Let vector a <2,-3,4> and vector b <-3,4,7>

Compute 3a - 2b

and

Use calculus to determine the values of
θ where r = cosθ has a slope of 0.

for the first one i just didnt know if to take the magnitude then multiply or multiply each number then get the magnitude...

and still dont exactly know how to do 2

2. Originally Posted by tiga killa
for the first one i just didnt know if to take the magnitude then multiply or multiply each number then get the magnitude...
no.

let $c$ be a scalar, and $\vec a = \left< a_1, a_2 \right>$ be a 2-dimensional vector, then $c \vec a = c \left< a_1, a_2 \right> = \left< c a_1, ca_2 \right>$

Let $\vec b = \left< b_1, b_2 \right>$ be another vector. we define $\vec a + \vec b$ as follows:

$\vec a + \vec b = \left< a_1, a_2 \right> + \left< b_1, b_2 \right> = \left< a_1 + b_1, a_2 + b_2 \right>$

that is, we just add corresponding components

those are some basic definitions you must know. now can you get anywhere?
and still dont exactly know how to do 2
you do not know how to differentiate cosine?

you haven't answered the question of whether we should be thinking of polor coordinates here or not.

3. I get the first one..

and yes the 2nd one is using polar coordinates

4. Originally Posted by tiga killa
I get the first one..

and yes the 2nd one is using polar coordinates
ok, the "hard" way is to use the formula $\frac {dy}{dx} = \frac {\frac {dr}{d \theta} \sin \theta + r \cos \theta}{\frac {dr}{d \theta} \cos \theta - r \sin \theta}$, then set the expression to 0 and solve. you can see how it is derived here

Alternatively, begin with $r = \cos \theta$

multiply through by $r$, to get

$r^2 = r \cos \theta$

now we know $r^2 = x^2 + y^2$ and $x = r \cos \theta$ in polar coordinates. thus we have

$x^2 + y^2 = x$

now you can differentiate implicitly to find $\frac {dy}{dx}$, then set it equal to zero to find $x$ and $y$, and hence the points you seek