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$\text{a}=2\text{u}+\text{v}-\mathbf{\text{w}},$ where $\text{u}=\text{i}-\text{k},$ $\text{v}=2\text{j},$ and $\text{w}=\text{i}-\text{j}$
Determine whether $\overrightarrow{AB}$ and $\overrightarrow{PQ}$ are equivalent vectors, where $A\left(1,1,1\right),B\left(3,3,3\right),P\left(1,4,5\right),$ and $Q\left(3,6,7\right).$
Equivalent vectors
Determine whether the vectors $\overrightarrow{AB}$ and $\overrightarrow{PQ}$ are equivalent, where $A\left(1,4,1\right),$ $B\left(\mathrm{-2},2,0\right),$ $P\left(2,5,7\right),$ and $Q\left(\mathrm{-3},2,1\right).$
For the following exercises, find vector $\text{u}$ with a magnitude that is given and satisfies the given conditions.
$\text{v}=\u27e87,\mathrm{-1},3\u27e9,$ $\Vert \text{u}\Vert =10,$ $\text{u}$ and $\text{v}$ have the same direction
$\text{u}=\u27e8\frac{70}{\sqrt{59}},-\frac{10}{\sqrt{59}},\frac{30}{\sqrt{59}}\u27e9$
$\text{v}=\u27e82,4,1\u27e9,$ $\Vert \text{u}\Vert =15,$ $\text{u}$ and $\text{v}$ have the same direction
$\text{v}=\u27e82\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,1\u27e9,$ $\Vert \text{u}\Vert =2,$ $\text{u}$ and $\text{v}$ have opposite directions for any $t,$ where $t$ is a real number
$\text{u}=\u27e8-\frac{4}{\sqrt{5}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,-\frac{4}{\sqrt{5}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,-\frac{2}{\sqrt{5}}\u27e9$
$\text{v}=\u27e83\phantom{\rule{0.2em}{0ex}}\text{sinh}\phantom{\rule{0.2em}{0ex}}t,0,3\u27e9,$ $\Vert \text{u}\Vert =5,$ $\text{u}$ and $\text{v}$ have opposite directions for any $t,$ where $t$ is a real number
Determine a vector of magnitude $5$ in the direction of vector $\overrightarrow{AB},$ where $A(2,1,5)$ and $B(3,4,\mathrm{-7}).$
$\u27e8\frac{5}{\sqrt{154}},\frac{15}{\sqrt{154}},-\frac{60}{\sqrt{154}}\u27e9$
Find a vector of magnitude $2$ that points in the opposite direction than vector $\overrightarrow{AB},$ where $A(\mathrm{-1},\mathrm{-1},1)$ and $B(0,1,1).$ Express the answer in component form.
Consider the points $A\left(2,\alpha ,0\right),B\left(0,1,\beta \right),$ and $C\left(1,1,\beta \right),$ where $\alpha $ and $\beta $ are negative real numbers. Find $\alpha $ and $\beta $ such that $\Vert \overrightarrow{OA}-\overrightarrow{OB}+\overrightarrow{OC}\Vert =\Vert \overrightarrow{OB}\Vert =4.$
$\alpha =\text{\u2212}\sqrt{7},$ $\beta =\text{\u2212}\sqrt{15}$
Consider points $A\left(\alpha ,0,0\right),B\left(0,\beta ,0\right),$ and $C\left(\alpha ,\beta ,\beta \right),$ where $\alpha $ and $\beta $ are positive real numbers. Find $\alpha $ and $\beta $ such that $\Vert \stackrel{\u2014}{OA}+\stackrel{\u2014}{OB}\Vert =\sqrt{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\Vert \stackrel{\u2014}{OC}\Vert =\sqrt{3}.$
Let $P\left(x,y,z\right)$ be a point situated at an equal distance from points $A\left(1,\mathrm{-1},0\right)$ and $B\left(\mathrm{-1},2,1\right).$ Show that point $P$ lies on the plane of equation $\mathrm{-2}x+3y+z=2.$
Let $P\left(x,y,z\right)$ be a point situated at an equal distance from the origin and point $A\left(4,1,2\right).$ Show that the coordinates of point $P$ satisfy the equation $8x+2y+4z=21.$
The points $A,B,$ and $C$ are collinear (in this order) if the relation $\Vert \overrightarrow{AB}\Vert +\Vert \overrightarrow{BC}\Vert =\Vert \overrightarrow{AC}\Vert $ is satisfied. Show that $A(5,3,\mathrm{-1}),$ $B(\mathrm{-5},\mathrm{-3},1),$ and $C(\mathrm{-15},\mathrm{-9},3)$ are collinear points.
Show that points $A(1,0,1),$ $B(0,1,1),$ and $C(1,1,1)$ are not collinear.
[T] A force $\text{F}$ of $50\phantom{\rule{0.2em}{0ex}}\text{N}$ acts on a particle in the direction of the vector $\overrightarrow{OP},$ where $P(3,4,0).$
a. $\text{F}=\u27e830,40,0\u27e9;$ b. $53\text{\xb0}$
[T] A force $\text{F}$ of $40\phantom{\rule{0.2em}{0ex}}\text{N}$ acts on a box in the direction of the vector $\overrightarrow{OP},$ where $P(1,0,2).$
If $\text{F}$ is a force that moves an object from point ${P}_{1}\left({x}_{1},{y}_{1},{z}_{1}\right)$ to another point ${P}_{2}\left({x}_{2},{y}_{2},{z}_{2}\right),$ then the displacement vector is defined as $\mathbf{\text{D}}=\left({x}_{2}-{x}_{1}\right)\text{i}+\left({y}_{2}-{y}_{1}\right)\mathbf{\text{j}}+\left({z}_{2}-{z}_{1}\right)\mathbf{\text{k}}.$ A metal container is lifted $10$ m vertically by a constant force $\text{F}.$ Express the displacement vector $\text{D}$ by using standard unit vectors.
