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Q 35.

Expert-verifiedFound in: Page 801

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Find the norm of the vector. $v=\left(\frac{1}{3},\frac{1}{3},-\frac{1}{3}\right)$

The norm of the vector $v=\left(\frac{1}{3},\frac{1}{3},-\frac{1}{3}\right)$ is $\sqrt{\frac{1}{3}}$.

We have given vector $v=\left(\frac{1}{3},\frac{1}{3},-\frac{1}{3}\right)$.

$||v||=\sqrt{{a}^{2}+{b}^{2}+{c}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(\frac{1}{3}\right)}^{2}+{\left(\frac{1}{3}\right)}^{2}+{\left(\frac{1}{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{9}+\frac{1}{9}+\frac{1}{9}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{3}{9}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{1}{3}}$

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