math20c Meesue Yoo fall 09 quiz1

# math20c Meesue Yoo fall 09 quiz1 - Quiz 1 Solution for MATH...

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Unformatted text preview: Quiz 1 Solution for MATH 20C, 2009 − → 1. For R = (1, 4, 3), ﬁnd the point P such that v = P R has components < 3, −2, 3 > and sketch v. Ans. Let P = (a, b, c). Then − → v = P R =< 1 − a, 4 − b, 3 − c >=< 3, −2, 3 > So, ⇒ 1 − a = 3, a = −2, 4 − b = −2, b = 6, 3−c=3 c=0 P = (−2, 6, 0). 2. Find a vector parametrization for the line which passes through (−2, 0, −2) and (4, 3, 7). Ans. The direction vector of the line is (or is parallel to ) v =< 4 − (−2), 3 − 0, 7 − (−2) >=< 6, 3, 9 > . (You could take the direction vector < 2, 1, 3 >). Then the line equation is r(t) =< −2, 0, −2 > +t < 6, 3, 9 >=< −2 + 6t, 3t, −2 + 9t > So, the vector parametrization for the given line is x = −2 + 6t x = −2 + 2t y= 3t y= t or z = −2 + 9t z = −2 + 3t 3. Determine whether the two vectors are orthogonal and if not, ﬁnd the cosine of the angle between the vectors. < 1, 1, 5 >, < 1, −1, 5 > Ans. To check the orthogonality of the two vectors, we check if the dot product of the two vectors is zero or not ; < 1, 1, 5 > • < 1, −1, 5 >= 1 − 1 + 25 = 25 ￿= 0. Since the dot product is nonzero, the given two vectors are not orthogonal. To calculate the cosine of the angle between the vectors, we note that v • w = ||v || · ||w|| cos θ. √ √ ⇒ 25 = 1 + 1 + 25 · 1 + 1 + 25 cos θ, 25 . 27 where θ is the angle between the two given vectors. Hence, cos θ = ...
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