A.(–6, 0), (–2, 0), and (0, 0)
B.(–6, 0), (–2, 0), and (4, 0)
C. (–4, 0), (0, 0), and (2, 0)
D.(–4, 0), (–2, 0), and (0, 0)
A. 78
B. 33
C. 90
D. 66
-3y=2x-6
2(x + 4) = x + 13
A. −3
B. −1
C. 5
D. 7
Answer: hello some part of your question is missing
Let v=〈−2,5〉 in R^2,and let y=〈0,3,−2〉 in R^3.
Find a unit vector u in R^2 such that u is perpendicular to v. How many such vectors are there?
answer:
One(1) unit vector ( < 5/√29, 2 /√29 > ) perpendicular to 〈−2,5〉
Step-by-step explanation:
let
u = < x , y > ∈/R^2 be perpendicular to v = < -2, 5 > ------ ( 1 )
hence :
-2x + 5y = 0
-2x = -5y
x = 5/2 y
back to equation 1
u = < 5/2y, y >
∴ || u || = y/2 √29
∧
u = < 5 /2 y * 2 / y√29 , y*2 / y√29 >
= < 5/√29, 2 /√29 > ( unit vector perpendicular to < -2, 5 > )