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The position vector of $C$ is $-3\vec i - 2\vec j + p\vec k$.

Show that, if $p$ is a variable, then the locus of $C$ is a straight line and find the two unit direction vectors along this line.

Intuitively I think that the locus will be along the unit $k$ vector. But I am not able to figure out how to go about proving this.

Can you guys point me in the right direction? Thanks for your help.

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    I see your point, I'll try to incorporate the comment in an answer.2011-07-26

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Adding Andre's comment as answer to close this question.

Let $p_1, p_2$ be distinct points with vectors $\vec c_1, \vec c_2$.

Then,

$ \vec c - \vec c_1 = (p - p_1) \vec k $

And, $ \begin{align} \vec c_2 - \vec c_1 &= (p_2 - p_1) \vec k \\ &= \dfrac{(p_2 - p_1)}{(p - p_1)} (p - p_1) \vec k \\ &= \dfrac{(p_2 - p_1)}{(p - p_1)} (\vec c - \vec c_1) \\ &= \lambda (\vec c - \vec c_1) \end{align} $

Hence, $\vec c, \vec c_1, \vec c_2$ are parallel with a common point. Thus points $p, p_1, p_2$ are collinear. Hence locus of $C$ is a straight line for variable $p$.