TUGAS VEKTOR

 Problem 1

Find the sum of the vectors $\displaystyle \overrightarrow{u}=⟨2,-1⟩$ and $\displaystyle \overrightarrow{v}=⟨3,5⟩$

Solution:
$\displaystyle \overrightarrow{u}+\overrightarrow{v}=⟨u_{1}+v_{1},u_{2}+v_{2}⟩=⟨2+3,-1+5⟩$

$\displaystyle \overrightarrow{u}+\overrightarrow{v}=⟨5,4⟩$

Problem 2

Given vectors $\displaystyle \overrightarrow{u}=⟨2,-1,3⟩$   $\displaystyle \overrightarrow{v}=⟨3,5,0⟩$    $\displaystyle \overrightarrow{w}=⟨-2,0,3⟩$

$\displaystyle \alpha =2\qquad \beta =-2\qquad \gamma =4$
Find $\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w} =$

Solution:

$\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w}=2⟨2,-1,3⟩+(-2)⟨3,5,0⟩+4⟨-2,0,3⟩=⟨4,-2,6⟩+⟨-6,-10,0⟩+⟨-8,0,12⟩$

$\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w}=⟨-10,-12,18⟩$

Problem 3

Find the direction of the vector $\displaystyle \overrightarrow{v}=⟨3,5⟩$

Solution:

The direction of a vector is the angle that the vector forms with the positive direction of the axis $\displaystyle X$. We can calculate the direction by the relationships that exist in a right triangle.
$\displaystyle \tan \alpha =\frac{y}{x}\Longrightarrow \alpha =\arctan (\frac{y}{x})$ being $\displaystyle \overrightarrow{u}=⟨x,y⟩$
Then, we have

$\displaystyle \alpha =\arctan (\frac{5}{3})=1.0304$

Problem 4

Find the magnitude of the vector $\displaystyle \alpha \overrightarrow{u}$ if $\displaystyle \alpha =2$ and $\displaystyle \overrightarrow{u}=⟨-2,4,1⟩$

Solution:
Look at the image.

$\displaystyle \alpha \overrightarrow{u}=2\cdot⟨-2,4,1⟩ =⟨-4,8,2⟩$
$\displaystyle \left\vert \alpha \overrightarrow{u}\right\vert =\left\vert ⟨-4,8,2⟩\right\vert =\sqrt{(-4)^{2}+(8)^{2}+(2)^{2}}=\sqrt{16+64+4}$

$\displaystyle \left\vert \alpha \overrightarrow{u}\right\vert =\sqrt{84}=2\sqrt{21}$

Problem 5

Given two vectors $\displaystyle \overrightarrow{u}=⟨2,3,1⟩$ and $\displaystyle \overrightarrow{v}=⟨-2,5,1⟩.$
Find the magnitude of the vector $\displaystyle \overrightarrow{z}=2\overrightarrow{u}-3\overrightarrow{v}$

Solution:

$\displaystyle \overrightarrow{z}=2\overrightarrow{u}-3\overrightarrow{v}\Longrightarrow \overrightarrow{z}=2⟨2,3,1⟩-3⟨-2,5,1⟩=⟨4,6,2⟩+⟨6,-15,-3⟩=⟨10,-9,-1⟩$
The magnitude of $\displaystyle \overrightarrow{z} = \left\vert \overrightarrow{z}\right\vert =\sqrt{10^{2}+(-9)^{2}+(-1)^{2}}=\sqrt{100+81+1}=\sqrt{182}$

Problem 6

Find the magnitude of $\displaystyle \overrightarrow{v}$.

Solution:
The coordinates of each vector are:

$\displaystyle \overrightarrow{u}=⟨4,4⟩\qquad \overrightarrow{v}=⟨-4,3⟩\qquad \overrightarrow{w}=⟨-5,-4⟩$

$\displaystyle \left\vert \overrightarrow{u}\right\vert =\sqrt{4^{2}+4^{2}}=\sqrt{32}=4\sqrt{2}$

$\displaystyle \left\vert \overrightarrow{v}\right\vert =\sqrt{(-4)^{2}+3^{2}}=\sqrt{25}=5$

$\displaystyle \left\vert \overrightarrow{w}\right\vert =\sqrt{(-5)^{2}+(-4)^{2}}=\sqrt{41}$

Problem 7

Find the vector sum $\displaystyle \overrightarrow{AB}+\overrightarrow{CA}$, if $\displaystyle A=⟨2,0,-3⟩$, $\displaystyle B=⟨1,1,-2⟩$ and $\displaystyle C=⟨0,3,2⟩$

Solution:

$\displaystyle \overrightarrow{AB}=\overrightarrow{B-A}=⟨1,1,-2⟩-⟨2,0,-3⟩=⟨-1,1,1⟩$

$\displaystyle \overrightarrow{CA}=$ $\displaystyle \overrightarrow{A-C}=⟨2,0,-3⟩-⟨0,3,2⟩=⟨2,-3,-5⟩$

Now $\displaystyle \overrightarrow{AB}+\overrightarrow{CA}=⟨-1,1,1⟩+⟨2,-3,-5⟩=⟨1,-2,-4⟩$

Problem 8

Find the sum of the vectors $\displaystyle \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}$.

