What is the norm or $< 8, - 4, 9 >$ $< 8 ,-4 , 9 >$ ?

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Answer 1 · 2017-04-11T06:12:21

The norm is $= 12.69$ $=12.69$

The norm of a vector $\to a = < p, q, r >$ $veca= <p,q,r>$ is

$∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣ = \sqrt{p^{2} + q^{2} + r^{2}}$ $||veca||=sqrt(p^2+q^2+r^2)$

So, we have

$| | < 8, - 4, 9 > | | = \sqrt{8^{2} + {(- 4)}^{2} + 9^{2}}$ $||<8,-4,9>||=sqrt(8^2+(-4)^2+9^2)$

$= \sqrt{64 + 16 + 81}$ $=sqrt(64+16+81)$

$= \sqrt{161}$ $=sqrt(161)$

$= 12.69$ $=12.69$

What is the norm or < 8 ,-4 , 9 ><8,−4,9>< 8 ,-4 , 9 >?