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 > )
Answer:
48
Step-by-step explanation:
x = -2 y =3
3x² - 2xy²
3(-2)² -2(-2)(3)²
square first
3*4 -2(-2)*9
multiplty from left to right
12-2(-2)*9
multiply from left to right again
12 -(-36) subtract using rule for integers that double minus is plus
12 + 36
48