Question

100 PTS!! PLEASE HELP! OVER DUE

Answer 1

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

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 , or equivalently, and .

If and , we just need to substitute known values and solve for k:

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

?

So,

Answer 2

`-(1)/(2) \vec {c}`

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
`(4)/(4)=1=(-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=(-2)/(4)\\k=-(1)/(2)`

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

`kc_y=d_y`

`(-(1)/(2)) (4)` ?
`(-2)`

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

So,
`-(1)/(2) \vec {c} = \vec {d}`