Solve the following problems: 1. Find the measure of the angle between each pair of vectors: a. u=3i+2j and v=5i-j b. u = i + 7j and v=-i+j c. u=6i+j and v = i +8j 2. A convey... Solve the following problems: 1. Find the measure of the angle between each pair of vectors: a. u=3i+2j and v=5i-j b. u = i + 7j and v=-i+j c. u=6i+j and v = i +8j 2. A conveyor belt generates a force F = 5i-3j + 3k that moves a suitcase 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 between 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 $\vec{u}=3\vec{i}+2\vec{j}$ and $\vec{v}=5\vec{i}-\vec{j}$

The dot product of two vectors is related to the angle between them by the formula: $\vec{u} \cdot \vec{v} = ||\vec{u}|| \cdot ||\vec{v}|| \cdot \cos{\theta}$. First, find the dot product $\vec{u} \cdot \vec{v}$: $$ \vec{u} \cdot \vec{v} = (3)(5) + (2)(-1) = 15 - 2 = 13 $$ Next, find the magnitudes of $\vec{u}$ and $\vec{v}$: $$ ||\vec{u}|| = \sqrt{3^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} $$ $$ ||\vec{v}|| = \sqrt{5^2 + (-1)^2} = \sqrt{25 + 1} = \sqrt{26} $$ Now, find the cosine of the angle between the vectors: $$ \cos{\theta} = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \cdot ||\vec{v}||} = \frac{13}{\sqrt{13} \cdot \sqrt{26}} = \frac{13}{\sqrt{13} \cdot \sqrt{2} \cdot \sqrt{13}} = \frac{13}{13\sqrt{2}} = \frac{1}{\sqrt{2}} $$ Therefore, $\theta = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^{\circ}$

2. Find the angle between vectors $\vec{u}=\vec{i}+7\vec{j}$ and $\vec{v}=-\vec{i}+\vec{j}$ Find the dot product $\vec{u} \cdot \vec{v}$: $$ \vec{u} \cdot \vec{v} = (1)(-1) + (7)(1) = -1 + 7 = 6 $$ Next, find the magnitudes of $\vec{u}$ and $\vec{v}$: $$ ||\vec{u}|| = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50} $$ $$ ||\vec{v}|| = \sqrt{(-1)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} $$ Now, find the cosine of the angle between the vectors: $$ \cos{\theta} = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \cdot ||\vec{v}||} = \frac{6}{\sqrt{50} \cdot \sqrt{2}} = \frac{6}{\sqrt{100}} = \frac{6}{10} = \frac{3}{5} $$ Therefore, $\theta = \cos^{-1}\left(\frac{3}{5}\right) \approx 53.13^{\circ}$

3. Find the angle between vectors $\vec{u}=6\vec{i}+\vec{j}$ and $\vec{v}=\vec{i}+8\vec{j}$ Find the dot product $\vec{u} \cdot \vec{v}$: $$ \vec{u} \cdot \vec{v} = (6)(1) + (1)(8) = 6 + 8 = 14 $$ Next, find the magnitudes of $\vec{u}$ and $\vec{v}$: $$ ||\vec{u}|| = \sqrt{6^2 + 1^2} = \sqrt{36 + 1} = \sqrt{37} $$ $$ ||\vec{v}|| = \sqrt{1^2 + 8^2} = \sqrt{1 + 64} = \sqrt{65} $$ Now, find the cosine of the angle between the vectors: $$ \cos{\theta} = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \cdot ||\vec{v}||} = \frac{14}{\sqrt{37} \cdot \sqrt{65}} = \frac{14}{\sqrt{2405}} \approx \frac{14}{49.04} \approx 0.2855 $$ Therefore, $\theta = \cos^{-1}\left(\frac{14}{\sqrt{2405}}\right) \approx 73.41^{\circ}$

4. Find the work done by the conveyor belt The work done by a force along a straight line is given by $W = \vec{F} \cdot \vec{d}$, where $\vec{F}$ is the force and $\vec{d}$ is the displacement vector. First, find the displacement vector $\vec{d}$ from point P(1, 1, 1) to point Q(9, 4, 7): $$ \vec{d} = (9-1)\vec{i} + (4-1)\vec{j} + (7-1)\vec{k} = 8\vec{i} + 3\vec{j} + 6\vec{k} $$ Now, find the work done: $$ W = \vec{F} \cdot \vec{d} = (5\vec{i} - 3\vec{j} + 3\vec{k}) \cdot (8\vec{i} + 3\vec{j} + 6\vec{k}) = (5)(8) + (-3)(3) + (3)(6) = 40 - 9 + 18 = 49 $$ So, the work done is 49 Joules.

5. Find the area of the parallelogram spanned by vectors $\vec{u}=2\vec{i}+3\vec{j}+4\vec{k}$ and $\vec{v}=\vec{i}+2\vec{j}+3\vec{k}$ The area of the parallelogram is given by the magnitude of the cross product of the two vectors: $A = ||\vec{u} \times \vec{v}||$. First, find the cross product $\vec{u} \times \vec{v}$: $$ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 2 & 3 & 4 \ 1 & 2 & 3 \end{vmatrix} = (3\cdot3 - 4\cdot2)\vec{i} - (2\cdot3 - 4\cdot1)\vec{j} + (2\cdot2 - 3\cdot1)\vec{k} = (9-8)\vec{i} - (6-4)\vec{j} + (4-3)\vec{k} = \vec{i} - 2\vec{j} + \vec{k} $$ Now, find the magnitude of the cross product: $$ ||\vec{u} \times \vec{v}|| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} $$ So, the area of the parallelogram is $\sqrt{6}$ square units.

6. Find the area of the triangle with vertices A(3, -1, 2), B(1, -1, -3), and C(4, -3, 1)

The area of the triangle formed by three points A, B, and C is given by half the magnitude of the cross product of the vectors $\vec{AB}$ and $\vec{AC}$: $A = \frac{1}{2}||\vec{AB} \times \vec{AC}||$. First, find the vectors $\vec{AB}$ and $\vec{AC}$: $$ \vec{AB} = (1-3)\vec{i} + (-1-(-1))\vec{j} + (-3-2)\vec{k} = -2\vec{i} + 0\vec{j} - 5\vec{k} $$ $$ \vec{AC} = (4-3)\vec{i} + (-3-(-1))\vec{j} + (1-2)\vec{k} = \vec{i} - 2\vec{j} - \vec{k} $$ Now, find the cross product $\vec{AB} \times \vec{AC}$: $$ \vec{AB} \times \vec{AC} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ -2 & 0 & -5 \ 1 & -2 & -1 \end{vmatrix} = (0\cdot(-1) - (-5)\cdot(-2))\vec{i} - ((-2)\cdot(-1) - (-5)\cdot1)\vec{j} + ((-2)\cdot(-2) - 0\cdot1)\vec{k} = (0-10)\vec{i} - (2+5)\vec{j} + (4-0)\vec{k} = -10\vec{i} - 7\vec{j} + 4\vec{k} $$ Find the magnitude of the cross product: $$ ||\vec{AB} \times \vec{AC}|| = \sqrt{(-10)^2 + (-7)^2 + 4^2} = \sqrt{100 + 49 + 16} = \sqrt{165} $$ Finally, find the area of the triangle: $$ A = \frac{1}{2}||\vec{AB} \times \vec{AC}|| = \frac{1}{2}\sqrt{165} $$ So, the area of the triangle is $\frac{\sqrt{165}}{2}$ square units.