MHB Let L1 be the line through P and Q

[karush]

Consider the points $P(2,-1,5)$ and $Q(3,-3,8)$, let $L_1$ be the line trough $P$ and $Q$

(a) Show that $\overrightarrow{PQ}=\pmatrix{ 1\cr -2\cr 3\cr}$

$\overrightarrow{PQ}=\pmatrix{3\cr -3\cr 8\cr}-\pmatrix{2\cr -1\cr 5\cr}$

(b) The line $L_1$ may be represented by $r=\pmatrix{3\cr -3\cr 8\cr}+s\pmatrix{1\cr -2\cr 3\cr}$

i don't know this notation but it looks like $r=Q+\overrightarrow{PQ}$ so we are taking a point and adding a vector to it?

more ? to come on this...

 

[Plato2]

Re: let L1 be the line throught P and Q

Consider the points $P(2,-1,5)$ and $Q(3,-3,8)$, let $L_1$ be the line trough $P$ and $Q$

(a) Show that $\overrightarrow{PQ}=\pmatrix{ 1\cr -2\cr 3\cr}$

$\overrightarrow{PQ}=\pmatrix{3\cr -3\cr 8\cr}-\pmatrix{2\cr -1\cr 5\cr}$

(b) The line $L_1$ may be represented by $r=\pmatrix{3\cr -3\cr 8\cr}+s\pmatrix{1\cr -2\cr 3\cr}$

i don't know this notation but it looks like $r=Q+\overrightarrow{PQ}$ so we are taking a point and adding a vector to it?

more ? to come on this...

$r(s)=Q+s\overrightarrow{PQ}$

 

[karush]

Re: let L1 be the line throught P and Q

Consider the points $P(2,-1,5)$ and $Q(3,-3,8)$, let $L_1$ be the line trough $P$ and $Q$

(a) Show that $\overrightarrow{PQ}=\pmatrix{ 1\cr -2\cr 3\cr}$

$\overrightarrow{PQ}=\pmatrix{3\cr -3\cr 8\cr}-\pmatrix{2\cr -1\cr 5\cr}$

(b) The line $L_1$ may be represented by $r=\pmatrix{3\cr -3\cr 8\cr}+s\pmatrix{1\cr -2\cr 3\cr}$

which is $r(s)=\pmatrix{3\cr -3\cr 8\cr}+s\pmatrix{1\cr -2\cr 3\cr}$

above is from OP
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(i) What information does the vector $\pmatrix{3\cr -3\cr 8\cr}$ give about $L_1$.
my question on this is I thot this was pointQ or can this be also a vector from $0,0,0$.

(i) Write down another vector representation for $L_1$ using $\pmatrix{3\cr -3\cr 8\cr}$
all I could come up with was $r(s)=\pmatrix{2\cr -1\cr 5\cr}-s\pmatrix{3\cr -3\cr 8\cr}=L_1$ but since $L_1$ is a line direction is not considered.

 

[Prove It]

Re: let L1 be the line throught P and Q

Consider the points $P(2,-1,5)$ and $Q(3,-3,8)$, let $L_1$ be the line trough $P$ and $Q$

(a) Show that $\overrightarrow{PQ}=\pmatrix{ 1\cr -2\cr 3\cr}$

$\overrightarrow{PQ}=\pmatrix{3\cr -3\cr 8\cr}-\pmatrix{2\cr -1\cr 5\cr}$

(b) The line $L_1$ may be represented by $r=\pmatrix{3\cr -3\cr 8\cr}+s\pmatrix{1\cr -2\cr 3\cr}$

i don't know this notation but it looks like $r=Q+\overrightarrow{PQ}$ so we are taking a point and adding a vector to it?

more ? to come on this...

It helps if you consider the relationship between vectors and lines. Vectors are defined by their direction and their magnitude. Lines are defined by their direction, their position, and are of infinite magnitude. That means that a line is really an infinitely long vector that is positioned somewhere.

If we know that the line passes through the points \displaystyle \begin{align*} P = (2, -1, 5) \end{align*} and \displaystyle \begin{align*} Q = (3, -3, 8) \end{align*}, then to get the equation of a line going through those points, we need a vector which goes in the same direction, so \displaystyle \begin{align*} \overrightarrow{PQ} = (1, -2, 3) \end{align*}, then you need to make it infinitely long, so multiply by a parameter \displaystyle \begin{align*} s \end{align*} giving \displaystyle \begin{align*} s(1, -2, 3) = (s, -2s, 3s) \end{align*}, and finally we need to position it somewhere, because this vector would be defined to go through the origin. The points that it goes through tells us how much the vector needs to be moved in the direction of each of the axes, so we would add that many to each of the components, so if we move it according to point Q, we get \displaystyle \begin{align*} L_1 = (s + 3, -2s -3, 3s + 8) \end{align*}.