If vector A =i+2j-k and vec A cross vec B =3i-j+5k. find vec B

Let B = a i + b j + c k. Then the cross product of A = i + 2j - k with B is

A × B = ( i + 2j - k ) × ( a i + b j + c k )

A × B  = a ( i × i ) + 2a ( j × i ) - a ( k × i )

… … … + b ( i × j ) + 2b ( j × j ) - b ( k × j )

… … … + c ( i × k ) + 2c ( j × k ) - c ( k × k )

A × B = 0 - 2a k - a j

… … … + b k + 0 + b i

… … … - c j + 2c i - 0

A × B = (b + 2c) i + (-a - c) j + (b - 2a) k

So we have

3 i - j + 5 k = (b + 2c) i + (c - a) j + (b - 2a) k

which gives us the system of equations,

{ b + 2c = 3

{ -a - c = -1

{ -2a + b = 5

Solve for a, b, and c.

• Eliminate c from the first two equations:

(b + 2c) + 2 (-a - c) = 3 + 2 (-1)

-2a + b = 1

But -2a + b = 5, and 5 ≠ 1, so there is no such vector B that satisfies the cross product!

d) the number of neutrons plus the number of protons

double the distance = divide strength by 4