MHB O=(0,0,0) a=(1,-2,3) b=(-3,4,2)

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The discussion revolves around vector notation and calculations involving points A, B, and O in a three-dimensional space. Participants clarify that the notation $B\hat{A}O$ likely refers to the angle between vectors BA and AO, which can be calculated using the law of cosines. There is confusion regarding the coordinates of point C, with suggestions that a typo may exist in the problem statement, as the calculations do not yield consistent results. The equations derived from equating coordinates lead to a system involving unknowns k and t, but participants struggle to resolve these equations correctly. Ultimately, the angle $BAO$ is computed, highlighting the importance of distinguishing between vector orientations in calculations.
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(a) (i) $\vec{B}-\vec{A}=\overrightarrow{AB}$

(ii) not sure what the notation $B\hat{A}O$ means?

(b) well if I understand this, if s=2 the one point will be
$\pmatrix {-3 \\ 4 \\ 2}$
and the other will be
$\pmatrix {-11 \\ 16 \\ 0}$
(c) and (d) frankly I am clueless
 
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karush said:
(a) (i) $\vec{B}-\vec{A}=\overrightarrow{AB}$
I have not encountered denoting vectors by one point, as in $\vec{B}$. Here $O$ is the origin, so you can say that $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$. I assumed you have verified this equality for concrete coordinates of $A$ and $B$.

karush said:
(ii) not sure what the notation $B\hat{A}O$ means?
Probably this means angle $BAO$. You can use the law of cosines to find it.

karush said:
(b) well if I understand this, if s=2 the one point will be
$\pmatrix {-3 \\ 4 \\ 2}$
and the other will be
$\pmatrix {-11 \\ 16 \\ 0}$
No, the first point you get when $s=0$ and the second when $s=2$.

karush said:
(c) and (d) frankly I am clueless
$L_2$ has equation $\pmatrix {x \\ y \\ z}=\overrightarrow{OA}+t\overrightarrow{OB}$. Equating this with the coordinates of $C$ gives you a system of 3 equations with 2 unknowns.

Similarly, when you replace $\pmatrix {x \\ y \\ z}$ with the coordinates of $C$ in (d), you'll get a system in $p$.
 
Evgeny.Makarov said:
$L_2$ has equation $\pmatrix {x \\ y \\ z}=\overrightarrow{OA}+t\overrightarrow{OB}$. Equating this with the coordinates of $C$ gives you a system of 3 equations with 2 unknowns.
If $C$ is $\pmatrix{k \\ -k \\ -5}$ and so
$\pmatrix{k \\ -k \\ -5}=\pmatrix{1 \\ -2 \\ 3}
+t\pmatrix{-3 \\ 4 \\ 2}$

I presume the $2$ unknowns are $k$ and $-k$ even tho they are the same variable $k$. also don't see where the 3 equations really are. Its probably so obvious that I don't see it.

however if {x, y, 5} = {1, -2, 3}+{-3, 4, 2}

then

{x, y, 5} = {-2, 2, 5}

so k=-2 and -k=2 ? if t=1
 
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karush said:
If $C$ is $\pmatrix{k \\ -k \\ -5}$ and so
$\pmatrix{k \\ -k \\ -5}=\pmatrix{1 \\ -2 \\ 3}
+t\pmatrix{-3 \\ 4 \\ 2}$

I presume the $2$ unknowns are $k$ and $-k$ even tho they are the same variable $k$. also don't see where the 3 equations really are.
The unknowns are $k$ and $t$. The first equation is $k=1+t(-3)$. The other two equations result from equating the second and third coordinates.
 
Evgeny.Makarov said:
The unknowns are $k$ and $t$. The first equation is $k=1+t(-3)$. The other two equations result from equating the second and third coordinates.

ok from $-5=3+t(2)$ we get $t=-4$

so plug that in for $t$

$k=1+(-4)(-3)=1+12=13$
$-k=-2+(-4)(4)=-2-16=-18$

this doesn't solve?

for (a)(ii) find $B\hat{A}O$ I used
$\displaystyle \cos{\theta} = \frac{\vec{AB}\cdot\vec{OA}}{|\vec{AB}|\ |\vec{OA}|}=
\frac{-19}{\sqrt{742}}$

$\displaystyle cos^{-1}\left(\frac{-19}{\sqrt{742}}\right)\approx 134^o$
 
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karush said:
ok from $-5=3+t(2)$ we get $t=-4$

so plug that in for $t$

$k=1+(-4)(-3)=1+12=13$
$-k=-2+(-4)(4)=-2-16=-18$

this doesn't solve?
Looks like it. Maybe there is a typo in the question and $C$ should have coordinates $(k,-k,5)$ instead of $(k,-k,-5)$.

karush said:
for (a)(ii) find $B\hat{A}O$ I used
$\displaystyle \cos{\theta} = \frac{\vec{AB}\cdot\vec{OA}}{|\vec{AB}|\ |\vec{OA}|}=
\frac{-19}{\sqrt{742}}$

$\displaystyle cos^{-1}\left(\frac{-19}{\sqrt{742}}\right)\approx 134^o$
Angle $BAO$ is the angle between $\vec{AB}$ and $\vec{AO}$, not $\vec{OA}$. Otherwise you get the complement of angle $BAO$ to $180^\circ$. Correspondingly, I get the same value of the cosine, but with the opposite sign.
 
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