c = u × v = ∣ i j k u 1 u 2 u 3 v 1 v 2 v 3 ∣ = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k c=u\times v=\begin{vmatrix}
i & j& k \\
u_1 & u_2 & u_3\\
v_1 & v_2 & v_3
\end{vmatrix}=(u_2v_3-u_3v_2)i-(u_1v_3-u_3v_1)j+(u_1v_2-u_2v_1)k c = u × v = ∣ ∣ i u 1 v 1 j u 2 v 2 k u 3 v 3 ∣ ∣ = ( u 2 v 3 − u 3 v 2 ) i − ( u 1 v 3 − u 3 v 1 ) j + ( u 1 v 2 − u 2 v 1 ) k
u ⋅ c = u 1 ( u 2 v 3 − u 3 v 2 ) − u 2 ( u 1 v 3 − u 3 v 1 ) + u 3 ( u 1 v 2 − u 2 v 1 ) = 0 u\cdot c=u_1(u_2v_3-u_3v_2)-u_2(u_1v_3-u_3v_1)+u_3(u_1v_2-u_2v_1)=0 u ⋅ c = u 1 ( u 2 v 3 − u 3 v 2 ) − u 2 ( u 1 v 3 − u 3 v 1 ) + u 3 ( u 1 v 2 − u 2 v 1 ) = 0
Which means u is perpendicular to c.
v ⋅ c = v 1 ( u 2 v 3 − u 3 v 2 ) − v 2 ( u 1 v 3 − u 3 v 1 ) + v 3 ( u 1 v 2 − u 2 v 1 ) = 0 v\cdot c=v_1(u_2v_3-u_3v_2)-v_2(u_1v_3-u_3v_1)+v_3(u_1v_2-u_2v_1)=0 v ⋅ c = v 1 ( u 2 v 3 − u 3 v 2 ) − v 2 ( u 1 v 3 − u 3 v 1 ) + v 3 ( u 1 v 2 − u 2 v 1 ) = 0
Which means v is perpendicular to c.
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