If #A= <-5 ,3 ,-3 ># and #B= <2 ,2 ,-7 >#, what is #A*B -||A|| ||B||#?

1 Answer
Steve M
Jun 16, 2018

# bb ul A * bb ul B - || bb ul A || \ || bb ul B || = 17 - sqrt(43)sqrt(57) #

Explanation:

We have:

# bb ul A = << -5, 3, -3 >># and # bb ul B = << 2, 2, -7 >>#

And so we compute the Scalar (or dot product):

# bb ul A * bb ul B= << -5, 3, -3 >> * << 2, 2, -7 >>#

# \ \ \ \ \ \ \ \ \ = (-5)(2) + (3)(6) + (-3)(-7) #

# \ \ \ \ \ \ \ \ \ = -10 + 6 + 21 #

# \ \ \ \ \ \ \ \ \ = 17 #

And we compute the vector norms (or magnitudes):

# || bb ul A || = || << -5, 3, -3 >> || #

# \ \ \ \ \ \ \ = sqrt( << -5, 3, -3 >> * << -5, 3, -3 >> ) #

# \ \ \ \ \ \ \ = sqrt( (-5)^2+ (3)^2 + (-3)^2 ) #

# \ \ \ \ \ \ \ = sqrt( 25 + 9 + 9) #

# \ \ \ \ \ \ \ = sqrt( 43 ) #

Similarly,

# || bb ul B || = || << 2, 2, -7 >> || #

# \ \ \ \ \ \ \ = sqrt( << 2, 2, -7 >> * << 2, 2, -7 >> ) #

# \ \ \ \ \ \ \ = sqrt( (2)^2+ (2)^2 + (-7)^2 ) #

# \ \ \ \ \ \ \ = sqrt( 4+4+49) #

# \ \ \ \ \ \ \ = sqrt( 57 ) #

So that:

# bb ul A * bb ul B - || bb ul A || \ || bb ul B || = 17 - sqrt(43)sqrt(57) #

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