Question

\displaystyle=\sqrt{{{3}{i}}}+\pi{j}+{c}{k},{v}={4}{i}−{j}−{k} where is a constant.

Trigonometric Functions
ANSWERED
asked 2021-02-02

\(\displaystyle=\sqrt{{{3}{i}}}+\pi{j}+{c}{k},{v}={4}{i}−{j}−{k}\), where is a constant.
(a) Compute \(‖u‖\), \(‖v‖\), and \(u \times v\) for the given vector in R3.
(b) Verify the Cauchy-Schwarz inequality for the given pair of vector.

Answers (1)

2021-02-03

Here \(u=\sqrt{3i}+\pi j+ck,\ v=4i−j−k\), where c is a constant.

(a) \(\displaystyle{\left|{\left|{u}\right|}\right|}=\sqrt{{\sqrt{{3}}^{{{2}}}+π^{{{2}}}+{c}^{{{2}}}=\sqrt{{{3}+π^{{{2}}}+{c}^{{{2}}}}}}}\)
\(\displaystyle{\left|{\left|{v}\right|}\right|}=\sqrt{{{4}^{{{2}}}+{1}+{1}}}=\sqrt{{{18}}}\)
and
\(\displaystyle{u}\cdot{v}={\left(\sqrt{{3}}{i}+π{j}+{c}{k}\right)}\cdot{\left({4}{i}−{j}−{k}\right)}={4}\sqrt{{3}}-π-{c}.\)
(b) \(\displaystyle{\left|{u}\cdot{v}\right|}={\left|{4}\sqrt{{3}}-π-{c}\right|}\leq\sqrt{{{3}+π^{{{2}}}+{c}^{{{2}}}}}\sqrt{{18}}={\left|{\left|{u}\right|}\right|}{\left|{\left|{v}\right|}\right|}.\)

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