$\mathbf{\text{D}}=10\text{k}$
A box is pulled $4$ yd horizontally in the x -direction by a constant force $\text{F}.$ Find the displacement vector in component form.
The sum of the forces acting on an object is called the resultant or net force . An object is said to be in static equilibrium if the resultant force of the forces that act on it is zero. Let ${\text{F}}_{1}=\u27e810,6,3\u27e9,$ ${\text{F}}_{2}=\u27e80,4,9\u27e9,$ and ${\text{F}}_{3}=\u27e810,\mathrm{-3},\mathrm{-9}\u27e9$ be three forces acting on a box. Find the force ${\text{F}}_{4}$ acting on the box such that the box is in static equilibrium. Express the answer in component form.
${\text{F}}_{4}=\u27e8\mathrm{-20},\mathrm{-7},\mathrm{-3}\u27e9$
[T] Let ${\text{F}}_{k}=\u27e81,k,{k}^{2}\u27e9,$ $k=1\text{,...},n$ be $n$ forces acting on a particle, with $n\ge 2.$
The force of gravity $\text{F}$ acting on an object is given by $\text{F}=m\text{g},$ where m is the mass of the object (expressed in kilograms) and $\text{g}$ is acceleration resulting from gravity, with $\Vert \mathbf{\text{g}}\Vert =9.8$ $\text{N/kg}.$ A 2-kg disco ball hangs by a chain from the ceiling of a room.
a. $\text{F}=\mathrm{-19.6}\text{k},$ $\Vert \mathbf{\text{F}}\Vert =19.6$ N; b. $\mathbf{\text{T}}=19.6\text{k},$ $\Vert \mathbf{\text{T}}\Vert =19.6$ N
A 5-kg pendant chandelier is designed such that the alabaster bowl is held by four chains of equal length, as shown in the following figure.
[T] A 30-kg block of cement is suspended by three cables of equal length that are anchored at points $P(\mathrm{-2},0,0),$ $Q\left(1,\sqrt{3},0\right),$ and $R\left(1,\text{\u2212}\sqrt{3},0\right).$ The load is located at $S\left(0,0,\mathrm{-2}\sqrt{3}\right),$ as shown in the following figure. Let ${\text{F}}_{1},$ ${\text{F}}_{2},$ and ${\text{F}}_{3}$ be the forces of tension resulting from the load in cables $RS,QS,$ and $PS,$ respectively.
a. $\text{F}=\mathrm{-294}\text{k}$ N; b. ${\text{F}}_{1}=\u27e8-\frac{49\sqrt{3}}{3},49,\mathrm{-98}\u27e9,$ ${\text{F}}_{2}=\u27e8-\frac{49\sqrt{3}}{3},\mathrm{-49},\mathrm{-98}\u27e9,$ and ${\text{F}}_{3}=\u27e8\frac{98\sqrt{3}}{3},0,\mathrm{-98}\u27e9$ (each component is expressed in newtons)
Two soccer players are practicing for an upcoming game. One of them runs 10 m from point A to point B . She then turns left at $90\text{\xb0}$ and runs 10 m until she reaches point C . Then she kicks the ball with a speed of 10 m/sec at an upward angle of $45\text{\xb0}$ to her teammate, who is located at point A . Write the velocity of the ball in component form.
Let $\mathbf{\text{r}}(t)=\u27e8x(t),y(t),z(t)\u27e9$ be the position vector of a particle at the time $t\in [0,T],$ where $x,y,$ and $z$ are smooth functions on $[0,T].$ The instantaneous velocity of the particle at time $t$ is defined by vector $\text{v}(t)=\u27e8x\text{\u2032}(t),y\text{\u2032}(t),z\text{\u2032}(t)\u27e9,$ with components that are the derivatives with respect to $t,$ of the functions x , y , and z , respectively. The magnitude $\Vert \text{v}(t)\Vert $ of the instantaneous velocity vector is called the speed of the particle at time t. Vector $\text{a}(t)=\u27e8x\text{\u2033}(t),y\text{\u2033}(t),z\text{\u2033}(t)\u27e9,$ with components that are the second derivatives with respect to $t,$ of the functions $x,y,$ and $z,$ respectively, gives the acceleration of the particle at time $t.$ Consider $\mathbf{\text{r}}(t)=\u27e8\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,2t\u27e9$ the position vector of a particle at time $t\in [0,30],$ where the components of $\text{r}$ are expressed in centimeters and time is expressed in seconds.
a. $\text{v}(1)=\u27e8\mathrm{-0.84},0.54,2\u27e9$ (each component is expressed in centimeters per second); $\Vert \text{v}(1)\Vert =2.24$ (expressed in centimeters per second); $\text{a}(1)=\u27e8\mathrm{-0.54},\mathrm{-0.84},0\u27e9$ (each component expressed in centimeters per second squared);
b.
[T] Let $\mathbf{\text{r}}(t)=\u27e8t,2{t}^{2},4{t}^{2}\u27e9$ be the position vector of a particle at time $t$ (in seconds), where $t\in [0,10]$ (here the components of $\text{r}$ are expressed in centimeters).
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