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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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Not "along", but certainly "parallel"... –  J. M. Jul 25 '11 at 5:40
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Details depend on how definitions have been given. But take distinct $p_1$, $p_2$, and general $p$, giving vectors $v_1$, $v_2$, $v$. Compare $v_2-v_1$ with $v-v_1$. You can even take $p_1=0$, $p_2=1$. –  André Nicolas Jul 25 '11 at 5:49
    
Got it! $v_2 - v_1 = \lambda (v - v_1)$. Thus these vectors are $\parallel$ with a common point, and hence points are collinear. Locus is a straight line. Thanks @Andre. –  mathguy80 Jul 25 '11 at 6:03
    
@mathguy80: It is perfectly normal for people who ask questions to post answers, and to accept the answer. If you feel like doing it, and have trouble with the TeX, I or someone else can fix things. –  André Nicolas Jul 25 '11 at 6:08
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"Can you guys point me in the right direction?" Pun intended? –  Gerry Myerson Jul 25 '11 at 6:16
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1 Answer 1

up vote 1 down vote accepted

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 $p$.

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