Esercizi sul prodotto vettoriale

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Esercizi sul prodotto vettoriale

Teoria del Prodotto Vettoriale

Il prodotto vettoriale è un’operazione algebrica che prende due vettori in uno spazio tridimensionale e restituisce un nuovo vettore che è perpendicolare ai due vettori originali. È uno strumento fondamentale in fisica e ingegneria, utilizzato per calcolare momenti, forze e aree.

Definizione

Se abbiamo due vettori a e b in uno spazio tridimensionale, il prodotto vettoriale è definito come:

\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \, \mathbf{n}
a×b=∣a∣∣b∣sin⁡(θ) n\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \, \mathbf{n}

dove:

  • |\mathbf{a}|∣a∣|\mathbf{a}| e |\mathbf{b}|∣b∣|\mathbf{b}| sono le lunghezze (norme) dei vettori,
  • \thetaθ\theta è l’angolo tra i due vettori,
  • \mathbf{n}n\mathbf{n} è un versore che segue la regola della mano destra.

In forma componente, se:

\mathbf{a} = (a_1, a_2, a_3) \quad \text{e} \quad \mathbf{b} = (b_1, b_2, b_3)
a=(a1,a2,a3)eb=(b1,b2,b3)\mathbf{a} = (a_1, a_2, a_3) \quad \text{e} \quad \mathbf{b} = (b_1, b_2, b_3)

allora il prodotto vettoriale può essere calcolato utilizzando il determinante di una matrice:

\mathbf{a} \times \mathbf{b} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{vmatrix}
a×b=∣ijka1a2a3b1b2b3∣\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}

Dove \mathbf{i}, \mathbf{j}, \mathbf{k}i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} sono i versori delle coordinate x, y e z.

Proprietà del Prodotto Vettoriale

  1. Anticommutatività: \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})a×b=−(b×a)\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})
  2. Distributività: \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}a×(b+c)=a×b+a×c\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}
  3. Associatività rispetto alla moltiplicazione per uno scalare: k(\mathbf{a} \times \mathbf{b}) = (k\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (k\mathbf{b})k(a×b)=(ka)×b=a×(kb)k(\mathbf{a} \times \mathbf{b}) = (k\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (k\mathbf{b})

Esercizi sul Prodotto Vettoriale

Esercizio 1: Calcolo del Prodotto Vettoriale

Calcola il prodotto vettoriale dei vettori:

\mathbf{a} = (2, 3, 4) \quad e \quad \mathbf{b} = (1, 0, -1)
a=(2,3,4)eb=(1,0,−1)\mathbf{a} = (2, 3, 4) \quad e \quad \mathbf{b} = (1, 0, -1)

Svolgimento:

Utilizziamo la formula del determinante per calcolare il prodotto vettoriale:

\begin{align*}
\mathbf{a} \times \mathbf{b} &= 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4\\
1 & 0 & -1
\end{vmatrix}\\
&= 
\mathbf{i}\begin{vmatrix}
3 & 4\\
0 & -1
\end{vmatrix}
- 
\mathbf{j}\begin{vmatrix}
2 & 4\\
1 & -1
\end{vmatrix}
+ 
\mathbf{k}\begin{vmatrix}
2 & 3\\
1 & 0
\end{vmatrix}\\
&= 
\mathbf{i}(3(-1) - 4(0)) - 
\mathbf{j}(2(-1) - 4(1)) + 
\mathbf{k}(2(0) - 3(1))\\
&= 
-3\mathbf{i} - (-2 - 4)\mathbf{j} - 3\mathbf{k}\\
&= 
-3\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}\\
&= (-3, 6, -3)
\end{align*}
a×b=∣ijk23410−1∣=i∣340−1∣−j∣241−1∣+k∣2310∣=i(3(−1)−4(0))−j(2(−1)−4(1))+k(2(0)−3(1))=−3i−(−2−4)j−3k=−3i+6j−3k=(−3,6,−3)\begin{align*} \mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4\\ 1 & 0 & -1 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} 3 & 4\\ 0 & -1 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 2 & 4\\ 1 & -1 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 2 & 3\\ 1 & 0 \end{vmatrix}\\ &= \mathbf{i}(3(-1) - 4(0)) - \mathbf{j}(2(-1) - 4(1)) + \mathbf{k}(2(0) - 3(1))\\ &= -3\mathbf{i} - (-2 - 4)\mathbf{j} - 3\mathbf{k}\\ &= -3\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}\\ &= (-3, 6, -3) \end{align*}

Risultato:

Il prodotto vettoriale \mathbf{a} \times \mathbf{b} = (-3, 6, -3)a×b=(−3,6,−3)\mathbf{a} \times \mathbf{b} = (-3, 6, -3).

