calculus vector of planes ..help,,,. 10 pts for best answer?

1) PQ has vector v = (6, 6) - (4, 4) = <2, 2>.
(a) The head for v' is at (1, 5) + v = (3, 7).
(b) The head for v0 is at (0, 0) + v = (2, 2).
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2) u = r v + s w = r <6, 1> + s <1, 3>
==> <4, -5> = <6r + s, r + 3s>

Equate like entries:
4 = 6r + s
-5 = r + 3s

Since r = -3s - 5, substituting this into the first equation yields
4 = 6(-3s - 5) + s
==> 4 = -17s - 30
==> s = -2

So, r = -3(-2) - 5 = 1.

Hence, <4, -5> = 1 * <6, 1> + (-2) <1, 3>.
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3) v = (0, -3, -1) - (2, -5, 6) = <-2, 2, -7>.

Similarly, compute v1, v2, v3, v4 in the same manner and see what matches v.
v1 = <-2, -10, 8>
v2 = <-2, 2, -10>
v3 = <-2, 2, -7>
v4 = <-2, 2, -7>.

Only v3 and v4 match v.
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I hope this helps!

1. even in your additional details you did not define v.
if v is the vector from P to Q it's clearly <2,2>
a) (1,5) + <2,2> = (3,7)
b) (2,2)

2. rv + sw = u
∙ ∙ r <6, 1> + s <1, 3> = <4, -5> ∙ ∙ ∙ ∙ ∙ ∙ so we get a system of linear eq
6r + 1s = 4
1r + 3s = -5 ∙ ∙ ∙ ∙ ∙ subtract, 3eq1 – eq2
17r ∙ ∙ ∙ = 17
∙ ∙ ∙ ∙ ∙r = 1
and then
6 + s = 4
∙ ∙ ∙ s = -2

so 1 <6, 1> – 2 <1, 3> = <4, -5>