightarrow{AB}$ for some number $k$.
• Test with components: $\overrightarrow{AB}=\left(\begin{array}{c}a\\ b\end{array}\right)$ and $\overrightarrow{AC}=\left(\begin{array}{c}c\\ d\end{array}\right)$ are parallel when $ad-bc=0$.
• The scalar (dot) product of $\vec{u}=\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)$ and $\vec{v}=\left(\begin{array}{c}{x}_2\\ y_2\end{array}\right)$ is $\vec{u}\cdot \vec{v}=x_1{x}_2+y_1{y}_2$.
• $\vec{u}$ and $\vec{v}$ (non-zero) are orthogonal (perpendicular) when $\vec{u}\cdot \vec{v}=0$.
• Applications: show a triangle has a right angle (dot product of two sides is zero), find unknowns that make points collinear or vectors perpendicular, identify rectangles and squares.

Formulas to know

Common misconceptions

• "The dot product is a vector." No: it is a number (a scalar).
• "Orthogonal vectors must have equal length." No: only the dot product matters.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Show that $P(-1,-3)$, $Q(1,1)$ and $R(3,5)$ are collinear.

   Show answer

   $\overrightarrow{PQ}=\left(\begin{array}{c}2\\ 4\end{array}\right)$, $\overrightarrow{PR}=\left(\begin{array}{c}4\\ 8\end{array}\right)=2\overrightarrow{PQ}$. Parallel with common point $P$: collinear.

   Common mistake: Forgetting to mention the common point.

2. Find the point $C$ on the x-axis such that $\angle ACB=90^{\circ }$, where $A(0,4)$ and $B(8,4)$.

   Show answer

   $\overrightarrow{CA}\cdot \overrightarrow{CB}=(-c)(8-c)+16=0$: $c^2-8c+16=0$, $c=4$. $C(4,0)$.

   Common mistake: Setting $\overrightarrow{AC}\cdot \overrightarrow{AB}=0$ (right angle at $A$ instead).

3. Find the dot product of $\vec{a}=3\vec{i}-2\vec{j}$ and $\vec{b}=4\vec{i}+6\vec{j}$ and say whether they are perpendicular.

   Show answer

   $12-12=0$: perpendicular.

   Common mistake: Multiplying $\vec{i}$ components with $\vec{j}$ components.

4. Write a vector orthogonal to $\left(\begin{array}{c}a\\ b\end{array}\right)$ and prove it.

   Show answer

   $\left(\begin{array}{c}-b\\ a\end{array}\right)$: dot product $-ab+ba=0$.

   Common mistake: Writing $\left(\begin{array}{c}b\\ a\end{array}\right)$.

5. Find $x$ so that $\left(\begin{array}{c}x\\ 3\end{array}\right)$ and $\left(\begin{array}{c}x\\ -12\end{array}\right)$ are orthogonal.

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   $x^2-36=0$, $x=\pm 6$.

   Common mistake: Giving only $x=6$.

6. Is $ABCD$ with $A(1,0)$, $B(4,1)$, $C(3,4)$, $D(0,3)$ a rectangle?

   Show answer

   $\overrightarrow{AB}=\left(\begin{array}{c}3\\ 1\end{array}\right)=\overrightarrow{DC}$ (parallelogram) and $\overrightarrow{AB}\cdot \overrightarrow{AD}=3(-1)+1(3)=0$: yes, a rectangle (in fact a square: both sides $\sqrt{10}$).

   Common mistake: Checking only one pair of sides.

7. Show that triangle $A(2,1)$, $B(6,3)$, $C(4,7)$ is right-angled and find where.

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   $\overrightarrow{BA}=\left(\begin{array}{c}-4\\ -2\end{array}\right)$, $\overrightarrow{BC}=\left(\begin{array}{c}-2\\ 4\end{array}\right)$: $8-8=0$, right angle at $B$.

   Common mistake: Testing only one vertex and stopping.

8. Find $k$ such that $A(1,k)$, $B(3,5)$, $C(7,13)$ are collinear.

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   $\overrightarrow{BC}=\left(\begin{array}{c}4\\ 8\end{array}\right)$, gradient 2; $\overrightarrow{AB}=\left(\begin{array}{c}2\\ 5-k\end{array}\right)$ must have $5-k=4$, $k=1$.

