Let $u$ $bbu$ , $v$ $bbv$ , $w$ $bb(w)$ be vectors. How can we prove $u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$ $bbu xx (bbv xx bb(w)) = (bbu * bb(w)) bbv - (bbu * bbv) bb(w)$ ?

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Let $u$ $bbu$ , $v$ $bbv$ , $w$ $bb(w)$ be vectors. How can we prove $u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$ $bbu xx (bbv xx bb(w)) = (bbu * bb(w)) bbv - (bbu * bbv) bb(w)$ ?

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Answer 1 · 2018-03-07T21:44:11

See explanation...

Explanation:

There may be a better way, but here's a direct method:

$< v_{1}, v_{2}, v_{3} > \times < w_{1}, w_{2}, w_{3} >$ $< v_1, v_2, v_3> xx < w_1, w_2, w_3>$

$=abs((hat(i), hat(j), hat(k)), (v_1, v_2, v_3), (w_1, w_2, w_3))$

$=< v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1>$

$< u_1, u_2, u_3> xx < v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1>$

$=abs((hat(i), hat(j), hat(k)), (u_1, u_2, u_3), (v_2w_3-v_3w_2, v_3w_1-v_1w_3, v_1w_2-v_2w_1))$

$=< u_2(v_1w_2-v_2w_1)-u_3(v_3w_1-v_1w_3), u_3(v_2w_3-v_3w_2)-u_1(v_1w_2-v_2w_1), u_1(v_3w_1-v_1w_3)-u_2(v_2w_3-v_3w_2)>$

$=< u_2v_1w_2-u_2v_2w_1-u_3v_3w_1+u_3v_1w_3, u_3v_2w_3-u_3v_3w_2-u_1v_1w_2+u_1v_2w_1, u_1v_3w_1-u_1v_1w_3-u_2v_2w_3+u_2v_3w_2>$

$(< u_1, u_2, u_3> * < w_1, w_2, w_3>)< v_1, v_2, v_3>$

$=(u_1w_1+u_2w_2+u_3w_3)< v_1, v_2, v_3>$

$=< u_1v_1w_1+u_2v_1w_2+u_3v_1w_3, u_1v_2w_1+u_2v_2w_2+u_3v_2w_3, u_1v_3w_1+u_2v_3w_2+u_3v_3w_3>$

$(< u_1, u_2, u_3 > * < v_1, v_2, v_3>)< w_1, w_2, w_3>$

$=(u_1v_1+u_2v_2+u_3v_3)< w_1, w_2, w_3>$

$=< u_1v_1w_1+u_2v_2w_1+u_3v_3w_1, u_1v_1w_2+u_2v_2w_2+u_3v_3w_2, u_1v_1w_3+u_2v_2w_3+u_3v_3w_3>$

So:

$(bb(u) * bb(w))bb(v) - (bb(u) * bb(v))bb(w)$

$=< color(red)(cancel(color(black)(u_1v_1w_1)))+u_2v_1w_2+u_3v_1w_3, u_1v_2w_1+color(green)(cancel(color(black)(u_2v_2w_2)))+u_3v_2w_3, u_1v_3w_1+u_2v_3w_2+color(blue)(cancel(color(black)(u_3v_3w_3)))> - < color(red)(cancel(color(black)(u_1v_1w_1)))+u_2v_2w_1+u_3v_3w_1, u_1v_1w_2+color(green)(cancel(color(black)(u_2v_2w_2)))+u_3v_3w_2, u_1v_1w_3+u_2v_2w_3+color(blue)(cancel(color(black)(u_3v_3w_3)))>$

$=< u_2v_1w_2-u_2v_2w_1-u_3v_3w_1+u_3v_1w_3, u_3v_2w_3-u_3v_3w_2-u_1v_1w_2+u_1v_2w_1, u_1v_3w_1-u_1v_1w_3-u_2v_2w_3+u_2v_3w_2>$

$=bb(u) xx (bb(v) xx bb(w))$

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Let $u$ $bbu$ , $v$ $bbv$ , $w$ $bb(w)$ be vectors. How can we prove $u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$ $bbu xx (bbv xx bb(w)) = (bbu * bb(w)) bbv - (bbu * bbv) bb(w)$ ?

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Let $u$ $bbu$ , $v$ $bbv$ , $w$ $bb(w)$ be vectors. How can we prove $u \times (v \times w) = (u \cdot w) v - (u \cdot v) w$ $bbu xx (bbv xx bb(w)) = (bbu * bb(w)) bbv - (bbu * bbv) bb(w)$ ?