Esercizio 2: Calcolo dell’Area di un Parallelogramma

Calcola l’area del parallelogramma formato dai vettori:

\mathbf{x} = (4, 5, 0) \quad e \quad \mathbf{y} = (1, -2, 3)
x=(4,5,0)ey=(1,−2,3)\mathbf{x} = (4, 5, 0) \quad e \quad \mathbf{y} = (1, -2, 3)

Svolgimento:

L’area del parallelogramma formato dai due vettori è data dalla norma del loro prodotto vettoriale:

A = |\mathbf{x} \times \mathbf{y}|
A=∣x×y∣A = |\mathbf{x} \times \mathbf{y}|

Calcoliamo prima il prodotto vettoriale:

\begin{align*}
\mathbf{x} \times \mathbf{y} &= 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
4 & 5 & 0\\
1 & -2 & 3
\end{vmatrix}\\
&= 
\begin{vmatrix}
5 & 0\\
-2 & 3
\end{vmatrix}\,\textbf{i}
-
\begin{vmatrix}
4 & 0\\
1 & 3
\end{vmatrix}\,\textbf{j}
+
\begin{vmatrix}
4 & 5\\
1 & -2
\end{vmatrix}\,\textbf{k}\\
&= (5(-2) - (0)(3))\,\textbf{i}
- (4(3) - (0)(1))\,\textbf{j}
+ (4(-2) - (5)(1))\,\textbf{k}\\
&= (-10)\,\textbf{i}-12\,\textbf{j}+(-8-5)\,\textbf{k}\\
&= (-10)\,\textbf{i}-12\,\textbf{j}+(-13)\,\textbf{k}\\
&= (-10,-12,-13)
\end{align*}
x×y=∣ijk4501−23∣=∣50−23∣ i−∣4013∣ j+∣451−2∣ k=(5(−2)−(0)(3)) i−(4(3)−(0)(1)) j+(4(−2)−(5)(1)) k=(−10) i−12 j+(−8−5) k=(−10) i−12 j+(−13) k=(−10,−12,−13)\begin{align*} \mathbf{x} \times \mathbf{y} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4 & 5 & 0\\ 1 & -2 & 3 \end{vmatrix}\\ &= \begin{vmatrix} 5 & 0\\ -2 & 3 \end{vmatrix}\,\textbf{i} - \begin{vmatrix} 4 & 0\\ 1 & 3 \end{vmatrix}\,\textbf{j} + \begin{vmatrix} 4 & 5\\ 1 & -2 \end{vmatrix}\,\textbf{k}\\ &= (5(-2) - (0)(3))\,\textbf{i} - (4(3) - (0)(1))\,\textbf{j} + (4(-2) - (5)(1))\,\textbf{k}\\ &= (-10)\,\textbf{i}-12\,\textbf{j}+(-8-5)\,\textbf{k}\\ &= (-10)\,\textbf{i}-12\,\textbf{j}+(-13)\,\textbf{k}\\ &= (-10,-12,-13) \end{align*}

Ora calcoliamo la norma del prodotto vettoriale:

|\mathbf{x} \times \mathbf{y}| = 
\sqrt{-10^2 + (-12)^2 + (-13)^2}= 
\sqrt {100 + 144 +169}= 
\sqrt {413}.
∣x×y∣=−102+(−12)2+(−13)2=100+144+169=413.|\mathbf{x} \times \mathbf{y}| = \sqrt{-10^2 + (-12)^2 + (-13)^2}= \sqrt {100 + 144 +169}= \sqrt {413}.

Risultato:

L’area del parallelogramma è A = |\sqrt {413}|A=∣413∣A = |\sqrt {413}|.

Esercizio 3: Prodotto Vettoriale in Fisica

Due forze agiscono su un oggetto:

  • Forza F₁ con modulo di 10 N nella direzione dell’asse x.
  • Forza F₂ con modulo di 15 N nella direzione dell’asse y.

Calcola il momento torcentale rispetto all’origine generato da queste forze se i punti di applicazione sono rispettivamente P₁(1,0) e P₂(0,2).

