In parallelogram $ABCD$, let $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AD}$. Write $\vec{AC}$ and $\vec{DB}$ in terms of $\vec{u}$ and $\vec{v}$.

$\vec{AC} = \vec{u} + \vec{v}$ and $\vec{DB} = \vec{u} - \vec{v}$

A displacement of $3$ km east is followed by $4$ km north. Find the resultant displacement.

$5$ km at about $53.1^\circ$ north of east

In triangle $ABC$, $M$ is the midpoint of $BC$. Write $\vec{AM}$ in terms of $\vec{AB}$ and $\vec{AC}$.

$\vec{AM} = \frac{1}{2}\left(\vec{AB} + \vec{AC}\right)$

$|\vec{u}| = 6$, $|\vec{v}| = 8$ and the angle between them is $60^\circ$. Find $|\vec{u} + \vec{v}|$ and $|\vec{u} - \vec{v}|$.

$|\vec{u} + \vec{v}| = 2\sqrt{37} \approx 12.17$ and $|\vec{u} - \vec{v}| = 2\sqrt{13} \approx 7.21$

$\vec{a}$ has length $7$ and points east. $\vec{b}$ has length $5$ and points $40^\circ$ west of north. Find the length and direction of $\vec{a} - \vec{b}$.

About $10.9$ units, $20.6^\circ$ south of east

Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Adding and Subtracting Vectors

A vector has a length and a direction, and nothing else. Two arrows with the same length and the same direction are the same vector, wherever you draw them. That freedom to slide arrows around is what makes adding them work.

How It Works

Adding: tip to tail. To find $\vec{u} + \vec{v}$ , draw $\vec{u}$ , then start $\vec{v}$ at the tip of $\vec{u}$ . The sum, called the resultant, runs from the tail of $\vec{u}$ to the tip of $\vec{v}$ . Put the two tails together instead and you get the parallelogram law: the sum is the diagonal from the shared tail. Same answer, since the opposite sides of a parallelogram are equal vectors.

Subtracting: add the opposite. $-\vec{v}$ has the length of $\vec{v}$ and points the other way, so $\vec{u} - \vec{v} = \vec{u} + (-\vec{v})$ . With tails together, $\vec{u} - \vec{v}$ is the other diagonal of the parallelogram, from the tip of $\vec{v}$ to the tip of $\vec{u}$ .

Scalar multiples. $k\vec{v}$ stretches $\vec{v}$ by $|k|$ , and flips it when $k < 0$ . Vectors combine by the usual rules of algebra, so you can expand brackets and collect like terms.

Length and direction of a resultant. Draw the triangle tip to tail. If the angle between $\vec{u}$ and $\vec{v}$ (tails together) is $\theta$ , the angle inside the triangle is $180^\circ - \theta$ . The cosine law gives the length of the resultant and the sine law gives its direction. When the vectors are perpendicular, Pythagoras and tangent do the job.

Most mistakes come from using $\theta$ itself in the cosine law. Sketch the triangle every time and read the angle off the sketch. State a direction as an angle from a named reference, such as $20^\circ$ south of east.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

Easy
In parallelogram $ABCD$ , let $\vec{u} = \vec{AB}$ and $\vec{v} = \vec{AD}$ . Write $\vec{AC}$ and $\vec{DB}$ in terms of $\vec{u}$ and $\vec{v}$ .
Show solution
1. Tip to tail: $\vec{AC} = \vec{AB} + \vec{BC}$ , and $\vec{BC} = \vec{AD}$ since opposite sides are equal vectors. So $\vec{AC} = \vec{u} + \vec{v}$ .
2. $\vec{DB} = \vec{DA} + \vec{AB} = -\vec{v} + \vec{u}$ .
Answer: $\vec{AC} = \vec{u} + \vec{v}$ and $\vec{DB} = \vec{u} - \vec{v}$
Easy
A displacement of $3$ km east is followed by $4$ km north. Find the resultant displacement.
Show solution
1. The two vectors are perpendicular, so the length is $\sqrt{3^2 + 4^2} = 5$ km.
2. The angle from east is $\tan^{-1}\left(\frac{4}{3}\right) \approx 53.1^\circ$ .
Answer: $5$ km at about $53.1^\circ$ north of east
Medium
In triangle $ABC$ , $M$ is the midpoint of $BC$ . Write $\vec{AM}$ in terms of $\vec{AB}$ and $\vec{AC}$ .
Show solution
1. $\vec{AM} = \vec{AB} + \vec{BM} = \vec{AB} + \frac{1}{2}\vec{BC}$ .
2. $\vec{BC} = \vec{BA} + \vec{AC} = \vec{AC} - \vec{AB}$ .
3. So $\vec{AM} = \vec{AB} + \frac{1}{2}\vec{AC} - \frac{1}{2}\vec{AB}$ .
Answer: $\vec{AM} = \frac{1}{2}\left(\vec{AB} + \vec{AC}\right)$
Medium
$|\vec{u}| = 6$ , $|\vec{v}| = 8$ and the angle between them is $60^\circ$ . Find $|\vec{u} + \vec{v}|$ and $|\vec{u} - \vec{v}|$ .
Show solution
1. Tip to tail, the angle inside the triangle for $\vec{u} + \vec{v}$ is $180^\circ - 60^\circ = 120^\circ$ .
2. $|\vec{u} + \vec{v}|^2 = 36 + 64 - 2(6)(8)\cos 120^\circ = 100 + 48 = 148$ .
3. For $\vec{u} - \vec{v}$ , tails together, the difference closes the triangle opposite the $60^\circ$ angle: $|\vec{u} - \vec{v}|^2 = 100 - 96\cos 60^\circ = 52$ .
Answer: $|\vec{u} + \vec{v}| = 2\sqrt{37} \approx 12.17$ and $|\vec{u} - \vec{v}| = 2\sqrt{13} \approx 7.21$
Test level
$\vec{a}$ has length $7$ and points east. $\vec{b}$ has length $5$ and points $40^\circ$ west of north. Find the length and direction of $\vec{a} - \vec{b}$ .
Show solution
1. $-\vec{b}$ has length $5$ and points $40^\circ$ east of south, which is $50^\circ$ south of east.
2. Tip to tail, $\vec{a}$ then $-\vec{b}$ : the angle inside the triangle is $180^\circ - 50^\circ = 130^\circ$ .
3. Cosine law: $|\vec{a} - \vec{b}|^2 = 49 + 25 - 2(7)(5)\cos 130^\circ \approx 119.0$ , so $|\vec{a} - \vec{b}| \approx 10.91$ .
4. Sine law for the angle $\theta$ between $\vec{a}$ and the resultant: $\sin\theta = \frac{5\sin 130^\circ}{10.91} \approx 0.351$ , so $\theta \approx 20.6^\circ$ .
5. The resultant turns from east toward $-\vec{b}$ , which is south.
Answer: About $10.9$ units, $20.6^\circ$ south of east

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Martian Lab teaches this in the Geometric Vectors lesson of its MCV4U Geometric Vectors unit. Watch it, then take the Geometric Vectors practice tests to see if it stuck.

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MCV4U: Adding and Subtracting Vectors

How It Works

5 Worked Examples

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