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[SOLVED] Vectors in a box

karush

Well-known member
Jan 31, 2012
2,720
The following diagram show a solid figure ABCDEFGH. Each of the six faces is a parallelogram

View attachment 1034
The coordinates of $A$ and $B$ are $A(7, -3, -5), B(17, 2, 5)$

a) Find (i) $\vec{AB}$ and (ii) $|AB|$

$\vec{AB}=A+B=(24, -1, 0)$

$|AB|=\sqrt{(17-7)^2+(-3-2)^2+(-5-5)^2}=15$

was assuming that $A$ and $B$ have origin of zero.
there are $7$ more question to this problem but thot I would see if this starting out right since the rest of it is built on this.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: vectors in a box

No, [tex]\displaystyle \begin{align*} \mathbf{AB} \end{align*}[/tex] is NOT [tex]\displaystyle \begin{align*} \mathbf{A} + \mathbf{B} \end{align*}[/tex]. Rather

[tex]\displaystyle \begin{align*} \mathbf{AB} &= \mathbf{AO} + \mathbf{OB} \\ &= -\mathbf{OA} + \mathbf{OB} \\ &= \mathbf{OB} - \mathbf{OA} \end{align*}[/tex]
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: vectors in a box

You have found the magnitude correctly, but to write the vector $\vec{AB}$, you want to use:

\(\displaystyle \vec{AB}=\left\langle x_B-x_A,y_B-y_A,z_B-z_A \right\rangle\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: vectors in a box

The following diagram show a solid figure ABCDEFGH. Each of the six faces is a parallelogram

View attachment 1034
The coordinates of $A$ and $B$ are $A(7, -3, -5), B(17, 2, 5)$

a) Find (i) $\vec{AB}$ and (ii) $|AB|$

$\vec{AB}=A+B=(24, -1, 0)$

$|AB|=\sqrt{(17-7)^2+(-3-2)^2+(-5-5)^2}=15$

was assuming that $A$ and $B$ have origin of zero.
there are $7$ more question to this problem but thot I would see if this starting out right since the rest of it is built on this.
You have found the magnitude correctly, but to write the vector $\vec{AB}$, you want to use:

\(\displaystyle \vec{AB}=\left\langle x_B-x_A,y_B-y_A,z_B-z_A \right\rangle\)
No, to work out the magnitude correctly it should either be [tex]\displaystyle \begin{align*} \sqrt{ (17 - 7)^2 + [2 - (-3)]^2 + [5 -(-5)]^2} \end{align*}[/tex] or [tex]\displaystyle \begin{align*} \sqrt{(7 - 17)^2 + (-3 - 2)^2 + (-5 - 5)^2} \end{align*}[/tex]. I realise that you get the same result, but the method is wrong.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: vectors in a box

No, to work out the magnitude correctly it should either be [tex]\displaystyle \begin{align*} \sqrt{ (17 - 7)^2 + [2 - (-3)]^2 + [5 -(-5)]^2} \end{align*}[/tex] or [tex]\displaystyle \begin{align*} \sqrt{(7 - 17)^2 + (-3 - 2)^2 + (-5 - 5)^2} \end{align*}[/tex]. I realise that you get the same result, but the method is wrong.
Good catch...I should have said, "You have the correct magnitude" instead. Time for bed. (Sleepy)
 

karush

Well-known member
Jan 31, 2012
2,720
Re: vectors in a box

so $\vec{AB} = \langle 17-7, 2+3, 5+5 \rangle = \langle 10, 5, 10 \rangle$
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: vectors in a box

so $\vec{AB} = \langle 17-7, 2+3, 0+5 \rangle = \langle 10, 5, 5 \rangle$
No, you want:

\(\displaystyle \vec{AB}=\langle 17-7, 2-(-3), 5-(-5) \rangle=?\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: vectors in a box

Good catch...I should have said, "You have the correct magnitude" instead. Time for bed. (Sleepy)
Sweet dreams :)
 

karush

Well-known member
Jan 31, 2012
2,720
Re: vectors in a box

No, you want:

\(\displaystyle \vec{AB}=\langle 17-7, 2-(-3), 5-(-5) \rangle=?\)
$\vec{AB} = \langle 17-7, 2+3, 5+5 \rangle = \langle 10, 5, 10 \rangle$


there are 7 more ? on this
 
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