HW #3 Solutions - homework 03 – FONTENOT BRIAN – Due...

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Unformatted text preview: homework 03 – FONTENOT, BRIAN – Due: Feb 4 2008, 4:00 am 1 Question 1, chap 3, sect 4. part 1 of 3 10 points Consider the two vectors vector M = ( a, b ) = a ˆ ı + b ˆ and vector N = ( c, d ) = c ˆ ı + d ˆ , where a = 4, b = 4, c = 3, and d = − 3. a and c represent the x-displacement and b and d represent the y-displacement in a Cartesian xy co-ordinate system. Note: ˆ ı and ˆ represent unit vectors ( i.e. vectors of length 1) in the x and y directions, respectively. What is the magnitude of the vector prod- uct vector M × vector M ? Correct answer: 0 (tolerance ± 1 %). Explanation: The magnitude of the vector product of two vectors bardbl vector A × vector B bardbl = A B sin θ , where θ is the angle between the two vectors placed with their tails at the same point. When vector A and vector B are the same vector, θ = 0 ◦ . Since sin0 ◦ = 0, the vector product of a vector with itself is zero or 0. Question 2, chap 3, sect 4. part 2 of 3 10 points What is the magnitude of the vector prod- uct vector M × vector N ? Correct answer: 24 (tolerance ± 1 %). Explanation: Take the vector products of the x- and y- displacement of vector M and vector N individually vector M × vector N = ( a ˆ ı + b ˆ ) × ( c ˆ ı + d ˆ ) = a c (ˆ ı × ˆ ı ) + b c (ˆ × ˆ ı ) + a d (ˆ ı × ˆ ) + b d (ˆ × ˆ ) = a d − b c = (4) ( − 3) − (4) (3) = − 24 , so bardbl vector M × vector N bardbl = 24 . Since ˆ ı ⊥ ˆ , we have ˆ ı × ˆ ı = 0, ˆ × ˆ ı = +1, ˆ ı × ˆ = − 1, and ˆ × ˆ = 0. Note: sin0 ◦ = 0 and sin90 ◦ = 1. The magnitude of bardbl ˆ ı × ˆ bardbl = (1)(1) sin90 ◦ = 1 , and bardbl ˆ × ˆ ı bardbl = (1)(1) sin90 ◦ = 1 . The vector product of two vectors, when non-zero, is a vector. These two vector prod- ucts have direction given by ˆ ı × ˆ = ˆ k andˆ × ˆ ı = − ˆ k , where ˆ k is a unit vector pointing along the positive z-axis. Thus the vectors ˆ ı × ˆ and ˆ × ˆ ı point in opposite directions. The result is vector M × vector N = ( a d − b c ) ˆ k = − 24 ˆ k . Question 3, chap 3, sect 4. part 3 of 3 10 points What is the direction of the vector product vector M × vector N ? 1. none of these since its direction cannot be determined 2. in the xz plane, but not along x or z 3. along the y-axis 4. in the xy plane, but not along x or y 5. along the z-axis correct 6. along the x-axis 7. in the yz plane, but not along y or z Explanation: vector M × vector N points along the z-axis. It only has a displacement in the ± ˆ k direction. Question 4, chap 3, sect 4. part 1 of 2 10 points Vector vector A is 0 . 857 units long and points in the positive y direction. Vector vector B has a negative x component 4 . 35 units long, a homework 03 – FONTENOT, BRIAN – Due: Feb 4 2008, 4:00 am 2 positive y component 4 . 43 units long, and no z component....
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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HW #3 Solutions - homework 03 – FONTENOT BRIAN – Due...

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