# Math Help - Vectors #2

1. ## Vectors #2

The paths of two aeroplanes in an aerial display are simultaneously defined by the vectors:
r1(t) = (16 - 3t)i + tj + (3 + 2t)k
r2(t) = (3 + 2t)i + (1 + t)j + (11 - t)k

t represents time in minutes. Find:
a) the position vector of the first plane after one minute.
r1(1) = 13i + j + 5k

b) the unit vectors parallel to the flights of each of the two planes
The inclusion of t has thrown me off.

c) the acute angle between their lines of flight, correct to two decimal places

d) the point at which their two paths cross
I can do this for vectors in two dimensions, but not three.

e) the vector which represents the displacement between the two planes after t seconds

If you could walk me through this question, that would be great. Bear in mind that most of this is new to me.

Thanks.

2. ## Re: Vectors #2

I don't see why the inclusion of "t" should throw you off. For any vector \displaystyle \begin{align*} \mathbf{a} \end{align*}, its unit vector parallel to \displaystyle \begin{align*} \mathbf{a} \end{align*} is \displaystyle \begin{align*} \hat{\mathbf{a}} = \frac{\mathbf{a}}{ \left| \mathbf{a} \right| } \end{align*}. It doesn't matter if this vector happens to be a function of another variable...

3. ## Re: Vectors #2

For c) use the following equation: $cos \theta = \frac{Vector 1 (dot) Vector2}{|Vector 1| \times |Vector 2|}$

For d) to find the intersection of two vectors in 3D, first you must come up with the parametric equation of each vector, then substitute the x and y (or z) value of Vector 1 into that of Vector 2 respectively and that will yield two new equations of which each contains 2 unknown parameter values. Solve for one of the variables, and plug in the corresponding value back to the original Parametric, thus yielding the (x,y,z) coordinates of the intersection.

4. ## Re: Vectors #2

Originally Posted by Prove It
I don't see why the inclusion of "t" should throw you off. For any vector \displaystyle \begin{align*} \mathbf{a} \end{align*}, its unit vector parallel to \displaystyle \begin{align*} \mathbf{a} \end{align*} is \displaystyle \begin{align*} \hat{\mathbf{a}} = \frac{\mathbf{a}}{ \left| \mathbf{a} \right| } \end{align*}. It doesn't matter if this vector happens to be a function of another variable...
That's what I thought, but when I plug in the values I get a horrible fraction containing many "t"s that shouldn't be there.

5. ## Re: Vectors #2

Originally Posted by Fratricide
That's what I thought, but when I plug in the values I get a horrible fraction containing many "t"s that shouldn't be there.
The vectors parallel to the line of flight won't have a $t$ in them in this situation of linear motion. rewrite your $r_1, r_2$ as

$r_k(t)=\vec{r}_k(0) + \vec{v_k}t$

The vector $v_k$ is the vector that is parallel to your line of motion. (It's the velocity vector)