If #A= <8 ,-5 ,6 ># and #B= <7 ,1 ,0 >#, what is #A*B -||A|| ||B||#?

1 Answer
Steve M
Mar 14, 2018

# bb ul(A) * bb ul(B) - || bb ul(A) || \ || bb ul(B) || = 51 - 25sqrt(2)#

Explanation:

We start with the vectors #A# and #B#

# bb ul(A) = << 8,-5,6 >> #
# bb ul(B) = << 7,1,0 >> #

The dot (or scalar) product is given by:

# bb ul(A) * bb ul(B) = << 8,-5,6 >> * << 7,1,0 >> #
# \ \ \ \ \ \ \ \ \ = (8)(7) + (-5)(1) + (6)(0) #
# \ \ \ \ \ \ \ \ \ = 56 -5 + 0 #
# \ \ \ \ \ \ \ \ \ = 51 #

And the norms are given by:

# || bb ul(A) || = || << 8,-5,6 >> || #
# \ \ \ \ \ \ \ = sqrt(8^2+(-5)^2+6^2) #
# \ \ \ \ \ \ \ = sqrt(64+25+36) #
# \ \ \ \ \ \ \ = sqrt(125) #
# \ \ \ \ \ \ \ = 5 #

# || bb ul(B) || = || << 7,1,0 >> || #
# \ \ \ \ \ \ \ = sqrt(7^2+1^2+0^2) #
# \ \ \ \ \ \ \ = sqrt(49+1+0) #
# \ \ \ \ \ \ \ = sqrt(50) #
# \ \ \ \ \ \ \ = 5sqrt(2) #

So then:

# bb ul(A) * bb ul(B) - || bb ul(A) || \ || bb ul(B) || = 51 - 25sqrt(2)#

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