# Vector: Simple Line Question

• March 6th 2008, 07:26 AM
Simplicity
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 $O$ are:

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

Find a vector equation for the line $BC$.

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$\bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}$.
• March 6th 2008, 07:34 AM
Plato
That happens to be just $B + \frac{\mu }{3}\left( {C - B} \right)$.
• March 6th 2008, 07:44 AM
Simplicity
Quote:

Originally Posted by Plato
That happens to be just $B + \frac{\mu }{3}\left( {C - B} \right)$.

I can understand that the first column vector is the position vector but the second one near $\mu$, is it the postion vecots made into a vector or both position vectors dotted or crossed.

Fro example is it:

$\vec {BC}$ or $B . C$ or $B \bold x C$ or something else? (Worried)

EDIT: Are you allowed to just remove a common denominator from the second bracket near $\mu$ and leave the equation without it?
• March 6th 2008, 08:05 AM
Plato
I am not sure that I understand your question. But
$\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)
$
• March 6th 2008, 08:10 AM
Simplicity
Quote:

Originally Posted by Plato
I am not sure that I understand your question. But
$\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 $3$ that was removed from the bracket went as the answer was $
\bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}
$
.

I hope my question is clear. (Worried)
• March 6th 2008, 08:26 AM
Plato
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.