1. ## vectors urgent help

the position vector of point A is 2i + 3j + k and the position vector of point B is 4i - 5j + 21k.

a.) (i) show that vector AB = 2i - 8j + 20k
(ii) find the unti vector u in the direction of vector AB
(iii) show that u is perpendicular to vector OA.

Let S be the midpoin=t of [AB]. THe line L1 passes through S and its parrallel to vector OA.
b.) (i) find the position vector of S.
(ii) Write down the equation of L1.

The line L2 has equation r = (5i + 10j +10k) + s(-2i + 5j -3k).
c.) explain why L1 and L2 are not parallel.
d.) the lines L1 and L2 intersect at the point P. Find the position vector of P.

2. Originally Posted by miley_22
the position vector of point A is 2i + 3j + k and the position vector of point B is 4i - 5j + 21k.

a.) (i) show that vector AB = 2i - 8j + 20k
(ii) find the unti vector u in the direction of vector AB
(iii) show that u is perpendicular to vector OA.
The vector $\displaystyle \bold{AB} = \bold{B} - \bold{A} = (4i - 5j + 21k) - (2i + 3j + k) = 2i - 8j + 20k$

The unit vector is the vector in the same direction, but with unit length. So
$\displaystyle \hat{\bold{u}} = \frac{\bold{AB}}{|\bold{AB}|}$

To show that $\displaystyle \hat{\bold{u}}$ is perpendicular to $\displaystyle \bold{OA} = 2i + 3j + k$, use the dot product. The dot product of two perpendicular vectors is 0.

Originally Posted by miley_22
Let S be the midpoin=t of [AB]. THe line L1 passes through S and its parrallel to vector OA.
b.) (i) find the position vector of S.
(ii) Write down the equation of L1.
Now this is just silly to me. Vectors can be moved all over the place without changing them, just as long as their length and direction aren't changed.

I think what the problem is trying to say is "Place the tails of the vectors $\displaystyle \bold{A}$ and $\displaystyle \bold{B}$ at the origin. Then let S be the midpoint..." Only in this way is the wording unambiguous.

So let's assume the rewritten problem is right. The heads of $\displaystyle \bold{A}$ and $\displaystyle \bold{B}$ are at (2, 3, 1) and (4, -5, 21), respectively. The midpoint S of these points is (3, -1, 11). So the vector $\displaystyle \bold{OS}$ is
$\displaystyle \bold{OS} = 3i - j + 11k$

The vector form of a line in 3-D is
$\displaystyle \bold{r} = \bold{r_0} + \bold{A}t$
where $\displaystyle \bold{r}$ is the position vector of a point on the line, $\displaystyle \bold{r_0}$ is the position vector to a point that the line passes through, and $\displaystyle \bold{A}$ is a vector, and t is a scalar parameter generating the points along the line. (The line will be parallel to $\displaystyle \bold{A}$.)

So for L1 we want:
$\displaystyle \bold{r} = (3i - j + 11k) + (2i + 3j + k)t$

Originally Posted by miley_22
The line L2 has equation r = (5i + 10j +10k) + s(-2i + 5j -3k).
c.) explain why L1 and L2 are not parallel.
d.) the lines L1 and L2 intersect at the point P. Find the position vector of P.
For c) consider this: What was the vector that forced L1 to be parallel to $\displaystyle \bold{OA}$? Similarly what is the vector that L2 is parallel to? Once you have these answers, the cross product of two parallel vectors is 0. Is this true for the L1 and L2 vectors?

For d) you know that the (x, y, z) coordinate points of L1 are (3 + 2t, -1 + 3t, 11 + t). For L2 they are (5 - 2s, 10 + 5s, 10 - 3s). How can you find if they have a point in common? (ie. for what value of s and t are the two points the same?)

-Dan