Question

100 PTS!! PLEASE HELP! OVER DUE

Answer 1

Answer:

$-\frac{1}{2} \vec {c}$

Step-by-step explanation:

Look at the component form of each vector.

Note that vector c is <4,4> and vector d is <-2,-2>

If one imagined the line that contained each vector, the line for both would have a slope of 1, because $\frac{4}{4}=1=\frac{-2}{-2}$

Since they have the same slope they are parallel, but since they are in opposite directions, we often call them "anti-parallel" (simply meaning parallel, but in opposite directions).

If two vectors are parallel, one vector can be multiplied by a scalar to result in the other vector.  This means that there is some number "k", such that $k \vec{c} = \vec {d}$, or equivalently, $kc_x=d_x$ and $kc_y=d_y$.

If $kc_x=d_x$ and $kc_y=d_y$, we just need to substitute known values and solve for k:

$kc_x=d_x\\k(4)=(-2)\\k=\dfrac{-2}{4}\\k=-\frac{1}{2}$

Double checking that k works for the y-coordinates as well:

$kc_y=d_y$

$(-\frac{1}{2}) (4)$    ?    $(-2)$

$-2=-2 \text{ } \checkmark$

So, $-\frac{1}{2} \vec {c} = \vec {d}$