What is the cross product of $[3, 1, - 4]$ $[3, 1, -4]$ and $[3, - 4, 2]$ $[3, -4, 2]$ ?

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Answer 1 · 2016-11-17T14:59:54

The vector is $= ⟨ - 14, - 18, - 15 ⟩$ $=〈-14,-18,-15〉$

Let $\to u = ⟨ 3, 1, - 4 ⟩$ $vecu=〈3,1,-4〉$ and $\to v = ⟨ 3, - 4, 2 ⟩$ $vecv=〈3,-4,2〉$

The cross product is given by the determinant

$\to u$ $vecu$ x $\to v$ $vecv$ $= | (veci,vecj,veck), (3,1,-4), (3,-4,2) |$

$= veci| (1,-4), (-4,2) | -vecj | (3,-4), (3,2) | +veck | (3,1), (3,-4) |$

$=veci(2-16)+vecj(-6-12)+veck(-12-3)$

$=vecw=〈-14,-18,-15〉$

Verification, the dot products must de $0$

$vecu.vecw=〈3,1,-4〉.〈-14,-18,-15〉=(-42-18+60)=0$

$vecv.vecw=〈3,-4,2〉.〈-14,-18,-15〉=(-42+72-30)=0$

Therefore, $vecw$ is perpendicular to $vecu$ and $vecv$

What is the cross product of [3, 1, -4][3,1,−4][3, 1, -4] and [3, -4, 2] [3,−4,2][3, -4, 2] ?