dB (a) Show that is perpendicular to B. ds IBI = 1 → B. B = d → (B. B) = ds dB ➡2. B = ds dB → IB ds dB (b) Show that is

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Db A Show That Is Perpendicular To B Ds Ibi 1 B B D B B Ds Db 2 B Ds Db Ib Ds Db B Show That Is 1 (45.48 KiB) Viewed 46 times

dB (a) Show that is perpendicular to B. ds IBI = 1 → B. B = d → (B. B) = ds dB ➡2. B = ds dB → IB ds dB (b) Show that is perpendicular to T. ds dB d d B = TxN ⇒ (Tx N) = (Tx N) ds = ds ds dt dt = T'x = [(T²× N') Ir'(t)| Ir'(t)| T' = T'x + (Tx N') IT' = [0+ + (Tx N' Ir '(t)| dB → IT ds Deduce from parts (a) and (b) that dB ds is parallel to N. We have B = TxN ⇒ BIT and B N. Since TN, B, T, and N form ---Select--- dB ---Select--- SO is ---Select--- ds to N. (c) = · = d (Tx N). dt [(T ² × N) + (T × N' Ir'(t)| = Tx N' Ir' (t)| Ir' (t)| dB set of vectors in the three-dimensional space R³. From parts (a) and (b), is ds