What is the angle between $< 6, 9, 1 >$ $<6,9,1>$ and $< 5, 6, 7 >$ $<5,6,7>$ ?

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What is the angle between $< 6, 9, 1 >$ $<6,9,1>$ and $< 5, 6, 7 >$ $<5,6,7>$ ?

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Answer 1 · 2016-03-10T20:31:13

$α = 37^{o}$ $alpha=37^o$

Explanation:

$\to A = < 6, 9, 1 > \to B = < 5, 6, 7 >$ $vec A= <6,9,1> " "vec B= <5,6,7>$
$1 - find magnitude of the vector \to A and \to B$ $1-"find magnitude of the vector " vec A and vec B$
$| | A | | = \sqrt{6^{2} + 9^{2} + 1^{2}} | | A | | = \sqrt{36 + 81 + 1} = \sqrt{118}$ $||A||=sqrt(6^2+9^2+1^2)" "||A||=sqrt(36+81+1)=sqrt118$
$| | B | | = \sqrt{5^{2} + 6^{2} + 7^{2}} | | B | | = \sqrt{25 + 36 + 49} = \sqrt{110}$ $||B||=sqrt(5^2+6^2+7^2)" "||B||=sqrt(25+36+49)=sqrt110$
$2 - find dot product of A and B$ $2-"find dot product of A and B"$
$A . B = 6 \cdot 5 + 9 \cdot 6 + 1 \cdot 7 = 30 + 54 + 7 = 91$ $A.B=6*5+9*6+1*7=30+54+7=91$
$3 - use the formula of dot product$ $3-"use the formula of dot product"$
$A . B = | | A | | \cdot | | B | | \cdot cos α$ $A.B=||A||*||B||*cos alpha$
$91 = \sqrt{118} \cdot \sqrt{110} \cdot cos α$ $91=sqrt 118*sqrt 110*cos alpha$
$cos α = \frac{91}{\sqrt{118 \cdot 110}}$ $cos alpha=91/sqrt(118*110)$
$cos α = \frac{91}{113, 93}$ $cos alpha =91/(113,93)$
$cos α = 0, 798736066005$ $cos alpha=0,798736066005$
$α = 37^{o}$ $alpha=37^o$

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What is the angle between $< 6, 9, 1 >$ $<6,9,1>$ and $< 5, 6, 7 >$ $<5,6,7>$ ?

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Explanation:

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