Solution:

$\displaystyle \overrightarrow{u}=⟨4,4⟩$
$\displaystyle \overrightarrow{v}=⟨-4,3⟩$
$\displaystyle \overrightarrow{w}=⟨-5,-4⟩$

$\displaystyle \overrightarrow{u}+\overrightarrow{v}=⟨0,7⟩$
$\displaystyle \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=\left( \overrightarrow{u}+\overrightarrow{v}\right) +\overrightarrow{w}=⟨0,7⟩+⟨-5,-4⟩=⟨-5,3⟩$

$\displaystyle \overrightarrow{u}+\overrightarrow{v}+\overrightarrow{w}=⟨-5,3⟩$

Problem 9

Using triangle method graph $\displaystyle \overrightarrow{u}-\overrightarrow{v}$

Solution:
First off, we must change the orientation of the vector $\displaystyle \overrightarrow{v}$

We get the vector $\displaystyle -\overrightarrow{v}$ and then we apply the triangle rule - $\displaystyle \overrightarrow{u}+(-\overrightarrow{v})$

Note that the yellow vector represents the sum $\displaystyle \overrightarrow{u}+(-\overrightarrow{v})=\overrightarrow{u}-\overrightarrow{v}$

Problem 10

Given three vectors $\displaystyle \overrightarrow{u}=⟨2,3,1⟩$, $\displaystyle \overrightarrow{v}=⟨-2,5,1⟩$ and $\displaystyle \overrightarrow{z}=\overrightarrow{u}-2\overrightarrow{v}$. Find the magnitude of the vector $\displaystyle 3\overrightarrow{z}$

Solution:

$\displaystyle \overrightarrow{z}=\overrightarrow{u}-2\overrightarrow{v}\Longrightarrow \overrightarrow{z}=⟨2,3,1⟩-2⟨-2,5,1⟩=⟨6,-7,-1⟩$

$\displaystyle 3\overrightarrow{z}=⟨18,-21,-3⟩$
$\displaystyle \left\vert 3\overrightarrow{z}\right\vert =\sqrt{18^{2}+(-21)^{2}+(-3)^{2}}=\sqrt{324+441+9}=\sqrt{774}=3\sqrt{86}$

Problem 11

Find the resultant vector of $\displaystyle \overrightarrow{u}-2\overrightarrow{v}+3\overrightarrow{w}$.

Solution:

$\displaystyle \overrightarrow{u}=⟨-4,3⟩$
$\displaystyle \overrightarrow{v}=⟨1,-2⟩$
$\displaystyle \overrightarrow{w}=⟨1,4⟩$

$\displaystyle \overrightarrow{u}-2\overrightarrow{v}+3\overrightarrow{w}=⟨-4,3⟩-2⟨1,-2⟩+3⟨1,4⟩=⟨-3,19⟩$

Problem 12

Find the vector $\displaystyle \left( \left\vert \overrightarrow{u}\right\vert +\left\vert \overrightarrow{v}\right\vert-\left\vert \overrightarrow{w}\right\vert \right) \left( \overrightarrow{w}-\overrightarrow{u}\right)$

Solution:

$\displaystyle \left\vert \overrightarrow{u}\right\vert =\sqrt{(-4)^{2}+3^{2}}=5$
$\displaystyle \left\vert \overrightarrow{v}\right\vert =\sqrt{(1)^{2}+(-2)^{2}}=\sqrt{5}$
$\displaystyle \left\vert \overrightarrow{w}\right\vert =\sqrt{(1)^{2}+(4)^{2}}=\sqrt{17}$

$\displaystyle \overrightarrow{w}-\overrightarrow{u}=⟨1,4⟩-⟨-4,3⟩=⟨5,1⟩$

$\displaystyle \left( \left\vert \overrightarrow{u}\right\vert +\left\vert \overrightarrow{v}\right\vert -\left\vert \overrightarrow{w}\right\vert \right) \left( \overrightarrow{w}-\overrightarrow{u}\right) =(5+\sqrt{5}-\sqrt{17})⟨5,1⟩$

Problem 13

Find the direction of the vector
$\displaystyle -\left\vert \overrightarrow{u}\right\vert \cdot \overrightarrow{u}$.
$\displaystyle \overrightarrow{u}=⟨-2,3⟩$

Solution:

$\displaystyle \left\vert \overrightarrow{u}\right\vert =\sqrt{(-2)^{2}+3^{2}}=\sqrt{13}$

