# 2.2 Vectors in three dimensions  (Page 8/14)

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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 $\stackrel{\to }{AB}$ and $\stackrel{\to }{PQ}$ 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 $\stackrel{\to }{AB}$ and $\stackrel{\to }{PQ}$ are equivalent, where $A\left(1,4,1\right),$ $B\left(-2,2,0\right),$ $P\left(2,5,7\right),$ and $Q\left(-3,2,1\right).$

For the following exercises, find vector $\text{u}$ with a magnitude that is given and satisfies the given conditions.

$\text{v}=⟨7,-1,3⟩,$ $‖\text{u}‖=10,$ $\text{u}$ and $\text{v}$ have the same direction

$\text{u}=⟨\frac{70}{\sqrt{59}},-\frac{10}{\sqrt{59}},\frac{30}{\sqrt{59}}⟩$

$\text{v}=⟨2,4,1⟩,$ $‖\text{u}‖=15,$ $\text{u}$ and $\text{v}$ have the same direction

$\text{v}=⟨2\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,2\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,1⟩,$ $‖\text{u}‖=2,$ $\text{u}$ and $\text{v}$ have opposite directions for any $t,$ where $t$ is a real number

$\text{u}=⟨-\frac{4}{\sqrt{5}}\text{sin}\phantom{\rule{0.2em}{0ex}}t,-\frac{4}{\sqrt{5}}\text{cos}\phantom{\rule{0.2em}{0ex}}t,-\frac{2}{\sqrt{5}}⟩$

$\text{v}=⟨3\phantom{\rule{0.2em}{0ex}}\text{sinh}\phantom{\rule{0.2em}{0ex}}t,0,3⟩,$ $‖\text{u}‖=5,$ $\text{u}$ 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 $\stackrel{\to }{AB},$ where $A\left(2,1,5\right)$ and $B\left(3,4,-7\right).$

$⟨\frac{5}{\sqrt{154}},\frac{15}{\sqrt{154}},-\frac{60}{\sqrt{154}}⟩$

Find a vector of magnitude $2$ that points in the opposite direction than vector $\stackrel{\to }{AB},$ where $A\left(-1,-1,1\right)$ and $B\left(0,1,1\right).$ 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 $‖\stackrel{\to }{OA}-\stackrel{\to }{OB}+\stackrel{\to }{OC}‖=‖\stackrel{\to }{OB}‖=4.$

$\alpha =\text{−}\sqrt{7},$ $\beta =\text{−}\sqrt{15}$

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 $‖\stackrel{—}{OA}+\stackrel{—}{OB}‖=\sqrt{2}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}‖\stackrel{—}{OC}‖=\sqrt{3}.$

Let $P\left(x,y,z\right)$ be a point situated at an equal distance from points $A\left(1,-1,0\right)$ and $B\left(-1,2,1\right).$ Show that point $P$ lies on the plane of equation $-2x+3y+z=2.$

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 $‖\stackrel{\to }{AB}‖+‖\stackrel{\to }{BC}‖=‖\stackrel{\to }{AC}‖$ is satisfied. Show that $A\left(5,3,-1\right),$ $B\left(-5,-3,1\right),$ and $C\left(-15,-9,3\right)$ are collinear points.

Show that points $A\left(1,0,1\right),$ $B\left(0,1,1\right),$ and $C\left(1,1,1\right)$ 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 $\stackrel{\to }{OP},$ where $P\left(3,4,0\right).$

1. Express the force as a vector in component form.
2. Find the angle between force $\text{F}$ and the positive direction of the x -axis. Express the answer in degrees rounded to the nearest integer.

a. $\text{F}=⟨30,40,0⟩;$ b. $53\text{°}$

[T] A force $\text{F}$ of $40\phantom{\rule{0.2em}{0ex}}\text{N}$ acts on a box in the direction of the vector $\stackrel{\to }{OP},$ where $P\left(1,0,2\right).$

1. Express the force as a vector by using standard unit vectors.
2. Find the angle between force $\text{F}$ and the positive direction of the x -axis.

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 ${\text{F}}_{1}=⟨10,6,3⟩,$ ${\text{F}}_{2}=⟨0,4,9⟩,$ and ${\text{F}}_{3}=⟨10,-3,-9⟩$ 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.

${\text{F}}_{4}=⟨-20,-7,-3⟩$

[T] Let ${\text{F}}_{k}=⟨1,k,{k}^{2}⟩,$ $k=1\text{,...},n$ be $n$ forces acting on a particle, with $n\ge 2.$

1. Find the net force $\text{F}=\sum _{k=1}^{n}{F}_{k}.$ Express the answer using standard unit vectors.
2. Use a computer algebra system (CAS) to find n such that $‖\mathbf{\text{F}}‖<100.$

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 $‖\mathbf{\text{g}}‖=9.8$ $\text{N/kg}.$ A 2-kg disco ball hangs by a chain from the ceiling of a room.

