Dot Product Practice: Applications and Problem Solving
Malik drags a sled 20 feet across a flat driveway. When he pulls the rope at an angle, only part of his pull moves the sled forward. What if he pulls the rope straight up? Does that pull help at all? The dot product answers questions like this, and it also gives you angles, perpendicular tests, heights of triangles and more. On a test, the hard part is recognizing which question a word problem is really asking. This page trains that skill.
The two formulas you always have
Everything on this page comes from two ways of computing the same number.
Some problems call the dot product the scalar product and the magnitude the modulus, and write vectors with parentheses like $(3, -2)$. These mean the same things as $\langle 3, -2\rangle$ and $\|\vec u\|$ here.
The sign of the dot product tells you the type of angle, because magnitudes are never negative.
| $\vec u\cdot\vec v$ | $\cos\theta$ | Angle $\theta$ |
|---|---|---|
| positive | positive | acute ($0^\circ \le \theta < 90^\circ$) |
| zero | zero | right ($90^\circ$) |
| negative | negative | obtuse ($90^\circ < \theta \le 180^\circ$) |
Calculate the scalar product of the vectors $\vec{u} = (3, -2)$ and $\vec{v} = (-1, 5)$.
Calculate the scalar product of the vectors $\vec{u} = (3, -2)$ and $\vec{v} = (-1, 5)$.
Given that the moduli of two vectors are $|\vec{a}| = 4$ and $|\vec{b}| = 6$, and that the angle between them is $60^\circ$, determine their scalar product.
Given that the moduli of two vectors are $|\vec{a}| = 4$ and $|\vec{b}| = 6$, and that the angle between them is $60^\circ$, determine their scalar product.
Find the modulus of the vector $\vec{w} = (-5, 12)$.
Find the modulus of the vector $\vec{w} = (-5, 12)$.
Testing and building perpendicular vectors
Two different questions sound alike, so separate them.
Determine, justifying your answer with a mathematical operation, whether the vectors $\vec{p} = (4, 6)$ and $\vec{q} = (-3, 2)$ are orthogonal.
Determine, justifying your answer with a mathematical operation, whether the vectors $\vec{p} = (4, 6)$ and $\vec{q} = (-3, 2)$ are orthogonal.
Calculate the value of the parameter $k$ so that the vectors $\vec{u} = (k, 3)$ and $\vec{v} = (2, -4)$ are perpendicular.
Calculate the value of the parameter $k$ so that the vectors $\vec{u} = (k, 3)$ and $\vec{v} = (2, -4)$ are perpendicular.
Find the coordinates of a vector that is perpendicular to $\vec{v} = (-7, 2)$ and has exactly the same modulus.
Find the coordinates of a vector that is perpendicular to $\vec{v} = (-7, 2)$ and has exactly the same modulus.
The angle between two vectors
Solving the angle formula for $\theta$ gives
$$\theta = \cos^{-1}\!\left(\frac{\vec u\cdot\vec v}{\|\vec u\|\,\|\vec v\|}\right).$$Your calculator's $\cos^{-1}$ returns a value from $0^\circ$ to $180^\circ$, which is exactly the range of an angle between two vectors. Keep exact values when you can. If $\cos\theta$ is $0$, $\tfrac12$ or $\tfrac{\sqrt3}{2}$, name the angle exactly ($90^\circ$, $60^\circ$, $30^\circ$). If the dot product is $0$, you can say $90^\circ$ without computing any magnitudes.
Calculate the angle between the vectors $\vec{a} = (1, \sqrt{3})$ and $\vec{b} = (-\sqrt{3}, 1)$.
Calculate the angle between the vectors $\vec{a} = (1, \sqrt{3})$ and $\vec{b} = (-\sqrt{3}, 1)$.
To stabilize a telecommunications antenna mast, two cables anchored at the same point on the mast and tied to the ground are tensioned. The vectors representing the…
To stabilize a telecommunications antenna mast, two cables anchored at the same point on the mast and tied to the ground are tensioned. The vectors representing the direction of the cables from the upper anchor to the ground are $\vec{T_1} = (-4, -5)$ and $\vec{T_2} = (6, -3)$. Calculate the angle of separation formed by the two cables at the upper tie point.
Distance and right triangles
The vector from $A$ to $B$ is $\overrightarrow{AB} = \langle x_B - x_A,\ y_B - y_A\rangle$. Its magnitude is the distance between the points: $AB = \|\overrightarrow{AB}\|$. This is the distance formula in vector language, so you do not need a dot product for it.
