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This one is weird for me. It says let vector A = a(8.0i-6.0j), where a is constant. Determine the TWO values of a that makes vector A a unit vector. = / i tried to solve for a but the i and j is messing up my head. how do i solve a?

2007-03-12 00:51:57 · 5 answers · asked by biscuits 2 in Science & Mathematics Mathematics

5 answers

A=8ai-6aj
sqrt(64a^2+36a^2)=1
sqrt(100a^2)=1=>10a=1=>a=1/10; a=-1/10

2007-03-12 01:04:25 · answer #1 · answered by djin 2 · 0 0

A=8ai-6aj
sqrt(64a^2+36a^2)=1
sqrt(100a^2)=1=>10a=1=>a=1/10

2007-03-12 08:07:26 · answer #2 · answered by hbj 2 · 0 1

The given vector is

A= 8a i - 6a j.

A unit vector is a vector whose length, (or magnitude) is unity or 1).

magnitude of the given vector = sqrt [ (8a)^2 + (6a)^2]

= sqrt [64 (a^2) + 36 (a^2)]

Now for this vector to be a unit vector, its magnitude should e equal to 1.

i.e. sqrt [64 (a^2) + 36 (a^2)] = 1
Squaring both sides,

64 (a^2) + 36 (a^2) =1

100 (a^2) = 1

(a^2) = 1/100

a = -1/10 or 1/10.

2007-03-12 08:46:00 · answer #3 · answered by gotcha 2 · 0 0

vector A = a(8.0i-6.0j),

first the notation instead of minus -
it should probably be a comma : a(8.0i , 6.0j)

ok , then the norm of a is sqrt(64 + 36) = 10
so the unit vectors are +a/10 or -a/10

c'est tout mon cher.

2007-03-12 08:04:33 · answer #4 · answered by gjmb1960 7 · 0 0

(8^2 + 6^2)^(1/2)
i.e. 10, -10....

2007-03-12 07:55:21 · answer #5 · answered by slimy dude 2 · 0 0

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