What is the norm of $< 7, - 2, - 1 >$ $< 7 , -2, -1 >$ ?

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Answer 1 · 2016-08-02T20:38:58

$| | A | | = 3 \sqrt{6} unit$ $||A||=3sqrt 6" unit"$

$A = < 7, - 2, - 1 >$ $A= <7,-2,-1>$

$| | A | | = \sqrt{7^{2} + {(- 2)}^{2} + {(- 1)}^{2}}$ $||A||=sqrt(7^2+(-2)^2+(-1)^2)$

$| A | ∣ = \sqrt{49 + 4 + 1}$ $|A||=sqrt(49+4+1)$

$| | A | | = \sqrt{54}$ $||A||=sqrt54$

$| | A | | = \sqrt{9 \cdot 6}$ $||A||=sqrt(9*6)$

$| | A | | = 3 \sqrt{6} unit$ $||A||=3sqrt 6" unit"$

What is the norm of < 7 , -2, -1 ><7,−2,−1>< 7 , -2, -1 >?