1. Find the measure of the angle between each pair of the vector a. U = 3i + 2j, V = 5i - j b. U = i + j, V = -i + j c. U = 6i + j, V = i + j 2. A conveyor belt generat... 1. Find the measure of the angle between each pair of the vector a. U = 3i + 2j, V = 5i - j b. U = i + j, V = -i + j c. U = 6i + j, V = i + j 2. A conveyor belt generates a force F = 5i - 3j + 3k that moves an object from point P(1, 1, 1) to point 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 U and V are vectors represented as 2i + 3j + 4k and i + 2j + 3k, then find the area of the parallelogram spanned by them. 4. Find the area of the triangle whose vertices are A(3, -1, 2), B(1, -1, -3), and C(4, -3, 1). Read more

Steps to Solve

1. Find the angle between vectors for part a

The formula for the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by:

$$ \cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} $$

First calculate the dot product for $\mathbf{u} = 3\mathbf{i} + 2\mathbf{j}$ and $\mathbf{v} = 5\mathbf{i} - \mathbf{j}$:

$$ \mathbf{u} \cdot \mathbf{v} = (3)(5) + (2)(-1) = 15 - 2 = 13 $$

Next, find the magnitudes of $\mathbf{u}$ and $\mathbf{v}$:

$$ ||\mathbf{u}|| = \sqrt{(3)^2 + (2)^2} = \sqrt{9 + 4} = \sqrt{13} $$ $$ ||\mathbf{v}|| = \sqrt{(5)^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} $$

Now, calculate $\cos{\theta}$:

$$ \cos{\theta} = \frac{13}{\sqrt{13} \cdot \sqrt{26}} = \frac{13}{\sqrt{13} \cdot \sqrt{2 \cdot 13}} = \frac{13}{13\sqrt{2}} = \frac{1}{\sqrt{2}} $$

So, $\theta = \arccos{\frac{1}{\sqrt{2}}} = \frac{\pi}{4}$ radians or $45^\circ$.

2. Find the angle between vectors for part b

The formula for the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by:

$$ \cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} $$

First calculate the dot product for $\mathbf{u} = \mathbf{i} + \mathbf{j}$ and $\mathbf{v} = -\mathbf{i}$:

$$ \mathbf{u} \cdot \mathbf{v} = (1)(-1) + (1)(0) = -1 $$

Next, find the magnitudes of $\mathbf{u}$ and $\mathbf{v}$:

$$ ||\mathbf{u}|| = \sqrt{(1)^2 + (1)^2} = \sqrt{1 + 1} = \sqrt{2} $$ $$ ||\mathbf{v}|| = \sqrt{(-1)^2 + (0)^2} = \sqrt{1} = 1 $$

Now, calculate $\cos{\theta}$:

$$ \cos{\theta} = \frac{-1}{\sqrt{2} \cdot 1} = \frac{-1}{\sqrt{2}} $$

So, $\theta = \arccos{\frac{-1}{\sqrt{2}}} = \frac{3\pi}{4}$ radians or $135^\circ$.

3. Find the angle between vectors for part c

The formula for the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is given by:

$$ \cos{\theta} = \frac{\mathbf{u} \cdot \mathbf{v}}{||\mathbf{u}|| \cdot ||\mathbf{v}||} $$

First calculate the dot product for $\mathbf{u} = 6\mathbf{i} + \mathbf{j}$ and $\mathbf{v} = \mathbf{i} + 8\mathbf{j}$:

$$ \mathbf{u} \cdot \mathbf{v} = (6)(1) + (1)(8) = 6 + 8 = 14 $$

Next, find the magnitudes of $\mathbf{u}$ and $\mathbf{v}$:

$$ ||\mathbf{u}|| = \sqrt{(6)^2 + (1)^2} = \sqrt{36 + 1} = \sqrt{37} $$ $$ ||\mathbf{v}|| = \sqrt{(1)^2 + (8)^2} = \sqrt{1 + 64} = \sqrt{65} $$

Now, calculate $\cos{\theta}$:

$$ \cos{\theta} = \frac{14}{\sqrt{37} \cdot \sqrt{65}} = \frac{14}{\sqrt{2405}} $$

So, $\theta = \arccos{\frac{14}{\sqrt{2405}}} \approx 73.74^\circ$.

4. Calculate the work done by the conveyor belt

The work done $W$ by a force $\mathbf{F}$ over a displacement $\mathbf{d}$ is given by:

$$ W = \mathbf{F} \cdot \mathbf{d} $$

The force is $\mathbf{F} = 5\mathbf{i} - 3\mathbf{j} + 3\mathbf{k}$. The displacement vector $\mathbf{d}$ is from point $P(1, 1, 1)$ to point $Q(9, 4, 7)$, so:

$$ \mathbf{d} = (9-1)\mathbf{i} + (4-1)\mathbf{j} + (7-1)\mathbf{k} = 8\mathbf{i} + 3\mathbf{j} + 6\mathbf{k} $$

The work done is:

$$ W = (5)(8) + (-3)(3) + (3)(6) = 40 - 9 + 18 = 49 $$

So, the work done is 49 Joules.

5. Find the area of the parallelogram

Given vectors $\mathbf{u} = 2\mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$ and $\mathbf{v} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$, the area of the parallelogram spanned by them is the magnitude of their cross product:

$$ A = ||\mathbf{u} \times \mathbf{v}|| $$

First, find the cross product:

$$ \mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 3 & 4 \ 1 & 2 & 3 \end{vmatrix} = (9-8)\mathbf{i} - (6-4)\mathbf{j} + (4-3)\mathbf{k} = \mathbf{i} - 2\mathbf{j} + \mathbf{k} $$

Then, find the magnitude:

$$ A = ||\mathbf{i} - 2\mathbf{j} + \mathbf{k}|| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6} $$

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

6. Find the area of the triangle

Given vertices $A(3, -1, 2)$, $B(1, -1, -3)$, and $C(4, -3, 1)$, we first find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$:

$$ \overrightarrow{AB} = (1-3)\mathbf{i} + (-1-(-1))\mathbf{j} + (-3-2)\mathbf{k} = -2\mathbf{i} + 0\mathbf{j} - 5\mathbf{k} $$ $$ \overrightarrow{AC} = (4-3)\mathbf{i} + (-3-(-1))\mathbf{j} + (1-2)\mathbf{k} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} $$

The area of the triangle is half the magnitude of the cross product of these vectors:

$$ A = \frac{1}{2} ||\overrightarrow{AB} \times \overrightarrow{AC}|| $$

Find the cross product:

$$ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -2 & 0 & -5 \ 1 & -2 & -1 \end{vmatrix} = (0-10)\mathbf{i} - (2-(-5))\mathbf{j} + (4-0)\mathbf{k} = -10\mathbf{i} - 7\mathbf{j} + 4\mathbf{k} $$

Then, find the magnitude:

$$ ||\overrightarrow{AB} \times \overrightarrow{AC}|| = \sqrt{(-10)^2 + (-7)^2 + (4)^2} = \sqrt{100 + 49 + 16} = \sqrt{165} $$

So, the area of the triangle is:

$$ A = \frac{1}{2} \sqrt{165} $$