What is the angle between #<7 , -2 , -4 > # and # < 5 , -2 , 1 > #?

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Answer 1 · 2016-11-12T03:56:11

#theta ~~ 0.69" radians" #

Compute the dot-product:

#< 7, -2, -4 > * < 5, -2, 1 > = (7)(5) + (-2)(-2) + (-4)(1)#

#< 7, -2, -4 > * < 5, -2, 1 > = 35#

Compute the magnitude of both vectors:

#|< 7, -2, -4 >| = sqrt(7^2 + (-2)^2 + (-4)^2)#

#|< 7, -2, -4 >| = sqrt(69)#

#|< 5, -2, 1 >| = sqrt(5^2 + (-2)^2 + 1^2)#

#|< 5, -2, 1 >| = sqrt(30)#

The other way to compute the dot-product is:

#< 7, -2, -4 > * < 5, -2, 1 > = |< 7, -2, -4 >||< 5, -2, 1 >|cos(theta)#

where #theta# is the angle between the two vectors.

Substitute the values that we have computed:

#35 = sqrt(69)sqrt(30)cos(theta)#

Solve for #theta#:

#theta = cos^-1(35/sqrt({69}{30}))#

#theta ~~ 0.69" radians" #