$\displaystyle -\left\vert \overrightarrow{u}\right\vert \cdot \overrightarrow{u}=-\sqrt{13}⟨-2,3⟩=⟨2\sqrt{13},-3\sqrt{13}⟩$

The vector direction is
$\displaystyle \alpha =\arctan (\frac{-3\sqrt{13}}{2\sqrt{13}})=\arctan (\frac{-3}{2})$

Problem 14

Find the vector $\displaystyle \left( \frac{\left\vert \overrightarrow{u}\right\vert +\left\vert \overrightarrow{v}\right\vert }{\left\vert \overrightarrow{w}\right\vert }\right) \left( \overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right)$

Solution:
$\displaystyle \overrightarrow{u}=⟨4,2⟩$
$\displaystyle \overrightarrow{v}=⟨0,5⟩$
$\displaystyle \overrightarrow{w}=⟨-5,1⟩$

$\displaystyle \left\vert \overrightarrow{u}\right\vert =\sqrt{4^{2}+2^{2}}=\sqrt{20}=2\sqrt{5}$
$\displaystyle \left\vert \overrightarrow{v}\right\vert =\sqrt{0^{2}+5^{2}}=5$
$\displaystyle \left\vert \overrightarrow{w}\right\vert =\sqrt{(-5)^{2}+1^{2}}=\sqrt{26}$

$\displaystyle \left( \frac{\left\vert \overrightarrow{u}\right\vert +\left\vert \overrightarrow{v}\right\vert }{\left\vert \overrightarrow{w}\right\vert }\right) \left( \overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}\right) =\left( \frac{2\sqrt{5}+5}{\sqrt{26}}\right) (⟨4,2⟩-⟨0,5⟩+⟨-5,1⟩)=\left( \frac{2\sqrt{5}+5}{\sqrt{26}}\right) ⟨-1,-2⟩$

Problem 15

The coordinates of the points A, B and C are
$\displaystyle A=(-1,1,2)$, $\displaystyle B=(1,-2,3)$, $\displaystyle C=(-2,1,1)$.
Find the vector sum of $\displaystyle \overrightarrow{u}+\overrightarrow{v}-\overrightarrow{w}$
if $\displaystyle \overrightarrow{u}=\overrightarrow{AB},\overrightarrow{v}=\overrightarrow{CB},\overrightarrow{w}=\overrightarrow{CA}$

Solution:

$\displaystyle \overrightarrow{u}=\overrightarrow{AB}=⟨1,-2,3⟩-⟨-1,1,2⟩=⟨2,-3,1⟩$

$\displaystyle \overrightarrow{v}=\overrightarrow{CB}=⟨1,-2,3⟩-⟨-2,1,1⟩=⟨3,-3,2⟩$

$\displaystyle \overrightarrow{w}=\overrightarrow{CA}=⟨-1,1,2⟩-⟨-2,1,1⟩=⟨1,0,1⟩$

$\displaystyle \overrightarrow{u}+\overrightarrow{v}-\overrightarrow{w}=⟨2,-3,1⟩+⟨3,-3,2⟩-⟨1,0,1⟩=⟨4,-6,2⟩$

Problem 16

Given points A, B and C and constants $\displaystyle \alpha, \beta, \gamma$ such that $\displaystyle A=(-1,2,1)$, $\displaystyle B=(0,2,-3)$, $\displaystyle C=(1,2,-1)$ and $\displaystyle \alpha=-2$, $\displaystyle \beta=2$, $\displaystyle \gamma =3$

$\displaystyle \overrightarrow{u}=\overrightarrow{CB}$
$\displaystyle \overrightarrow{v}=\overrightarrow{AC}$
$\displaystyle \overrightarrow{w}=\overrightarrow{BA}$
Find the vector $\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w}$

Solution:

$\displaystyle \overrightarrow{u}=\overrightarrow{CB}=⟨0,2,-3⟩-⟨1,2,-1⟩=⟨-1,0,-2⟩$
$\displaystyle \overrightarrow{v}=\overrightarrow{AC}=⟨1,2,-1⟩-⟨-1,2,1⟩=⟨2,0,-2⟩$
$\displaystyle \overrightarrow{w}=\overrightarrow{BA}=⟨-1,2,1⟩-⟨0,2,-3⟩=⟨-1,0,4⟩$

$\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w} =-2⟨-1,0,-2⟩+2⟨2,0,-2⟩+3⟨-1,0,4⟩=$

$\displaystyle =⟨2,0,4⟩+⟨4,0,-4⟩+⟨-3,0,12⟩=⟨3,0,12⟩$

$\displaystyle \alpha \overrightarrow{u}+\beta \overrightarrow{v}+\gamma \overrightarrow{w}=⟨3,0,12⟩$