Svolgimento:

Il momento torcentale MMM è dato dal prodotto vettoriale tra il raggio e la forza applicata:

M = r × F.
M=r×F.M = r × F.

Calcoliamo i momenti torcentali separatamente per ciascuna forza.

  1. Per la forza F₁ applicata nel punto P₁(1,0):

    F₁ = (10,0)
    F1=(10,0)F₁ = (10,0)
    r₁ = (1,0)
    r1=(1,0)r₁ = (1,0)

    Calcoliamo il momento:

    M₁ = r₁ × F₁ =
(1,0) × (10,0) =
|i j| |1 0| |10 0| =
(0)i + (-10)j + (0)k =
M₁ = (0,-10,0).
    M1=r1×F1=(1,0)×(10,0)=∣ij∣∣10∣∣100∣=(0)i+(−10)j+(0)k=M1=(0,−10,0).M₁ = r₁ × F₁ = (1,0) × (10,0) = |i j| |1 0| |10 0| = (0)i + (-10)j + (0)k = M₁ = (0,-10,0).

  2. Per la forza F₂ applicata nel punto P₂(0,2):

    F₂ = (0,15)
    F2=(0,15)F₂ = (0,15)
    r₂ = (0,2)
    r2=(0,2)r₂ = (0,2)

    Calcoliamo il momento:

    M₂ = r₂ × F₂ =
(0,2) × (0,15) =
|i j| |0 2| |0 15| =
(-30)i + (0)j + (0)k =
M₂ = (-30,0,0).
    M2=r2×F2=(0,2)×(0,15)=∣ij∣∣02∣∣015∣=(−30)i+(0)j+(0)k=M2=(−30,0,0).M₂ = r₂ × F₂ = (0,2) × (0,15) = |i j| |0 2| |0 15| = (-30)i + (0)j + (0)k = M₂ = (-30,0,0).

Ora sommiamo i momenti torcentali:

M_{totale}= M₁ + M₂ = (0,-10,0)+(-30,0,0)=(-30,-10,0).
Mtotale=M1+M2=(0,−10,0)+(−30,0,0)=(−30,−10,0).M_{totale}= M₁ + M₂ = (0,-10,0)+(-30,0,0)=(-30,-10,0).

Risultato:

Il momento torcentale totale generato dalle forze è M_{totale}=(-30,-10,0).Mtotale=(−30,−10,0).M_{totale}=(-30,-10,0).

English version

Vector Product Exercises

Vector Product Theory

The vector product is an algebraic operation that takes two vectors in a three-dimensional space and returns a new vector that is perpendicular to the two original vectors. It is a fundamental tool in physics and engineering, used to calculate moments, forces, and areas.

Definition

If we have two vectors a and b in a three-dimensional space, the vector product is defined as:

\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \, \mathbf{n}
a×b=∣a∣∣b∣sin⁡(θ) n\mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin(\theta) \, \mathbf{n}

where:

  • |\mathbf{a}|∣a∣|\mathbf{a}| and |\mathbf{b}|∣b∣|\mathbf{b}| are the lengths (norms) of the vectors,
  • \thetaθ\theta is the angle between the two vectors,
  • \mathbf{n}n\mathbf{n} is a versor that follows the right-hand rule.

In component form, if:

\mathbf{a} = (a_1, a_2, a_3) \quad \text{e} \quad \mathbf{b} = (b_1, b_2, b_3)
a=(a1,a2,a3)eb=(b1,b2,b3)\mathbf{a} = (a_1, a_2, a_3) \quad \text{e} \quad \mathbf{b} = (b_1, b_2, b_3)

then the cross product can be computed using the determinant of a matrix:

\mathbf{a} \times \mathbf{b} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
a_1 & a_2 & a_3 \\
b_1 & b_2 & b_3
\end{vmatrix}
a×b=∣ijka1a2a3b1b2b3∣\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}

Where \mathbf{i}, \mathbf{j}, \mathbf{k}i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} are the unit vectors of the x, y and z coordinates.

