2. A conveyor belt generates a force $$ \vec{F}=5 \mathbf{i}-3 \mathbf{j}+3 \mathbf{k} $$ that moves a suitcase from point $$ \mathrm{P}(1,1,1) $$ to Point $$ \mathrm{Q}(9,4,7) $$ along a straight line. Find the work done by the conveyor belt. The distance is measured in meters and the force is measured in newtons. 3. If $$ \mathbf{u} $$ and $$ \mathbf{v} $$ are vectors represented as $$ 2 \mathbf{i}+3 \mathbf{j}+4 \mathbf{k} $$ and $$ \mathbf{i}+2 \mathbf{j}+3 \mathbf{k} $$, then find the area of the parallelogram spanned between them. Find the area of the triangle whose vertices are $$ \mathrm{A}(3,-1,2), \mathrm{B}(1,-1,-3) $$ and $$ \mathrm{C}(4,-3,1) $$. 5. Find the area of the quadrilateral whose vertices are $$ \mathrm{A}(1,0), \mathrm{B}(3,5), \mathrm{C}(7,2) $$ and $$ \mathrm{D}(5,-2) $$. A car mechanic applies a force of 800 N to a wrench to loosen a bolt. She applies the force perpendicular to the arm of the wrench. The distance from the bolt to her hand is 0.40 m . What is the magnitude of the torque applied?

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UpStudy AI · Accepted Answer

1. The work done by the conveyor belt is given by  
$$
W = \vec{F}\cdot\vec{d},
$$  
where the force is  
$$
\vec{F} = 5\,\mathbf{i} - 3\,\mathbf{j} + 3\,\mathbf{k},
$$  
and the displacement vector $$ \vec{d} $$ from point $$ \mathrm{P}(1,1,1) $$ to point $$ \mathrm{Q}(9,4,7) $$ is  
$$
\vec{d} = \big(9-1,\; 4-1,\; 7-1\big) = (8,\,3,\,6).
$$

Taking the dot product we obtain:
$$
W = 5\cdot8 + (-3)\cdot3 + 3\cdot6 = 40 - 9 + 18 = 49.
$$

Thus, the work done is  
$$
W = 49\ \text{Joules}.
$$

---

2. For vectors  
$$
\mathbf{u} = 2\,\mathbf{i} + 3\,\mathbf{j} + 4\,\mathbf{k} \quad \text{and} \quad \mathbf{v} = \mathbf{i} + 2\,\mathbf{j} + 3\,\mathbf{k},
$$  
the area of the parallelogram spanned by them is given by the magnitude of their cross product:
$$
\text{Area} = \left|\mathbf{u} \times \mathbf{v}\right|.
$$

First, compute the cross product:
$$
\mathbf{u} \times \mathbf{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
2 & 3 & 4 \
1 & 2 & 3 
\end{vmatrix}
= \mathbf{i}(3\cdot3-4\cdot2) - \mathbf{j}(2\cdot3-4\cdot1) + \mathbf{k}(2\cdot2-3\cdot1).
$$
Simplify each component:
$$
\begin{aligned}
\mathbf{i} &: \quad 3\cdot3 - 4\cdot2 = 9 - 8 = 1, \
\mathbf{j} &: \quad -(2\cdot3 - 4\cdot1) = -(6 - 4) = -2, \
\mathbf{k} &: \quad 2\cdot2 - 3\cdot1 = 4 - 3 = 1.
\end{aligned}
$$
Thus,
$$
\mathbf{u} \times \mathbf{v} = \langle 1,\,-2,\,1\rangle.
$$

Now take its magnitude:
$$
\left|\mathbf{u} \times \mathbf{v}\right| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}.
$$

Therefore, the area of the parallelogram is  
$$
\sqrt{6}.
$$

---

3. The area of the triangle with vertices  
$$
\mathrm{A}(3,-1,2), \quad \mathrm{B}(1,-1,-3), \quad \mathrm{C}(4,-3,1)
$$
can be computed using vectors. Form two sides of the triangle:
$$
\vec{AB} = \mathrm{B} - \mathrm{A} = (1-3,\; -1-(-1),\; -3-2) = (-2,\,0,\,-5),
$$
$$
\vec{AC} = \mathrm{C} - \mathrm{A} = (4-3,\; -3-(-1),\; 1-2) = (1,\,-2,\,-1).
$$

