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December 22, 2014

December 22, 2014

Posted by **Steve** on Sunday, November 2, 2008 at 2:39pm.

- Vectors -
**Damon**, Sunday, November 2, 2008 at 5:57pmA = i - 2 j so |A| = sqrt 5

B = -3 i + j so |B| = sqrt 10

The magnitude of the cross product A x B is |A| |B| sin T

where T is the angle between the vectors.

i j k

1 -2 0

-3 1 0 --> -5 k

so the magnitude of the cross product is 5

so

sqrt 5 * sqrt 10 * sin T = 5

5 sqrt 2 * sin T = 5

sin T = 1/sqrt 2

The component of A perpendicular to B, call it R, therefore has magnitude sqrt 5 /sqrt 2

We know that it is perpendicular to B so

B dot R has magnitude = 0

so

Rx Bx + RyBy = 0

Rx(-3)+ Ry(1) = 0

so Ry = 3 Rx

so call it

R = c Rx + 3c Ry where c is a constant

magnitude = c sqrt (10)

which has to be sqrt 5/sqrt 2

so c sqrt 10 = sqrt 5/sqrt 2

c = sqrt 5 /(sqrt 2*sqrt 2 * sqrt 5)

c = 1/2 but it could be negative

so our vector is

R = (1/2) i + (3/2) j

or

R = -(1/2) i - (3/2) j

looking at a sketch of the vectors, use the second one but both are valid.

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