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The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors. It is denoted by the symbol \( \times \).
In physics, torque (\( \vec{\tau} \)) can be calculated with the cross product of the position vector (\( \vec{r} \)) and the force vector (\( \vec{F} \)).
The formula for the cross product is:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]When applied to our problem:
Force vector: \( \vec{F} = (-8.0 \mathrm{~N}) \hat{\mathrm{i}} + (6.0 \mathrm{~N}) \hat{\mathrm{j}} \)
Position vector: \( \vec{r} = (3.0 \mathrm{~m}) \hat{\mathrm{i}} + (4.0 \mathrm{~m}) \hat{\mathrm{j}} \)These vectors are used to find the torque.
The calculation with respective components gives:
\[ \vec{\tau} = (3.0 \mathrm{~m}) \times (6.0 \mathrm{~N}) - (4.0 \mathrm{~m}) \times (-8.0 \mathrm{~N}) \]
\[ \vec{\tau} = 50.0 \mathrm{~mN} \hat{\mathrm{k}} \] This result shows the torque around the origin.

The dot product is another mathematical operation, signifying the product of two vectors' magnitudes and the cosine of the angle between them. It is symbolized by \( \cdot \) and given by:
\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \]In our problem, it's used to find the angle between the force vector and the position vector.
The force vector: \( \vec{F} \)
The position vector: \( \vec{r} \)
The dot product calculates as:
\[ \vec{r} \cdot \vec{F} = (3.0 \mathrm{~m}) \cdot (-8.0 \mathrm{~N}) + (4.0 \mathrm{~m}) \cdot (6.0 \mathrm{~N}) = 0 \]
Since the dot product is zero, the cosine of the angle is zero. Therefore, the angle between them is 90 degrees.

The magnitude of a vector is its length or size and can be calculated using the Pythagorean theorem.
For any vector \( \vec{V} = a \hat{\mathrm{i}} + b \hat{\mathrm{j}} \),
The magnitude \( |\vec{V}| \) is given by:
\[ |\vec{V}| = \sqrt{a^2 + b^2} \] For the position vector \( \vec{r} \):
\[ |\vec{r}| = \sqrt{(3.0 \mathrm{~m})^2 + (4.0 \mathrm{~m})^2} = 5.0 \mathrm{~m} \]
For the force vector \( \vec{F} \):
\[ |\vec{F}| = \sqrt{(-8.0 \mathrm{~N})^2 + (6.0 \mathrm{~N})^2} = 10.0 \mathrm{~N} \]
The magnitude thus provides information about the size of these vectors.

A force vector represents a force applied in a specific direction and has both magnitude and direction.
In mathematical terms, it is expressed as:
\[ \vec{F} = F_x \hat{\mathrm{i}} + F_y \hat{\mathrm{j}} \]
For our problem, the force vector is:
\( \vec{F} = (-8.0 \mathrm{~N}) \hat{\mathrm{i}} + (6.0 \mathrm{~N}) \hat{\mathrm{j}} \)
This identifies 8.0 N in the negative x-direction and 6.0 N in the positive y-direction.
Force vectors are significant in torque calculations because they show how force applied can cause rotational motion.

The position vector indicates a specific point in space relative to an origin. It also has both magnitude and direction.
It is expressed as:
\[ \vec{r} = r_x \hat{\mathrm{i}} + r_y \hat{\mathrm{j}} \] In our problem, the position vector is:
\( \vec{r} = (3.0 \mathrm{~m}) \hat{\mathrm{i}} + (4.0 \mathrm{~m}) \hat{\mathrm{j}} \)
This denotes a point 3.0 meters along the x-axis and 4.0 meters along the y-axis.
You use this vector during cross product calculations to compute torque as it shows the point of force application relative to the origin.