solving problem in Banesh Hoffmann's Vectors

[james snail]

1. The problem statement, all variables and given/known data
Exercise 1.3, chapter 3: if e[SUB]x and e[SUB]y lie along perpendicular lines, and e[SUB]x has a magnitude of 4 and e[SUB]y has a magnitude of 6, what are the components of the vector relative to the reference frame defined by the base vectors e[SUB]x and e[SUB]y?


2. Relevant equations this relates to preceding exercise 1.2 in which vector of magnitude 144 makes 45 degrees with each of two perpendicular lines. I solved this with pythagorean theorem coming up with component vectors 6 x square root of 2, (4 sets with differrng signs.)
also, per Hoffman, X + Y = V, and X= [I]V[SUB]xtimese[SUB]x, and Y = [I][V][SUB]ytimese[SUB]y.


3. The attempt at a solution since in 1.2 X and Y are each = 6 X square root of 2, I set V[SUB]xe[SUB]x = 6 X square root 2 thus 4e[SUB]x = 6X square root 2. Answer e[SUB]x = 3/2 square root of 2; similarly, e[SUB]y = square root 2. This is incorrect as per Hoffmann, correct answers are + or - 18 X square root 2 and + or - 12 square root 2.
will be most grateful to be shown how Hoffmann got his correct answers.

[Mark44]

This is very difficult to read. When you have x and X as variables, it's not a good idea to use X to indicate multiplication.

For subscripts, use [ sub] before and [ /sub] after the subscript (omit the leading spaces. For example, to get ey, I wrote e[ sub]y[ /sub], (again, omitting the leading spaces inside the brackets).

Also, welcome to Physics Forums!

[HallsofIvy]

1. The problem statement, all variables and given/known data
Exercise 1.3, chapter 3: if e[SUB]x and e[SUB]y lie along perpendicular lines, and e[SUB]x has a magnitude of 4 and e[SUB]y has a magnitude of 6, what are the components of the vector relative to the reference frame defined by the base vectors e[SUB]x and e[SUB]y?


2. Relevant equations this relates to preceding exercise 1.2 in which vector of magnitude 144 makes 45 degrees with each of two perpendicular lines. I solved this with pythagorean theorem coming up with component vectors 6 x square root of 2, (4 sets with differrng signs.)
also, per Hoffman, X + Y = V, and X= [I]V[SUB]xtimese[SUB]x, and Y = [I][V][SUB]ytimese[SUB]y.


3. The attempt at a solution since in 1.2 X and Y are each = 6 X square root of 2, I set V[SUB]xe[SUB]x = 6 X square root 2 thus 4e[SUB]x = 6X square root 2. Answer e[SUB]x = 3/2 square root of 2; similarly, e[SUB]y = square root 2. This is incorrect as per Hoffmann, correct answers are + or - 18 X square root 2 and + or - 12 square root 2.
will be most grateful to be shown how Hoffmann got his correct answers.
Since the given vector, of length 144, is at 45 degrees to the given basis vectors, its projection on each has length 144 cos(45)= 72\sqrt{2}. Since the length of e_x is 4, that is (72/4)\sqrt{2}= 18\sqrt{2} times e_x. Since the length of e_y is 6, the projection on it is (72/6)\sqrt{2}= 12\sqrt{2} so the projection is 12\sqrt{2} times e_y.

You are taking 4e_x when it is given that e_x already has length 4. Why are you multiplying by another 4? And what became of the 144 length of the vector itself.

[james snail]

rewriting per suggestions received;
exercise 1.3 of chapter 3: In Exercise 1.2 if ex and ey lie along nthe perpendicular lines, and ex has a magnitude of 4, and ey has a magnitude of 6, what are the components of the vector relative to the reference frame defined by the base vectors ex and ey?

o.k., this refers to previous exercise 1.2 in which vector of magnitude 144 makes 45 degrees with each of two perpendicular lines. It asks for component vectors, which I calculated, using Pythagorean theorem, as 6 times square root of 2 (four sets with differing + and - signs).

Hoffmann gives X as = product of Vx and ex; and Y as product Vyand ey. So in trying to solve 1.3 I substituted 4 for Vx and 6 for Vy, coming up with 4ex = 6 times square root of 2 and 6ey = same (as in 1.2. both X and Y are 6 times square root of 2.) The trouble is these do not yield correct answers, which Hoffmann gives as + or - 18 times square root of two for ex and + or - 12 times square root of 2 for ey. So please tell me how to achieve correct results.

[james snail]

finally realized my error arithmatic; could have solved with Pythagorean theorem if had squared the 144 vector magnitude in exercise 1.2; this would have resulted in 144/square root of 2 = to 72 times square root of 2 for the components X and Y.