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Q. 7

Expert-verifiedFound in: Page 823

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

If *u* and *v* are nonzero vectors in ${\mathrm{\mathbb{R}}}^{3}$, why do the equations role="math" localid="1649263352081" $u\xb7(u\times v)=0$ and $v\xb7(u\times v)=0$ tell us that the cross product is orthogonal to both *u* and *v*?

We prove that $u\xb7(u\times v)=0\mathrm{and}v\xb7(u\times v)=0$ tell us that the cross product is orthogonal to both *u* and *v.*

If *u* and *v* are nonzero vectors in ${\mathrm{\mathbb{R}}}^{3}$, why do the equations $u\xb7(u\times v)=0$ and $v\xb7(u\times v)=0$ tell us that the cross product is orthogonal to both *u* and *v*?

Let $u=({u}_{1},{u}_{2},{u}_{3})\mathrm{and}v=({v}_{1},{v}_{2},{v}_{3})$.

The determinant of a 3 × 3 matrix is

$u\times v=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ {\mathrm{u}}_{1}& {\mathrm{u}}_{2}& {\mathrm{u}}_{3}\\ {\mathrm{v}}_{1}& {\mathrm{v}}_{2}& {\mathrm{v}}_{3}\end{array}\right]$

$u\times v=({u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1})$

Now proving the equation $u\xb7(u\times v)=0$

$u\xb7(u\times v)=({u}_{1},{u}_{2},{u}_{3})\xb7({u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1})\phantom{\rule{0ex}{0ex}}u\xb7(u\times v)={u}_{1}({u}_{2}{v}_{3}-{u}_{3}{v}_{2})+{u}_{2}({u}_{3}{v}_{1}-{u}_{1}{v}_{3})+{u}_{3}({u}_{1}{v}_{2}-{u}_{2}{v}_{1})\phantom{\rule{0ex}{0ex}}u\xb7(u\times v)=0$

$v\xb7(u\times v)=({v}_{1},{v}_{2},{v}_{3})\xb7({u}_{2}{v}_{3}-{u}_{3}{v}_{2},{u}_{3}{v}_{1}-{u}_{1}{v}_{3},{u}_{1}{v}_{2}-{u}_{2}{v}_{1})\phantom{\rule{0ex}{0ex}}v\xb7(u\times v)={v}_{1}({u}_{2}{v}_{3}-{u}_{3}{v}_{2})+{v}_{2}({u}_{3}{v}_{1}-{u}_{1}{v}_{3})+{v}_{3}({u}_{1}{v}_{2}-{u}_{2}{v}_{1})\phantom{\rule{0ex}{0ex}}v\xb7(u\times v)=0$

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