1. ## Vectors

1. Find two unit vectors that make an angle of 60* with v = <3,4>

2. Find the cross product a X b for a = i + (e^t)k +(e^-t)k, b = 2i + (e^t)j - (e^-t)k

3. Use the scalar triple product to determine whether the points A(1,3,2), B(3,-1,6),C(5,2,0), and D(3,6,-4) lie on the same plane.

1. ?

2. I set it up as the matrices way..
| i j k |
| 1 e^t e^-t |
| 2 e^t -e^-t |

and came out with -2i - (-e^-t - 2e^-t)j + (-e^-t-2e^t)k

3)
| 1 3 2 |
| 3 -1 6 |
| 5 2 0 |
|3 6 -4 |

is this the way to do this problem? I end up with like
|0|
|-4| if i do this.

can someone help me with these? I think i am doing 2 and 3 incorrectly and 1 i do not know how.

thanks.

2. Hello, xfyz!

I'm still working on #1 . . . don't know why it's giving me trouble.

2. Find the cross product $\displaystyle a \times b$ for: .$\displaystyle a \:= \:i + e^tj + e^{-t}k,\;\;b \:= \:2i + e^tj - e^{-t}k$

I set up the matrix: .$\displaystyle \begin{vmatrix} i & j & k \\ 1 & e^t & e^{\text{-}t} \\ 2 & e^t & \text{-}e^{\text{-}t} \end{vmatrix}$

and came out with: .$\displaystyle -2i - \underbrace{(-e^{\text{-}t} - 2e^{\text{-}t})}_{\text{Combine!}}j + \underbrace{(-e^{\text{-}t} - 2e^t)}_{\text{Wrong!}}k$

We have: .$\displaystyle \begin{vmatrix}i & j & k \\ 1 & e^t & e^{\text{-}t} \\ 2 & e^t & \text{-}e^{\text{-}t} \end{vmatrix}$

. . $\displaystyle = \;i(\text{-}1-1) -j(\text{-}e^{\text{-}t}-2e^{\text{-}t}) + k(e^t - 2e^t)$

. . $\displaystyle = \;{\color{blue}-2i + 3e^{-t}j - e^tk}$

3. Use the scalar triple product to determine whether the points:
$\displaystyle A(1,3,2),\;B(3,-1,6),\;C(5,2,0),\;D(3,6,-4)$ lie on the same plane.
It took me a minute to remember what the scalar triple product is for.

Given three vectors $\displaystyle \vec u,\:\vec v,\:\vec w$, the scalar triple product, $\displaystyle u\cdot (v \times w)$ gives us
the volume of the parallelepiped ("slanted box") determined by the vectors.

Let: .$\displaystyle \begin{array}{ccccc}\vec{u} & = & \overrightarrow{AB} & = & \langle2,\text{-}4,4\rangle \\ \vec v & = & \overrightarrow{AC} & = & \langle 4,\text{-}1,\text{-}2\rangle \\ \vec{w} & = & \overrightarrow{AD} & = & \langle2,3,\text{-}6\rangle \end{array}$

The three vectors have a common point, $\displaystyle A.$
If the vectors point in three "different" directions (like the legs of a tripod),
. . the parallelepiped will have a volume (greater than zero).
If the vectors are in the same plane, then the "box" will have volume zero.

Scalar triple product: .$\displaystyle \begin{vmatrix}2 & \text{-}4 & 4 \\ 4 & \text{-}1 & \text{-}2 \\ 2 & 3 & \text{-}6 \end{vmatrix} \;=\;2(6+6) + 4(-24+4) + 4(12+2)$

. . $\displaystyle = \;2(12) + 4(-20) + 4(14) \;=\;24 - 80 + 56 \;=\;{\color{blue}0}$ . . . . There!

3. Originally Posted by xfyz
1. Find two unit vectors that make an angle of 60* with v = <3,4>
For vectors $\displaystyle u=(u_1,u_2)$ and $\displaystyle v=(v_1,v_2)$, the scalar product u.v can be calculated in two ways. It is equal to $\displaystyle u_1v_1+u_2v_2$ and it is also equal to $\displaystyle |u||v|\cos\theta$, where θ is the angle between u and v.

If $\displaystyle u=(x,y),\;v=(3,4)$, θ = 60° and u is a unit vector, then $\displaystyle u.v = 3x+4y$ (from the first of those two formulas), and $\displaystyle u.v = 5\times{\textstyle\frac12}$ (from the second one, because |u|=1, $\displaystyle |v|=\sqrt{3^2+4^2}=5$ and cos(60°)=1/2). Also, $\displaystyle x^2+y^2=1$ because u is a unit vector.

Thus $\displaystyle 3x+4y = 5/2$, from which $\displaystyle y = (5-6x)/8$. Substitute that value for y into the equation $\displaystyle x^2+y^2=1$ and you'll get a quadratic equation for x ...