Tutorial1_sol

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

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online