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Question Number 69620 by aseer imad last updated on 25/Sep/19

Commented by kaivan.ahmadi last updated on 26/Sep/19

b is answer since (1/5)(4,0,−3).(3,−1,4)=(1/5)(12+0−12)=0⇒b^→ is perpendicular to 3i−j+4k and ∣b^→ ∣=(√((1/(25))(16+9)))=1⇒b^→ is unit

bisanswersince15(4,0,−3).(3,−1,4)=15(12+0−12)=0⇒b→isperpendicularto3i−j+4kand∣b→∣=125(16+9)=1⇒b→isunit

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