If $A = <2 ,6 ,-8 >$ , $B = <-9 ,4 ,-2 >$ and $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2017-01-13T18:07:38

$50.67^o$

C would be <11,2,-6>

Now if the angle between A and C is $theta$ , then $cos theta =(A.C)/(|A| |C|)$

$=(11*2 + 2*6 + (-8)*(-6))/(sqrt(2^2 +6^2 +(-8)^2) sqrt(11^2 +2^2 +(-6)^2)$

$=(22+12+48)/(sqrt 104 sqrt 161)$

$=82/sqrt16744$

$=82/129.399$

$=0.634$

$theta =50.67^o$

If A = <2 ,6 ,-8 >A = <2 ,6 ,-8 >, B = <-9 ,4 ,-2 >B = <-9 ,4 ,-2 > and C=A-BC=A-B, what is the angle between A and C?