Vector Product Properties

  1. Anticommutativity: \mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})a×b=−(b×a)\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})
  2. Distributivity: \mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}a×(b+c)=a×b+a×c\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}
  3. Associativity under scalar multiplication: k(\mathbf{a} \times \mathbf{b}) = (k\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (k\mathbf{b})k(a×b)=(ka)×b=a×(kb)k(\mathbf{a} \times \mathbf{b}) = (k\mathbf{a}) \times \mathbf{b} = \mathbf{a} \times (k\mathbf{b})

Vector Product Exercises

Exercise 1: Calculating the Vector Product

Calculate the vector product of the vectors:

\mathbf{a} = (2, 3, 4) \quad and \quad \mathbf{b} = (1, 0, -1)
a=(2,3,4)andb=(1,0,−1)\mathbf{a} = (2, 3, 4) \quad and \quad \mathbf{b} = (1, 0, -1)

Procedure:

We use the determinant formula to calculate the vector product:

\begin{align*}
\mathbf{a} \times \mathbf{b} &=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 3 & 4\\
1 & 0 & -1
\end{vmatrix}\\
&=
\mathbf{i}\begin{vmatrix}
3 & 4\\
0 & -1
\end{vmatrix}
-
\mathbf{j}\begin{vmatrix}
2 & 4\\
1 & -1
\end{vmatrix}
+
\mathbf{k}\begin{vmatrix}
2 & 3\\
1 & 0
\end{vmatrix}\\
&=
\mathbf{i}(3(-1) - 4(0)) -
\mathbf{j}(2(-1) - 4(1)) +
\mathbf{k}(2(0) - 3(1))\\
&=
-3\mathbf{i} - (-2 - 4)\mathbf{j} - 3\mathbf{k}\\
&=
-3\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}\\
&= (-3, 6, -3)
\end{align*}
a×b=∣ijk23410−1∣=i∣340−1∣−j∣241−1∣+k∣2310∣=i(3(−1)−4(0))−j(2(−1)−4(1))+k(2(0)−3(1))=−3i−(−2−4)j−3k=−3i+6j−3k=(−3,6,−3)\begin{align*} \mathbf{a} \times \mathbf{b} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4\\ 1 & 0 & -1 \end{vmatrix}\\ &= \mathbf{i}\begin{vmatrix} 3 & 4\\ 0 & -1 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 2 & 4\\ 1 & -1 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 2 & 3\\ 1 & 0 \end{vmatrix}\\ &= \mathbf{i}(3(-1) - 4(0)) - \mathbf{j}(2(-1) - 4(1)) + \mathbf{k}(2(0) - 3(1))\\ &= -3\mathbf{i} - (-2 - 4)\mathbf{j} - 3\mathbf{k}\\ &= -3\mathbf{i} + 6\mathbf{j} - 3\mathbf{k}\\ &= (-3, 6, -3) \end{align*}

Result:

The vector product \mathbf{a} \times \mathbf{b} = (-3, 6, -3)a×b=(−3,6,−3)\mathbf{a} \times \mathbf{b} = (-3, 6, -3).

Exercise 2: Calculating the Area of ​​a Parallelogram

Calculate the area of ​​the parallelogram formed by the vectors:

\mathbf{x} = (4, 5, 0) \quad and \quad \mathbf{y} = (1, -2, 3)
x=(4,5,0)andy=(1,−2,3)\mathbf{x} = (4, 5, 0) \quad and \quad \mathbf{y} = (1, -2, 3)

Process:

The area of ​​the parallelogram formed by the two vectors is given by the norm of their vector product:

A = |\mathbf{x} \times \mathbf{y}|
A=∣x×y∣A = |\mathbf{x} \times \mathbf{y}|

Let’s first calculate the vector product:

