Tutorial1_sol

# Tutorial1_sol - MATH1211/10-11(1)/Tu1sol Department of...

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Unformatted text preview: MATH1211/10-11(1)/Tu1sol Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010–2011, 1st semester) Solution to Tutorial 1 1. From the equation 2 x- 3 y + 5 z = 30, we see that n = (2 ,- 3 , 5) is a normal vector to the plane, and (0 , , 6) is a point on the plane. Now a = (3 , 2 , 0) and b = (0 , 5 , 3) are non-parallel vectors and they are not having opposite directions, but both of them are orthogonal to n because a · n = 0 and b · n = 0. Hence a parametric equation for the plane is given by x ( s,t ) = (0 , , 6) + s a + t b , or x ( s,t ) = (0 , , 6) + s (3 , 2 , 0) + t (0 , 5 , 3). (Note that correct answer is not unique.) 2. k a- b k > k a + b k ⇐⇒ k a- b k 2 > k a + b k 2 ⇐⇒ ( a- b ) · ( a- b ) > ( a + b ) · ( a + b ) ⇐⇒ a · a + b · b- a · b- b · a > a · a + b · b + a · b + b · a ⇐⇒ > 4 a · b ⇐⇒ > a · b k a k k b k = cos θ, where θ is the angle between a and b . Since θ is between 0 and π , this implies that...
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## This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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Tutorial1_sol - MATH1211/10-11(1)/Tu1sol Department of...

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