Calculate the distance between the points $A(2, -1)$ and $B(5, 3)$ by first determining the components of the vector $\vec{AB}$.
Calculate the distance between the points $A(2, -1)$ and $B(5, 3)$ by first determining the components of the vector $\vec{AB}$.
Check, using exclusively the scalar product, whether the triangle defined by the vertices $A(1, 2)$, $B(4, 3)$ and $C(3, -1)$ is a right triangle. If so, indicate at…
Check, using exclusively the scalar product, whether the triangle defined by the vertices $A(1, 2)$, $B(4, 3)$ and $C(3, -1)$ is a right triangle. If so, indicate at which vertex the right angle is located.
Given the points $A(-2, 1)$ and $B(4, 5)$, find the coordinates of a point $C$ located on the x-axis such that the triangle $ABC$ is a right triangle with the right…
Given the points $A(-2, 1)$ and $B(4, 5)$, find the coordinates of a point $C$ located on the x-axis such that the triangle $ABC$ is a right triangle with the right angle at the vertex $C$. How many possible solutions exist for the point $C$?
Building a vector from conditions
Drag the point to see how the components of a vector relate to its length and quadrant.
Sometimes the conditions are a magnitude and a dot product with a known vector. The dot product condition gives a line in $x$ and $y$, and the magnitude gives a circle, so you get a system.
In a video game physics engine, a character located at the coordinate $P(10, 15)$ must fire a projectile in the direction of the vector $\vec{d} = (3, -4)$ at a constant…
In a video game physics engine, a character located at the coordinate $P(10, 15)$ must fire a projectile in the direction of the vector $\vec{d} = (3, -4)$ at a constant speed of $20$ distance units per frame. Determine the exact velocity vector $\vec{v}$ of the projectile, knowing that it must have the same direction and sense as $\vec{d}$ but with a modulus of $20$.
Find the coordinates of a vector $\vec{w} = (x, y)$ given that its modulus is $\sqrt{13}$ and that its scalar product with the vector $\vec{v} = (2, 3)$ results in $13$.
Find the coordinates of a vector $\vec{w} = (x, y)$ given that its modulus is $\sqrt{13}$ and that its scalar product with the vector $\vec{v} = (2, 3)$ results in $13$.
A point that makes a segment perpendicular to a line
A robot arm, a ramp or a sensor often has to move from a point $P$ to a line so that the path meets the line at a right angle. The recipe uses the line's direction vector.
An industrial robotic arm moves in a two-dimensional plane. Its gripper is initially located at the point $P(2, 5)$ and must move in a straight line to a point $Q$…
An industrial robotic arm moves in a two-dimensional plane. Its gripper is initially located at the point $P(2, 5)$ and must move in a straight line to a point $Q$ located on a conveyor belt, whose path is defined by the line $x - 2y + 4 = 0$. If the displacement vector $\vec{PQ}$ must be orthogonal to the direction of the conveyor belt, determine the exact coordinates of the point $Q$.
Work: how much of the force moves the object
The work done by a constant force $\vec F$ over a displacement $\vec d$ is a dot product:
$$W = \vec F\cdot\vec d = \|\vec F\|\,\|\vec d\|\cos\theta.$$Only the part of the force that points along the motion counts. If $\theta = 90^\circ$, then $\cos\theta = 0$ and the force does no work.
If the force is given in newtons and the distance in feet, the work comes out in newton-feet (N·ft). Always carry the units of the problem.
In physics, the work $W$ done by a constant force $\vec{F}$ when displacing an object along a displacement vector $\vec{d}$ is calculated using the scalar product…
In physics, the work $W$ done by a constant force $\vec{F}$ when displacing an object along a displacement vector $\vec{d}$ is calculated using the scalar product $W = \vec{F} \cdot \vec{d}$. If a crane applies a force of $\vec{F} = (150, -40)$ Newtons to move a container from point $A(2, 5)$ to point $B(12, 5)$ (coordinates expressed in feet), calculate the total work done.
Projections, heights and linear combinations
The idea behind work (how much of one vector lies along another) has a name: the projection.
The perpendicular leftover is the height of the triangle with base $\vec v$ and third side $\vec u$. That gives a way to find an area using only dot products.
Projections also help with linear combinations. To write $\vec c = \alpha\vec a + \beta\vec b$, you can always solve a $2\times 2$ system. When $\vec a\cdot\vec b = 0$ the dot product tells you something extra: $\vec a$ and $\vec b$ are perpendicular, so they form a perpendicular basis.
Calculate the orthogonal projection vector of the vector $\vec{u} = (4, -3)$ onto the vector $\vec{v} = (1, 2)$.