1. Find the force of gravity $\text{F}$ acting on the disco ball and find its magnitude.
2. Find the force of tension $\text{T}$ in the chain and its magnitude.
Express the answers using standard unit vectors.

a. $\text{F}=-19.6\text{k},$ $‖\mathbf{\text{F}}‖=19.6$ N; b. $\mathbf{\text{T}}=19.6\text{k},$ $‖\mathbf{\text{T}}‖=19.6$ 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.

1. Find the magnitude of the force of gravity acting on the chandelier.
2. Find the magnitudes of the forces of tension for each of the four chains (assume chains are essentially vertical). [T] A 30-kg block of cement is suspended by three cables of equal length that are anchored at points $P\left(-2,0,0\right),$ $Q\left(1,\sqrt{3},0\right),$ and $R\left(1,\text{−}\sqrt{3},0\right).$ The load is located at $S\left(0,0,-2\sqrt{3}\right),$ 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.

1. Find the gravitational force $\text{F}$ acting on the block of cement that counterbalances the sum ${\text{F}}_{1}+{\mathbf{\text{F}}}_{2}+{\mathbf{\text{F}}}_{3}$ of the forces of tension in the cables.
2. Find forces ${\text{F}}_{1},$ ${\text{F}}_{2},$ and ${\text{F}}_{3}.$ Express the answer in component form. a. $\text{F}=-294\text{k}$ N; b. ${\text{F}}_{1}=⟨-\frac{49\sqrt{3}}{3},49,-98⟩,$ ${\text{F}}_{2}=⟨-\frac{49\sqrt{3}}{3},-49,-98⟩,$ and ${\text{F}}_{3}=⟨\frac{98\sqrt{3}}{3},0,-98⟩$ (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 $90\text{°}$ 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 $45\text{°}$ to her teammate, who is located at point A . Write the velocity of the ball in component form. Let $\mathbf{\text{r}}\left(t\right)=⟨x\left(t\right),y\left(t\right),z\left(t\right)⟩$ be the position vector of a particle at the time $t\in \left[0,T\right],$ where $x,y,$ and $z$ are smooth functions on $\left[0,T\right].$ The instantaneous velocity of the particle at time $t$ is defined by vector $\text{v}\left(t\right)=⟨x\text{′}\left(t\right),y\text{′}\left(t\right),z\text{′}\left(t\right)⟩,$ with components that are the derivatives with respect to $t,$ of the functions x , y , and z , respectively. The magnitude $‖\text{v}\left(t\right)‖$ of the instantaneous velocity vector is called the speed of the particle at time t. Vector $\text{a}\left(t\right)=⟨x\text{″}\left(t\right),y\text{″}\left(t\right),z\text{″}\left(t\right)⟩,$ 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 $\mathbf{\text{r}}\left(t\right)=⟨\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,2t⟩$ the position vector of a particle at time $t\in \left[0,30\right],$ where the components of $\text{r}$ are expressed in centimeters and time is expressed in seconds.

1. Find the instantaneous velocity, speed, and acceleration of the particle after the first second. Round your answer to two decimal places.
2. Use a CAS to visualize the path of the particle—that is, the set of all points of coordinates $\left(\text{cos}\phantom{\rule{0.2em}{0ex}}t,\text{sin}\phantom{\rule{0.2em}{0ex}}t,2t\right),$ where $t\in \left[0,30\right].$

a. $\text{v}\left(1\right)=⟨-0.84,0.54,2⟩$ (each component is expressed in centimeters per second); $‖\text{v}\left(1\right)‖=2.24$ (expressed in centimeters per second); $\text{a}\left(1\right)=⟨-0.54,-0.84,0⟩$ (each component expressed in centimeters per second squared);

b. [T] Let $\mathbf{\text{r}}\left(t\right)=⟨t,2{t}^{2},4{t}^{2}⟩$ be the position vector of a particle at time $t$ (in seconds), where $t\in \left[0,10\right]$ (here the components of $\text{r}$ are expressed in centimeters).

1. Find the instantaneous velocity, speed, and acceleration of the particle after the first two seconds. Round your answer to two decimal places.
2. Use a CAS to visualize the path of the particle defined by the points $\left(t,2{t}^{2},4{t}^{2}\right),$ where $t\in \left[0,60\right].$

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