The area of the triangle is half the magnitude of the cross product of these two vectors:
$$
\text{Area} = \frac{1}{2} \left|\vec{AB} \times \vec{AC}\right|.
$$

Compute the cross product:
$$
\vec{AB} \times \vec{AC} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \
-2 & 0 & -5 \
1 & -2 & -1 
\end{vmatrix}.
$$

For each component:
$$
\begin{aligned}
\mathbf{i}: &\quad 0\cdot(-1) - (-5)\cdot(-2) = 0 - 10 = -10, \
\mathbf{j}: &\quad -\big[(-2)\cdot(-1) - (-5)\cdot1\big] = -\big[2 + 5\big] = -7, \
\mathbf{k}: &\quad (-2)\cdot(-2) - 0\cdot1 = 4 - 0 = 4.
\end{aligned}
$$
Thus,
$$
\vec{AB} \times \vec{AC} = \langle -10,\,-7,\,4\rangle.
$$

Now, compute its magnitude:
$$
\left|\vec{AB} \times \vec{AC}\right| = \sqrt{(-10)^2 + (-7)^2 + 4^2} = \sqrt{100 + 49 + 16} = \sqrt{165}.
$$

So, the area of the triangle is  
$$
\frac{1}{2}\sqrt{165}.
$$

---

4. The area of the quadrilateral with vertices  
$$
\mathrm{A}(1,0),\quad \mathrm{B}(3,5),\quad \mathrm{C}(7,2),\quad \mathrm{D}(5,-2)
$$
can be found using the shoelace formula. Arrange the vertices in order and use the formula:
$$
\text{Area} = \frac{1}{2}\left| \sum_{i=1}^{n} (x_i y_{i+1}) - \sum_{i=1}^{n} (y_i x_{i+1}) \right|,
$$
with the indices taken cyclically (i.e. $$ x_{n+1}=x_1 $$ and $$ y_{n+1}=y_1 $$).

For the vertices:
$$
\begin{array}{ccc}
x & y \
1 & 0 \
3 & 5 \
7 & 2 \
5 & -2 \
\end{array}
$$

Calculate the sums:
$$
\begin{aligned}
\text{Sum}_1 &= 1\cdot5 + 3\cdot2 + 7\cdot(-2) + 5\cdot0 = 5 + 6 - 14 + 0 = -3, \
\text{Sum}_2 &= 0\cdot3 + 5\cdot7 + 2\cdot5 + (-2)\cdot1 = 0 + 35 + 10 - 2 = 43.
\end{aligned}
$$

Thus, the area is
$$
\text{Area} = \frac{1}{2} \left| -3 - 43 \right| = \frac{1}{2}\times 46 = 23.
$$

So, the area of the quadrilateral is  
$$
23.
$$

---

5. The torque applied by the mechanic is given by  
$$
\tau = F\,r\,\sin\theta,
$$
where $$ F = 800\,\text{N} $$, $$ r = 0.40\,\text{m} $$, and since the force is perpendicular to the wrench, $$ \theta = 90^\circ $$ (implying $$ \sin\theta = 1 $$). Thus,
$$
\tau = 800 \times 0.40 \times 1 = 320\,\text{N}\cdot\text{m}.
$$

The magnitude of the torque applied is  
$$
320\,\text{N}\cdot\text{m}.
$$
1. The work done by the conveyor belt is 49 Joules. 2. The area of the parallelogram spanned by vectors **u** and **v** is √6. 3. The area of the triangle with vertices A(3,-1,2), B(1,-1,-3), and C(4,-3,1) is (1/2)√165. 4. The area of the quadrilateral with vertices A(1,0), B(3,5), C(7,2), and D(5,-2) is 23. 5. The magnitude of the torque applied by the mechanic is 320 N·m.

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