MHB O=(0,0,0) a=(1,-2,3) b=(-3,4,2)

[karush]

View attachment 1158

(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

 

[Evgeny.Makarov]

(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 $B$.

(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.

(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 $s=2$.

(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 $p$.

 

[karush]

$L_2$ 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

 

[Evgeny.Makarov]

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.

 

[karush]

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
$\displaystyle \cos{\theta} = \frac{\vec{AB}\cdot\vec{OA}}{|\vec{AB}|\ |\vec{OA}|}=
\frac{-19}{\sqrt{742}}$

$\displaystyle cos^{-1}\left(\frac{-19}{\sqrt{742}}\right)\approx 134^o$

 

[Evgeny.Makarov]

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 $(k,-k,-5)$.

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.