Solving problem in Banesh Hoffmann's Vectors

[james snail]

Homework Statement

Exercise 1.3, chapter 3: if ex and ey lie along 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?

Homework 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= Vxtimesex, and Y = [V]ytimesey.

The Attempt at a Solution

since in 1.2 X and Y are each = 6 X square root of 2, I set V_{xex = 6 X square root 2 thus 4ex = 6X square root 2. Answer ex = 3/2 square root of 2; similarly, ey = 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 e_y, I wrote e[ sub]y[ /sub], (again, omitting the leading spaces inside the brackets).

Also, welcome to Physics Forums!

 

[HallsofIvy]

Homework Statement

Exercise 1.3, chapter 3: if ex and ey lie along 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?

Homework 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= Vxtimesex, and Y = [V]ytimesey.

The Attempt at a Solution

since in 1.2 X and Y are each = 6 X square root of 2, I set V_{xex = 6 X square root 2 thus 4ex = 6X square root 2. Answer ex = 3/2 square root of 2; similarly, ey = 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.}

<sub>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.</sub>

 

[james snail]

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

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 V_x and e_x; and Y as product V_yand e_y. So in trying to solve 1.3 I substituted 4 for V_x and 6 for V_y, coming up with 4e_x = 6 times square root of 2 and 6e_y = 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 e_x and + or - 12 times square root of 2 for e_y. 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.