Answer:
[tex]-\frac{1}{2} \vec {c}[/tex]
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 [tex]\frac{4}{4}=1=\frac{-2}{-2}[/tex]
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 [tex]k \vec{c} = \vec {d}[/tex], or equivalently, [tex]kc_x=d_x[/tex] and [tex]kc_y=d_y[/tex].
If [tex]kc_x=d_x[/tex] and [tex]kc_y=d_y[/tex], we just need to substitute known values and solve for k:
[tex]kc_x=d_x\\k(4)=(-2)\\k=\dfrac{-2}{4}\\k=-\frac{1}{2}[/tex]
Double checking that k works for the y-coordinates as well:
[tex]kc_y=d_y[/tex]
[tex](-\frac{1}{2}) (4)[/tex] ? [tex](-2)[/tex]
[tex]-2=-2 \text{ } \checkmark[/tex]
So, [tex]-\frac{1}{2} \vec {c} = \vec {d}[/tex]