Calculate the orthogonal projection vector of the vector $\vec{u} = (4, -3)$ onto the vector $\vec{v} = (1, 2)$.
Given the vectors $\vec{a} = (3, -1)$ and $\vec{b} = (1, 3)$, express the vector $\vec{c} = (7, 1)$ as a linear combination of $\vec{a}$ and $\vec{b}$. Once solved,…
Given the vectors $\vec{a} = (3, -1)$ and $\vec{b} = (1, 3)$, express the vector $\vec{c} = (7, 1)$ as a linear combination of $\vec{a}$ and $\vec{b}$. Once solved, calculate the scalar product of $\vec{a}$ and $\vec{b}$ and interpret the result geometrically.
Calculate the exact area of the triangle determined by the vertices $A(1, 1)$, $B(5, 2)$ and $C(3, 6)$. To do this, use the orthogonal projection of the vector…
Calculate the exact area of the triangle determined by the vertices $A(1, 1)$, $B(5, 2)$ and $C(3, 6)$. To do this, use the orthogonal projection of the vector $\vec{AC}$ onto the vector $\vec{AB}$ to determine the height of the triangle, and the modulus of $\vec{AB}$ as the base.
Properties of the dot product and vector proofs
A vector proof has a standard pattern. Name the sides with vectors, express the diagonals or segments in terms of them, and show a dot product is $0$ (perpendicular) or a magnitude identity holds.
For a rhombus the extra condition is $\|\vec u\| = \|\vec v\|$ instead of $\vec u\cdot\vec v = 0$. The diagonals are $\vec u + \vec v$ and $\vec v - \vec u$, and you want to show that their dot product is $0$.
For the vectors $\vec{u}$ and $\vec{v}$, we know that $|\vec{u}| = 3$, $|\vec{v}| = 5$ and that they form an angle of $120^\circ$. Applying the properties of the scalar…
For the vectors $\vec{u}$ and $\vec{v}$, we know that $|\vec{u}| = 3$, $|\vec{v}| = 5$ and that they form an angle of $120^\circ$. Applying the properties of the scalar product, calculate the exact value of the expression $(2\vec{u} - \vec{v}) \cdot (\vec{u} + 3\vec{v})$.
Prove vectorially, using the properties of the scalar product, that the diagonals of any rhombus intersect perpendicularly. To do this, consider a rhombus generated by…
Prove vectorially, using the properties of the scalar product, that the diagonals of any rhombus intersect perpendicularly. To do this, consider a rhombus generated by two free vectors $\vec{u}$ and $\vec{v}$ that start from the same vertex and satisfy $|\vec{u}| = |\vec{v}|$.
Given that $\vec{u}$ and $\vec{v}$ are two unit vectors, determine the angle they must form with each other so that the sum vector $\vec{u} + \vec{v}$ has a modulus…
Given that $\vec{u}$ and $\vec{v}$ are two unit vectors, determine the angle they must form with each other so that the sum vector $\vec{u} + \vec{v}$ has a modulus equal to $\sqrt{3}$.
Angle conditions with unknowns
When a problem gives you an angle and asks for an unknown number or vector, write the angle formula as an equation and solve it.
Determine the possible values of the parameter $m$ so that the vectors $\vec{u} = (m, 1)$ and $\vec{v} = (2, m)$ form an angle of $30^\circ$.
Determine the possible values of the parameter $m$ so that the vectors $\vec{u} = (m, 1)$ and $\vec{v} = (2, m)$ form an angle of $30^\circ$.
Find the coordinates of all the vectors $\vec{a} = (x, y)$ that form an exact angle of $45^\circ$ with the vector $\vec{b} = (2, -2)$ and whose modulus is exactly half…
Find the coordinates of all the vectors $\vec{a} = (x, y)$ that form an exact angle of $45^\circ$ with the vector $\vec{b} = (2, -2)$ and whose modulus is exactly half the modulus of $\vec{b}$.
In the installation of a solar panel, the normal vector (perpendicular to the surface of the panel) is given by $\vec{n} = (k, \sqrt{3})$. The sun's rays strike the…
In the installation of a solar panel, the normal vector (perpendicular to the surface of the panel) is given by $\vec{n} = (k, \sqrt{3})$. The sun's rays strike the panel according to the vector $\vec{s} = (1, -1)$. Calculate the positive value of the parameter $k$ so that the angle of incidence (the angle formed between the normal vector $\vec{n}$ and the opposite vector to the sun's rays $-\vec{s}$) is $60^\circ$.