1. vectors

how do i do this please xx

2. These are very basic practice problems. You will have to show some familiarity with them before you can move on.

Hints:

ai) Dot Product
aii) Cross Product

b)
If Z = a + bi
Z^2 = (a^2 - b^2) + 2abi
Z-bar = a - bi (The conjugate)

c) Factor and use the quadratic formula.

3. Hello, Amanda!

Here's the first one . . .

a) Condider the following position vectors: . $\begin{array}{ccc}\vec{v} & = & i-j\\ \vec{w} & = & j - 2k\end{array}$

(1) Find the angle between the two vectors.
(2) Find a vector perpendicular to both vectors.
We have: . $\begin{array}{ccc}\vec{v} & = & \langle1,\,\text{-}1,\,0\rangle \\ \vec{w} & = & \langle 0,\,1,\,\text{-}2\rangle\end{array}$

(1) We're expected to know this formula: . $\cos\theta \;=\;\frac{|\vec{v}\bullet\vec{w}|}{|\vec{v}||\vec {w}|}$

So we have: . $\cos\theta \;=\;\frac{|(1)(0) + (\text{-}1)(1) + (0)(\text{-}2)|}{\sqrt{1^2+(\text{-}1)^2+0^2}\,\sqrt{0^2+1^2+(\text{-}2)^2}} \;=\;\frac{|0 -1 + 0|}{\sqrt{2}\,\sqrt{5}}\;=\;\frac{1}{\sqrt{10}}$

. . Therefore: . $\theta \;=\;\cos^{-1}\!\left(\frac{1}{\sqrt{10}}\right) \;\approx\;71.6^o$

(2) We're expect to know that a vector perpendicular to two vectors
. . .is the cross product of the two vectors.

We have: . $\vec{n} \;=\;\begin{vmatrix}i & j & k \\ 1 & \text{-}1 & 0 \\ 0 & 1 & \text{-}2\end{vmatrix} \;=\;i(2-0) - j(\text{-}2 -0) + k(1-0) \;=\;2i + 2j + k$

. . Therefore: . $\vec{n} \;=\;\langle 2,\,2,\,1\rangle$

4. thank you both for your help, though I am finding part B rather tricky
xx

5. Let $z = 2 - 3j$ and $s = 1 + 2j$

We want:
$\frac{2 + z}{1 + s}$

$= \frac{2 + (2 - 3j)}{1 + (1 + 2j)}$

$= \frac{2 + 2 - 3j}{1 + 1 + 2j}$

$= \frac{4 - 3j}{2 + 2j}$

Now simplify:
$= \frac{4 - 3j}{2 + 2j} \cdot \frac{2 - 2j}{2 - 2j}$

$= \frac{(4 - 3j)(2 - 2j)}{(2 + 2j)(2 - 2j)}$

$= \frac{8 - 14j + 6j^2}{4 - 4j^2}$

$= \frac{8 - 14j - 6}{4 + 4}$

$= \frac{2 - 14j}{8}$

$= \frac{1 - 7j}{4}$

-Dan

6. blimy didnt think it was as easy as that LOL! on b where it is Z^2 not sure on that#xx

7. Originally Posted by Amanda-UK
blimy didnt think it was as easy as that LOL! on b where it is Z^2 not sure on that#xx
Z = a + bi
$Z^{2} = (a+bi)^{2} = (a+bi)(a+bi) = (a^{2} - b^{2}) + 2abi$ I'm confident you can do it. Show us what you get.