# Thread: Vector: Simple Line Question

1. ## Vector: Simple Line Question

I partially understand the method but not fully so can someone please briefly go through the method. Help would be appreciated.

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Q: The positions vectors with respect to the origin $\displaystyle O$ are:

$\displaystyle B: -4 \bold i + 4 \bold j - \bold k$,
$\displaystyle C: 5 \bold i - 2 \bold j + 11 \bold k$.

Find a vector equation for the line $\displaystyle BC$.

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$\displaystyle \bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}$.

2. That happens to be just $\displaystyle B + \frac{\mu }{3}\left( {C - B} \right)$.

3. Originally Posted by Plato
That happens to be just $\displaystyle B + \frac{\mu }{3}\left( {C - B} \right)$.
I can understand that the first column vector is the position vector but the second one near $\displaystyle \mu$, is it the postion vecots made into a vector or both position vectors dotted or crossed.

Fro example is it:

$\displaystyle \vec {BC}$ or $\displaystyle B . C$ or $\displaystyle B \bold x C$ or something else?

EDIT: Are you allowed to just remove a common denominator from the second bracket near $\displaystyle \mu$ and leave the equation without it?

4. I am not sure that I understand your question. But
$\displaystyle \left\{ \begin{array}{l} B = - 4i + 4j - k \\ C = 5i - 2j + 11k \\ \end{array} \right.\quad \Rightarrow \quad C - B = 9i - 6j + 12k = 3\left( {3i - 2j + 4k} \right)$

5. Originally Posted by Plato
I am not sure that I understand your question. But
$\displaystyle \left\{ \begin{array}{l} B = - 4i + 4j - k \\ C = 5i - 2j + 11k \\ \end{array} \right.\quad \Rightarrow \quad C - B = 9i - 6j + 12k = 3\left( {3i - 2j + 4k} \right)$
I was wondering where the $\displaystyle 3$ that was removed from the bracket went as the answer was $\displaystyle \bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}$.

I hope my question is clear.

6. If D is the direction vector of a line the any scalar multiple of D is also a direction vector.
Given any to points B & C the line determined by them is B+t(C-B).
Because we can factor out a 3 in C-B is just makes the direction vector simplier.