\begin{align*}
\mathbf{x} \times \mathbf{y} &=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k}\\
4 & 5 & 0\\
1 & -2 & 3
\end{vmatrix}\\
&=
\begin{vmatrix}
5 & ​​0\\
-2 & 3
\end{vmatrix}\,\textbf{i}
-
\begin{vmatrix}
4 & 0\\
1 & 3
\end{vmatrix}\,\textbf{j}
+
\begin{vmatrix}
4 & 5\\
1 & -2
\end{vmatrix}\,\textbf{k}\\
&= (5(-2) - (0)(3))\,\textbf{i}
- (4(3) - (0)(1))\,\textbf{j}
+ (4(-2) - (5)(1))\,\textbf{k}\\
&= (-10)\,\textbf{i}-12\,\textbf{j}+(-8-5)\,\textbf{k}\\
&= (-10)\,\textbf{i}-12\,\textbf{j}+(-13)\,\textbf{k}\\
&= (-10,-12,-13)
\end{align*}
x×y=∣ijk4501−23∣=∣5​​0−23∣ i−∣4013∣ j+∣451−2∣ k=(5(−2)−(0)(3)) i−(4(3)−(0)(1)) j+(4(−2)−(5)(1)) k=(−10) i−12 j+(−8−5) k=(−10) i−12 j+(−13) k=(−10,−12,−13)\begin{align*} \mathbf{x} \times \mathbf{y} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 4 & 5 & 0\\ 1 & -2 & 3 \end{vmatrix}\\ &= \begin{vmatrix} 5 & ​​0\\ -2 & 3 \end{vmatrix}\,\textbf{i} - \begin{vmatrix} 4 & 0\\ 1 & 3 \end{vmatrix}\,\textbf{j} + \begin{vmatrix} 4 & 5\\ 1 & -2 \end{vmatrix}\,\textbf{k}\\ &= (5(-2) - (0)(3))\,\textbf{i} - (4(3) - (0)(1))\,\textbf{j} + (4(-2) - (5)(1))\,\textbf{k}\\ &= (-10)\,\textbf{i}-12\,\textbf{j}+(-8-5)\,\textbf{k}\\ &= (-10)\,\textbf{i}-12\,\textbf{j}+(-13)\,\textbf{k}\\ &= (-10,-12,-13) \end{align*}

Now let’s calculate the norm of the vector product:

|\mathbf{x} \times \mathbf{y}| =
\sqrt{-10^2 + (-12)^2 + (-13)^2}=
\sqrt {100 + 144 +169}=
\sqrt {413}.
∣x×y∣=−102+(−12)2+(−13)2=100+144+169=413.|\mathbf{x} \times \mathbf{y}| = \sqrt{-10^2 + (-12)^2 + (-13)^2}= \sqrt {100 + 144 +169}= \sqrt {413}.

Result:

The area of ​​the parallelogram is A = |\sqrt {413}|A=∣413∣A = |\sqrt {413}|.

Exercise 3: Vector Product in Physics

Two forces act on an object:

  • Force F₁ with a magnitude of 10 N in the direction of the x-axis.
  • Force F₂ with a magnitude of 15 N in the direction of the y-axis.

Calculate the torque with respect to the origin generated by these forces if the application points are respectively P₁(1,0) and P₂(0,2).

Procedure:

The torque MMM is given by the vector product of the radius and the applied force:

M = r × F.
M=r×F.M = r × F.

Let’s calculate the torques separately for each force.

  1. For the force F₁ applied at the point P₁(1,0):
F₁ = (10,0)
F1=(10,0)F₁ = (10,0)
r₁ = (1,0)
r1=(1,0)r₁ = (1,0)

Let’s calculate the torque:

M₁ = r₁ × F₁ =
(1,0) × (10,0) =
|i j| |1 0| |10 0| =
(0)i + (-10)j + (0)k =
M₁ = (0,-10,0).
M1=r1×F1=(1,0)×(10,0)=∣ij∣∣10∣∣100∣=(0)i+(−10)j+(0)k=M1=(0,−10,0).M₁ = r₁ × F₁ = (1,0) × (10,0) = |i j| |1 0| |10 0| = (0)i + (-10)j + (0)k = M₁ = (0,-10,0).
  1. For the force F₂ applied at the point P₂(0.2):
F₂ = (0.15)
F2=(0.15)F₂ = (0.15)
r₂ = (0.2)
r2=(0.2)r₂ = (0.2)

Let’s calculate the moment:

M₂ = r₂ × F₂ =
(0.2) × (0.15) =
|i j| |0 2| |0 15| =
(-30)i + (0)j + (0)k =
M₂ = (-30,0,0).
M2=r2×F2=(0.2)×(0.15)=∣ij∣∣02∣∣015∣=(−30)i+(0)j+(0)k=M2=(−30,0,0).M₂ = r₂ × F₂ = (0.2) × (0.15) = |i j| |0 2| |0 15| = (-30)i + (0)j + (0)k = M₂ = (-30,0,0).

Now let’s add the torques:

M_{total}= M₁ + M₂ = (0,-10,0)+(-30,0,0)=(-30,-10,0).
Mtotal=M1+M2=(0,−10,0)+(−30,0,0)=(−30,−10,0).M_{total}= M₁ + M₂ = (0,-10,0)+(-30,0,0)=(-30,-10,0).

Result:

The total torque generated by the forces is M_{total}=(-30,-10,0).Mtotal=(−30,−10,0).M_{total}=(-30,-10,0).

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