Sections 1.2 solutions

# We distribute through w v cv v 0 c w v and v v p cv

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Unformatted text preview: n be found on the opposite side of the plane that is pependicular to v. v 12. We solve: w cv  v  0. We distribute through: w  v cv  v  0 , c  w  v and v v p  cv  w  v v is then referred to as the &quot;projection&quot; of w on v. v 16. v  1, 1, . . , 1  9  3. So ū  1 1, 1, . . , 1 is a unit vector in the direction of v. 3 w  1 1, 1, 0, . . , 0 is perpendicular to v, with length 1. 2 vj 17. If v  1, 0, 1 then cos   v  ī  1 , cos    0 , cos   v  k  1 and v v v 2 2 the sum of the squares is 1 as advertised. 21. Using the suggestion v  w 2  v 2  2v  w  w 2 v 2  2 v w  w 2  v  w 2 and taking the 1 square roots on both sides, we have the triangle inequality v  w vw w 1 , sin   w 2 . Now 23. cos   w w cos   cos    cos  cos   sin  sin   w 1 v 1  w 2 v 2  w  v . wv wv wv (Actually I think it’s more impressive going the other way - deriving the trig formula from the dot product formula for cos  ). 2...
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