Finding Components of Perpendicular Vectors in a Cartesian System

[Johnny Blade]

Homework Statement

Two vectors (c and d) are in a xy Cartesian system and they are perpendicular to the "a" vector (ax=3.2, ay=1.6) and they have a module of 5 m. The vector c has a positive x component, and the other, d, has a negative x component.

b) What are the x and y components of the vector "c"?
c) What are the x and y components of the vector "d"?

Homework Equations

I found the slope of the c and d vectors. m=-2. With the slope I made this: (c-d)^2+(-2c+2d)^2=25

The Attempt at a Solution

I think I need to find another equation and use substitution with the above equation, but now I can't think of another equation.

 

[berkeman]

I'm not familiar with the term "module" in this context. Is that the magnitude of each of the two vectors c & d, or is it somehow their total length, end to end?

You are on the right track getting the slope of c & d. But I don't quite get this part:

"(c-d)^2+(-2c+2d)^2=25"

If the magnitude of c is 5, then you will have \sqrt{{c_x}^2 + {c_y}^2} = 5

And the 2nd equation comes from your knowing the slope, which is the ratio of c_x and c_y

Does that help?

 

[Johnny Blade]

Module might be a mistake on my part. I go at a French school, so I translated this problem with limited knowledge of English terms.

Module is like the d_R (Déplacement resultant). It's the distance between the start of the first vector and the end of the last vector.

Because the module of them is 5 wouldn't this (c-d)^2 + (-2c+2d)^2 = 25 be the right equation?

With the second equation I will find c, and then I would substitute it in the first equation to find the d.

 

[learningphysics]

Is the module the magnitude of a vector? I'm confused because you said the length of them is 5.

But you also said it is the length from the start of the first vector, to the end of the last...

 

[berkeman]

I'm still not tracking what you are saying, Johnny_Blade. The sketch that I drew for myself has the original vector going up and to the right, with the slope of 1/2. And the line that c and d are on is perpendicular, going through the origin with a slope of -2 as you said.

And I drew the vector c aong that line, starting at the origin and pointing down and right. I drew the d vector starting at the origin, pointing up and left along the same line as c. I labelled c with a length of 5, and I labelled d with a length of 5. Then the equations that I wrote in my first post would apply, and you can calculate the cartesian coordinates of the c and d vectors.

But if I'n not interpreting the problem statement and the term "module" correctly, then my sketch may not be correct.

 

[Johnny Blade]

I'll use it in a example.
\vec{c}=2m [E]

\vec{d}=3m [E]

\vec{d_R} (this is like the module, except the module is the quantity only.) =?

\vec{d_R}=\vec{c}+\vec{d}

\vec{d_R}=5m [E]

The module is 5m.

 

[learningphysics]

we know that c and d are in opposite directions...

Suppose I write \vec{c} = A\vec{u} where u is a unit vector in the direction of c... so the magnitude of c is A.

and suppose the magnitude of d is B.

so \vec{d} = -B\vec{u}

so \vec{c}+\vec{d} = (A - B)\vec{u}

Does that mean A-B is the module of c and d. (or B-A if B>A) ?

There seems to be infinite solutions for A-B = 5. or B-A = 5 etc...

 

[berkeman]

There seems to be infinite solutions for A-B = 5. or B-A = 5 etc...

I agree. If each is 5m long, then the problem is solvable. If the difference is 5m, there are infinitely many solutions to the problem as stated. Johnny -- can you maybe provide a pointer to a web page that defines the "module" of vectors? Even if it is in French, that might help.

 

[Johnny Blade]

I'll check at school tomorrow. Maybe I made a mistake somewhere in my notes or something.