What is the angle between $< 6, 3, - 6 >$ $<6,3,-6 >$ and $< 1, 7, - 3 >$ $<1,7,-3>$ ?

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What is the angle between $< 6, 3, - 6 >$ $<6,3,-6 >$ and $< 1, 7, - 3 >$ $<1,7,-3>$ ?

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Answer 1 · 2016-03-07T22:36:23

$θ ≅ 49^{o}$ $theta~=49^o$

Explanation:

$\to A \to < 6, 3, - 6 > \to B \to < 1, 7, - 3 >$ $vec A-><6,3,-6>" "vec B-><1,7,-3>$
$step-1: find ||A|| and ||B||$ $"step-1: find ||A|| and ||B||"$
$| | A | | = \sqrt{6^{2} + 3^{2} + {(- 6)}^{2}} = \sqrt{36 + 9 + 36} = \sqrt{81}$ $||A||=sqrt(6^2+3^2+(-6)^2)=sqrt(36+9+36)=sqrt81$
$| | B | | = \sqrt{1^{2} + 7^{2} + {(- 3)}^{2}} = \sqrt{1 + 49 + 9} = \sqrt{59}$ $||B||=sqrt(1^2+7^2+(-3)^2)=sqrt(1+49+9)=sqrt 59$
$step-2: find dot product A \cdot B$ $"step-2: find dot product "A*B$
$A \cdot B = (6 \cdot 1) + (3 \cdot 7) + (- 6 \cdot - 3) = 6 + 21 + 18 = 45$ $A*B=(6*1)+(3*7)+(-6*-3)=6+21+18=45$
$step-3: use the formula : A \cdot B = | | A | | \cdot | | B | | \cdot cos θ$ $"step-3: use the formula :" A*B=||A||*||B||*cos theta$

$45 = \sqrt{81} \cdot \sqrt{59} \cdot cos θ$ $45=sqrt81*sqrt59*cos theta$
$cos θ = \frac{45}{\sqrt{81 \cdot 59}}$ $cos theta=45/(sqrt(81*59))$
$cos θ = \frac{45}{\sqrt{4779}}$ $cos theta=45/sqrt4779$
$cos θ = 0, 6509$ $cos theta=0,6509$
$θ ≅ 49^{o}$ $theta~=49^o$

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What is the angle between $< 6, 3, - 6 >$ $<6,3,-6 >$ and $< 1, 7, - 3 >$ $<1,7,-3>$ ?

1 Answer

Explanation:

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What is the angle between <6,3,−6><6,3,-6 > and <1,7,−3><1,7,-3> ?

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What is the angle between $< 6, 3, - 6 >$ $<6,3,-6 >$ and $< 1, 7, - 3 >$ $<1,7,-3>$ ?