If $A = < 6, 1, 5 >$ $A = <6 ,1 ,5 >$ , $B = < 4, 6, - 2 >$ $B = <4 ,6 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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If $A = < 6, 1, 5 >$ $A = <6 ,1 ,5 >$ , $B = < 4, 6, - 2 >$ $B = <4 ,6 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Answer 1 · 2016-11-07T17:59:50

${52.84}^{o}$ $52.84^o$

Explanation:

C=A-B=<2,-5,7>
So we have to find angle between <6,1,5> and <2,-5,7>

If that angle is $θ$ $theta$ then Cos $θ = \frac{A . C}{| A | | C |}$ $theta = (A.C)/(|A| |C|$

= $\frac{6 \cdot 2 + 1 \cdot (- 5) + 5.7}{\sqrt{6^{2} + 1^{2} + 5^{2}} \sqrt{2^{2} + {(- 5)}^{2} + 7^{2}}}$ $(6*2 +1*(-5) +5.7)/(sqrt(6^2+1^2+5^2) sqrt(2^2 +(-5)^2 +7^2))$

= $\frac{42}{\sqrt{62} \sqrt{78}}$ $42/(sqrt62 sqrt78)$

= $\frac{21}{\sqrt{31} \sqrt{39}}$ $21/(sqrt31 sqrt39$ = 0.6039

$θ$ $theta$ = arccos 0.6039= ${52.84}^{o}$ $52.84^o$

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If $A = < 6, 1, 5 >$ $A = <6 ,1 ,5 >$ , $B = < 4, 6, - 2 >$ $B = <4 ,6 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?

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Explanation:

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Humanities

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If A=<6,1,5>A = <6 ,1 ,5 >, B=<4,6,−2>B = <4 ,6 ,-2 > and C=A−BC=A-B, what is the angle between A and C?

1 Answer

Explanation:

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If $A = < 6, 1, 5 >$ $A = <6 ,1 ,5 >$ , $B = < 4, 6, - 2 >$ $B = <4 ,6 ,-2 >$ and $C = A - B$ $C=A-B$ , what is the angle between A and C?