High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize High School Grade 12 Calculus Textbook > Vector Products

Cross Product Applications

Cross ProductPrevious Section
Equations of Lines in R2R^2 & R3R^3Next Section
Popular Courses
Find My Course
0:00 / 0:00

Cross Product Application: Area & Volume

The area of a parallelogram with sides u⃗\vec{u} and v⃗\vec{v} is A=∣∣u⃗×v⃗∣∣A=\left|\left|\vec{u}\times\vec{v}\right|\right|

The area of a triangle with sides u⃗\vec{u} and v⃗\vec{v} is A=12∣∣u⃗×v⃗∣∣A=\frac{1}{2}\left|\left|\vec{u}\times\vec{v}\right|\right|


The volume of a parallelopiped with sides u⃗\vec{u}, v⃗\vec{v}, and w⃗\vec{w} is V=∣u⃗⋅(v⃗×w⃗)∣V=\left|\vec{u}\cdot\left(\vec{v}\times\vec{w}\right)\right|

0:00 / 0:00

Example: Area of a Parallelogram

Find the area of the triangle with vertices A(1, 0, 1)A\left(1,\ 0,\ 1\right), B(−1, 2, −2)B\left(-1,\ 2,\ -2\right) and C(3, 1, 1)C\left(3,\ 1,\ 1\right).

AB→=B→−A→=[−1, 2, −2]−[1, 0, 1]=[−2, 2, −3]\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}=\left[-1,\ 2,\ -2\right]-\left[1,\ 0,\ 1\right]=\left[-2,\ 2,\ -3\right]
AC→=C→−A→=[3, 1, 1]−[1, 0, 1]=[2, 1, 0]\overrightarrow{AC}=\overrightarrow{C}-\overrightarrow{A}=\left[3,\ 1,\ 1\right]-\left[1,\ 0,\ 1\right]=\left[2,\ 1,\ 0\right]

The area of the triangle is A=12∣∣AB→×AC→∣∣A=\frac{1}{2}\left|\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|\right|
A=12∣∣[−2, 2, −3]×[2, 1, 0]∣∣\displaystyle A=\frac{1}{2}\left|\left|\left[-2,\ 2,\ -3\right]\times\left[2,\ 1,\ 0\right]\right|\right|
A=12∣∣[(2)(0)−(−3)(1), (−3)(2)−(−2)(0), (−2)(1)−(2)(2)]∣∣\displaystyle A=\frac{1}{2}\left|\left|\left[\left(2\right)\left(0\right)-\left(-3\right)\left(1\right),\ \left(-3\right)\left(2\right)-\left(-2\right)\left(0\right),\ \left(-2\right)\left(1\right)-\left(2\right)\left(2\right)\right]\right|\right|
A=12∣∣[3, −6, −6]∣∣\displaystyle A=\frac{1}{2}\left|\left|\left[3,\ -6,\ -6\right]\right|\right|
A=12(3)2+(−6)2+(−6)2\displaystyle A=\frac{1}{2}\sqrt{\left(3\right)^2+\left(-6\right)^2+\left(-6\right)^2}
A=129+36+36\displaystyle A=\frac{1}{2}\sqrt{9+36+36}
A=1281\displaystyle A=\frac{1}{2}\sqrt{81}
A=92\displaystyle A=\frac{9}{2}

Practice: Volume of a Parallelpiped

Find the volume of a parallelopiped defined by the sides u⃗=[2,3,−1]\vec{u}=[2,3,-1], v⃗=[0,1,1]\vec v=[0, 1, 1], and w⃗=[−1,−1,0]\vec w=[-1, -1, 0].
0:00 / 0:00

Cross Product Application: Torque

Torque represents the overall twisting effect of an applied force. It is measured in Newton meters

Suppose we have a fixed point MM with an extended arm. If a force applied to this arm causes the arm to rotate about the point MM, we call this effect torque.



τ⃗=r⃗×F⃗\vec{\tau}=\vec{r}\times\vec{F} and ∣∣τ⃗∣∣=∣∣r⃗∣∣∣∣F⃗∣∣sin⁡θ\left|\left|\vec{\tau}\right|\right|=||\vec{r}||||\vec{F}||\sin\theta
  • r⃗\vec r is the vector from the fixed point MM to the point in which the force is appled (∣r⃗∣|\vec r| is measured in m)
  • F⃗\vec F is the force that is applied (∣F⃗∣|\vec F| is measured in N)
  • θ\theta is the angle between the vectors r⃗\vec r and F⃗\vec F (0≤θ≤180°0\le\theta\le180\degree)

Example
A wrench 35cm long is used to tighten a bolt by applying a force of 30N at an angle of 60°60\degree to the end of the handle of the wrench. Find the magnitude of the torque.

∣∣τ⃗∣∣=∣∣r⃗∣∣∣∣F⃗∣∣sin⁡θ||\vec \tau||=||\vec r||||\vec F||\sin\theta
∣∣τ⃗∣∣=(0.35)(30)sin⁡(60°)||\vec \tau||=(0.35)(30)\sin(60\degree)
∣∣τ⃗∣∣=(0.35)(30)(32)||\vec \tau||=(0.35)(30)\left(\frac{\sqrt{3}}{2}\right)
∣∣τ⃗∣∣=5.2530≈9.09 Nm||\vec \tau||=5.25\sqrt {30}\approx9.09\ Nm
0:00 / 0:00

Example: Torque

You and your friend are trying to push open a gate that has a single hinge on one side. The gate starts opening if you apply a force of 60N perpendicular to the gate, at a point that is 60cm away from the hinge, parallel to the ground. If your friend is only able to push the gate with a force of 40N perpendicular to the door, where should they apply the force in order to open the gate?

The same amount of torque is required to open the gate.

You
Since you apply a force of 60N perpendicular to the gate, the angle between r⃗\vec r and F⃗\vec F is 90°90\degree :
∣∣τ⃗∣∣=∣∣r⃗∣∣∣∣F⃗∣∣sin⁡(90°)||\vec \tau||=||\vec r||||\vec F||\sin(90\degree)
∣∣τ⃗∣∣=(0.60)(60)sin⁡(90°)||\vec \tau||=(0.60)(60)\sin(90\degree)
∣∣τ⃗∣∣=36||\vec \tau||=36

Your friend
∣∣τ⃗∣∣=∣∣r⃗∣∣∣∣F⃗∣∣sin⁡(90°)||\vec \tau||=||\vec r||||\vec F||\sin(90\degree)
36=∣∣r⃗∣∣(40)sin⁡(90°)36=||\vec r||(40)\sin(90\degree)
∣∣r⃗∣∣=3640||\vec{r}||=\frac{36}{40}
∣∣r⃗∣∣=0.9\left|\left|\vec{r}\right|\right|=0.9

Therefore, your friend should push the door at a distance of 90 cm away from the hinge, parallel to the ground.