Question

Given \(\mathbf{A}=\mathbf{i}+\mathbf{j}-2 \mathbf{k}, \mathbf{B}=2 \mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{C}=\mathbf{j}-5 \mathbf{k}\): Let \(O\) be the tail of \(\mathbf{B}\) and let \(\mathbf{A}\) be a force acting at the head of \(\mathbf{B}\). Find the torque of \(\mathbf{A}\) about \(O ;\) about a line through \(O\) perpendicular to the plane of \(\mathbf{A}\) and \(\mathbf{B}\); about a line through \(O\) parallel to \(\mathbf{C}\).

Short answer

The torque \(\boldsymbol{\tau} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\). The torque about the perpendicular line is \sqrt{59}\ units and about the line parallel to \(\mathbf{C}\) is \frac{-8}{\sqrt{26}}\ units.

Step-by-step solution

- Find the torque \(\boldsymbol{\tau}\) of \(\mathbf{A}\) about \(O\)

The torque \(\boldsymbol{\tau}\) is given by the cross product \(\mathbf{r} \times \mathbf{F}\) where \(\mathbf{r}\) is the position vector from \(O\) to the point where the force \(\mathbf{A}\) is applied and \(\mathbf{F} = \mathbf{A}\). Since \(\mathbf{r}\) is \(\mathbf{B}\), the torque becomes \[\boldsymbol{\tau} = \mathbf{B} \times \mathbf{A}\].

- Calculate the cross product \[\mathbf{B} \times \mathbf{A}\]

First, write \(\mathbf{B} = 2 \mathbf{i} - \mathbf{j} + 3 \mathbf{k}\) and \(\mathbf{A} = \mathbf{i} + \mathbf{j} - 2 \mathbf{k}\). The cross product is calculated using the determinant form: \[ \mathbf{B} \times \mathbf{A} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & -1 & 3 \ 1 & 1 & -2 \end{vmatrix} \].

- Solve the determinant

Expanding the determinant, we get: \[ \mathbf{B} \times \mathbf{A} = \mathbf{i}((-1)(-2) - 3(1)) - \mathbf{j}((2)(-2) - 3(1)) + \mathbf{k}((2)(1) - (-1)(1)) \].This simplifies to: \[ \mathbf{B} \times \mathbf{A} = \mathbf{i}(2 - 3) - \mathbf{j}(-4 - 3) + \mathbf{k}(2 + 1) \]\[ \mathbf{B} \times \mathbf{A} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k} \].

- Determine the torque about the line through \(O\) perpendicular to the plane of \((\mathbf{A}, \mathbf{B})\)

A line through \(O\) perpendicular to the plane of \(\mathbf{A}\) and \(\mathbf{B}\) has a direction vector that is parallel to \(\mathbf{B} \times \mathbf{A} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\). The magnitude of this cross product vector represents the torque about the line.\[\text{Magnitude} = \sqrt{(-1)^2 + 7^2 + 3^2} = \sqrt{1 + 49 + 9} = \sqrt{59} \]. Hence, the torque is \sqrt{59}\ units.

- Determine the torque about the line through \(O\) parallel to \(\mathbf{C}\)

First, calculate \(\mathbf{u}_{\mathbf{C}}\), the unit vector of \(\mathbf{C} = \mathbf{j} - 5 \mathbf{k}\), as: \[ \mathbf{u}_{\mathbf{C}} = \frac{\mathbf{C}}{\| \mathbf{C} \|} = \frac{\mathbf{j} - 5 \mathbf{k}}{\sqrt{1 + 25}} = \frac{\mathbf{j} - 5 \mathbf{k}}{\sqrt{26}} \]. The component of \(\boldsymbol{\tau} = -\mathbf{i} + 7\mathbf{j} + 3\mathbf{k}\) along \(\mathbf{u}_{\mathbf{C}}\) is given by the dot product: \(\boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}}\).\[ \boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}} = (-1 \mathbf{i} + 7 \mathbf{j} + 3 \mathbf{k}) \cdot \left(\frac{0 \mathbf{i} + 1 \mathbf{j} - 5 \mathbf{k}}{\sqrt{26}}\right)\]\[ \boldsymbol{\tau} \cdot \mathbf{u}_{\mathbf{C}} = \frac{(0)(-1) + (7)(1) + (3)(-5)}{\sqrt{26}} = \frac{7 - 15}{\sqrt{26}} = \frac{-8}{\sqrt{26}} \]. Therefore, the torque about a line parallel to \(\mathbf{C}\) is \(\frac{-8}{\sqrt{26}}\) units.

Key concepts

Cross Product

In vector calculus, the cross product of two vectors produces a third vector that is perpendicular to the plane formed by the initial two vectors. It's often used in physics to determine the direction and magnitude of quantities like torque and magnetic force. The cross product of two vectors \(\textbf{A}\) and \(\textbf{B}\), denoted as \(\textbf{A} \times \textbf{B}\), can be calculated using a determinant matrix:
\[\textbf{A} \times \textbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ A_x & A_y & A_z \ B_x & B_y & B_z \ \end{vmatrix} \]
The resulting vector's components are derived by solving this determinant.

Torque

Torque, also known as the moment of force, is a measure of how much a force acting on an object causes that object to rotate. The magnitude of the torque \(\boldsymbol{\tau}\) depends on three factors: the magnitude of the force \(\textbf{F}\), the distance \(\textbf{r}\) from the point where the force is applied to the axis of rotation, and the angle between the force vector and the position vector. Mathematically, it is given by the cross product:
\[\boldsymbol{\tau} = \textbf{r} \times \textbf{F} \]
The direction of torque is perpendicular to the plane formed by \(\textbf{r}\) and \(\textbf{F}\), which complies with the right-hand rule.

Unit Vector

A unit vector is a vector with a magnitude of 1 and points in a specific direction. Unit vectors are used to specify directions and are particularly helpful in physics and engineering. To find the unit vector \(\textbf{u}_{\textbf{A}}\) of any vector \(\textbf{A}\), you divide the vector by its magnitude \(\textbf{|A|}\):
\[\textbf{u}_{\textbf{A}} = \frac{\textbf{A}}{|\textbf{A}|}\]
For example, if vector \(\textbf{A} = a\textbf{i} + b\textbf{j} + c\textbf{k}\), its magnitude \(|\textbf{A}|\) is calculated as:
\[\textbf{|A|} = \sqrt{a^2 + b^2 + c^2} \]
The unit vector is then:
\[\textbf{u}_{\textbf{A}} = \frac{a\textbf{i} + b\textbf{j} + c\textbf{k}}{\textbf{|A|}} \]

Dot Product

The dot product (or scalar product) of two vectors results in a scalar value and is a measure of how much one vector goes in the direction of another. The dot product of vectors \(\textbf{A}\) and \(\textbf{B}\), denoted as \(\textbf{A} \bullet \textbf{B}\), is calculated using their components:
\[\textbf{A} \bullet \textbf{B} = A_xB_x + A_yB_y + A_zB_z \]
Alternatively, it can be related to the magnitudes of the vectors and the cosine of the angle \(\theta\) between them:
\[\textbf{A} \bullet \textbf{B} = |\textbf{A}| |\textbf{B}| \cos \theta \]
This product provides insights into the alignment of the vectors: if \(\theta\) is 0 degrees (vectors are parallel), the dot product is maximized. If \(\theta\) is 90 degrees (vectors are perpendicular), the dot product is zero.