Vectors on straight line finding value of p The position vectors of points A and B relative to an origin O are given by
$$\overrightarrow{OA}=\pmatrix{p\\1\\1}\qquad\text{and}\qquad\overrightarrow{OB}=\pmatrix{4\\2\\p},$$
where $p$ is a constant. 
(i) In the case where OAB is a straight line, state the value of p and find the unit vector in the
direction of OA.[3] 
How to find the value of p(using a mathematical sense), i try to remove constant OB out giving(2) out....
 A: The points $O, A, B$ are collinear iff $OA$ and $OB$ are parallel (linearly dependent).
This is the case iff one is a scalar multiple of the other, and comparing the middle components gives that in this case, $$||\overline{OB}|| = 2 ||\overline{OA}||,$$ and so comparing, e.g., the last entry gives $p = 2$.
A perhaps more geometric solution would be to use the fact that the two vectors (in $\mathbb{R}^3$) are parallel iff their cross product is zero. Computing gives that
$$\overline{OA} \times \overline{OB} = \begin{pmatrix}p - 2 \\ 4 - p^2 \\ 2 p - 4\end{pmatrix},$$
and this is zero iff $p = 2$.
A: Vector OB is some  constant multiple of vector OA.
$$\vec{OA} = \lambda  \vec{OB}$$
After this it's very simple, you have three equations and two unknowns to solve.
A: The problem states they're colinear.  Thus there is a cA+B=0.
You know the components of A (p, 1, 1) and B (4, 2, p)
so (cP, c, c) + (4, 2, p) = (0, 0, 0)
then...
a) cP + 4 = 0
b) c+2 = 0
c) c + p = 0
c') c = -p
b') c = -2....    -p = -2...    p = 2
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Question though.... imagine the problem didn't specify they were colinear vectors.  Then I /think/ you couldn't find P, because you'd have to solve for three variables-- the coefficient of A, the coefficient of B, and the coordinate P.  Am I correct?