   Common mistake: Using a perpendicular condition.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $A(0,0)$, $B(3,1)$, $C(4,4)$, $D(1,3)$. What kind of quadrilateral is $ABCD$?	a. $\left(\begin{array}{c}8\\ -6\end{array}\right)$
   2. Find $y$ so that $A(1,2)$, $B(3,4)$, $C(5,y)$ are collinear.	b. 6
   3. Find a vector of length 10 perpendicular to $\left(\begin{array}{c}3\\ 4\end{array}\right)$.	c. $k=1$
   4. Find $k$ so that $A(1,k)$, $B(3,5)$ and $C(5,9)$ are collinear.	d. a rhombus (equal sides, perpendicular diagonals)

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   1 → d (a rhombus (equal sides, perpendicular diagonals)) · 2 → b (6) · 3 → a ($\left(\begin{array}{c}8\\ -6\end{array}\right)$) · 4 → c ($k=1$)

   • Step 1. $AB=BC=CD=DA=\sqrt{10}$.
     Step 2. Diagonals $\overrightarrow{AC}=\left(\begin{array}{c}4\\ 4\end{array}\right)$ and $\overrightarrow{BD}=\left(\begin{array}{c}-2\\ 2\end{array}\right)$: dot product $-8+8=0$. Since $AC\ne BD$, it is a rhombus, not a square.
   • The line has gradient 1, so $y=6$.
   • Step 1. $\left(\begin{array}{c}4\\ -3\end{array}\right)$ is perpendicular ($12-12=0$) and has length 5.
     Step 2. Double it for length 10: $\left(\begin{array}{c}8\\ -6\end{array}\right)$.
   • Step 1. $\overrightarrow{AB}=\left(\begin{array}{c}2\\ 5-k\end{array}\right)$ and $\overrightarrow{BC}=\left(\begin{array}{c}2\\ 4\end{array}\right)$ must be parallel.
     Step 2. $2\times 4-2(5-k)=0\Rightarrow 8-10+2k=0\Rightarrow k=1$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $90^{\circ }$  ·  $60^{\circ }$  ·  $A$  ·  acute
   1. If $\vec{u}\cdot \vec{v}>0$, the angle between them is .
   2. Triangle $A(0,0)$, $B(4,0)$, $C(0,3)$ has a right angle at .
   3. $\overrightarrow{AB}=\left(\begin{array}{c}2\\ -1\end{array}\right)$ and $\overrightarrow{BC}=\left(\begin{array}{c}1\\ 2\end{array}\right)$. The angle at $B$ is .
   Show answer

   1: acute · 2: $A$ · 3: $90^{\circ }$

   • The sign of the dot product follows the cosine of the angle.
   • $\overrightarrow{AB}\cdot \overrightarrow{AC}=0$.
   • $\overrightarrow{BA}\cdot \overrightarrow{BC}=(-2)(1)+(1)(2)=0$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. Find the angle between $\left(\begin{array}{c}2\\ 1\end{array}\right)$ and $\left(\begin{array}{c}1\\ 3\end{array}\right)$.	a. 0
   2. Find $\left(\begin{array}{c}-1\\ 4\end{array}\right)\cdot \left(\begin{array}{c}2\\ 0.5\end{array}\right)$.	b. $45^{\circ }$
   3. $|\vec{u}|=3$, $|\vec{v}|=4$ and $\vec{u}\perp \vec{v}$. Find $|\vec{u}+\vec{v}|$.	c. at $B$
   4. $A(1,2)$, $B(3,6)$, $C(5,5)$. Where is the right angle of triangle $ABC$?	d. 5

   Show answer

   1 → b ($45^{\circ }$) · 2 → a (0) · 3 → d (5) · 4 → c (at $B$)

   • Step 1. Dot product $=2+3=5$; lengths $\sqrt{5}$ and $\sqrt{10}$.
     Step 2. $\cos \theta =\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$, so $\theta =45^{\circ }$.
   • $-2+2=0$: they are orthogonal.
   • Step 1. $|\vec{u}+\vec{v}{|}^2=|\vec{u}{|}^2+2\vec{u}\cdot \vec{v}+|\vec{v}{|}^2=9+0+16$.
     Step 2. $|\vec{u}+\vec{v}|=\sqrt{25}=5$.
   • Step 1. $\overrightarrow{BA}=\left(\begin{array}{c}-2\\ -4\end{array}\right)$ and $\overrightarrow{BC}=\left(\begin{array}{c}2\\ -1\end{array}\right)$.
     Step 2. Dot product: $-4+4=0$, so the angle at $B$ is $90^{\circ }$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   4  ·  1  ·  2  ·  3
   1. The dot product of $\left(\begin{array}{c}a\\ b\end{array}\right)$ with itself is 25 and $a=3$. Then $b$ can be .
   2. Find $t$ so that $A(2,1)$, $B(4,t)$, $C(8,7)$ are collinear. →
   3. Find $k$ if $\left(\begin{array}{c}k\\ 2\end{array}\right)\parallel \left(\begin{array}{c}3\\ 6\end{array}\right)$. →
   Show answer

   1: 4 · 2: 3 · 3: 1

   • $9+b^2=25$.
   • Step 1. $\overrightarrow{AB}=\left(\begin{array}{c}2\\ t-1\end{array}\right)$.
     Step 2. $\overrightarrow{AC}=\left(\begin{array}{c}6\\ 6\end{array}\right)$: $12-6(t-1)=0$.
   • $6k-6=0$.

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