Help with finding parallel vectors

  • Thread starter Thread starter Aske
  • Start date Start date
  • Tags Tags
    Parallel Vectors
AI Thread Summary
To determine the values of n and m for vectors u and v to be parallel or perpendicular, one must use the properties of vector equations. For parallel vectors, u can be expressed as a scalar multiple of v, leading to three equations based on their components. For perpendicular vectors, the dot product of u and v must equal zero, resulting in another equation involving m and n. The solution involves isolating the unknowns and solving the resulting system of equations. Understanding these principles is crucial for effectively tackling the problem.
Aske
Messages
9
Reaction score
0

Homework Statement


The question states "let u= 2i+mj-10k and v=i-3j+nk, find the value of n and m such that u,v are parallel", the second part states asks the same but "u,v are perpendicular"


Homework Equations





The Attempt at a Solution


I attempted to use a dot product solution I guess, because vectors u+v should equal 0 when perpendicular. I'm lost on what to actually do. Thanks in advance.
 
Physics news on Phys.org
Remember the definition of the dot product:

\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z

As you correctly stated, for two perpendicular vectors, the dot product is 0.

Let two vectors, \vec w, \vec q be parallel. That means that they're both in the same direction, and therefore, they only differ by some scalar factor R (For instance, \vec q could be 2 times longer than \vec w (R=2) or it could be the same length, but anti-parallel (A negative R value of -1 would achieve that goal) or 2 times longer, but anti-parallel (R=-2)).

So in general, we can write: \vec q=R\vec w
Note that we've written a vector equation. That's actually 3 scalar equations in one. Simply solve for your three variables, R, m, n and you're done.

What's important is that you understand how we've identified parallel\anti-parallel vectors and perpendicular ones. Are these two points clear to you?
 
If two vectors

u=x1 i +y1 j + z1 k and v=x2 i +y2 j + z2 k

are parallel, then one of them is a multiple of the other

u=a*v (a is a scalar).

That means the same for all components:


x1=a*x2
y2=a*y2
z1=a*z2.

The two vectors are perpendicular if their dot product is 0 which means

x1x2 + y1y2+ z1z2 = 0.


ehild
 
RoyalCat said:
Remember the definition of the dot product:

\vec A \cdot \vec B = A_x B_x + A_y B_y + A_z B_z

As you correctly stated, for two perpendicular vectors, the dot product is 0.

Let two vectors, \vec w, \vec q be parallel. That means that they're both in the same direction, and therefore, they only differ by some scalar factor R (For instance, \vec q could be 2 times longer than \vec w (R=2) or it could be the same length, but anti-parallel (A negative R value of -1 would achieve that goal) or 2 times longer, but anti-parallel (R=-2)).

So in general, we can write: \vec q=R\vec w
Note that we've written a vector equation. That's actually 3 scalar equations in one. Simply solve for your three variables, R, m, n and you're done.

What's important is that you understand how we've identified parallel\anti-parallel vectors and perpendicular ones. Are these two points clear to you?
I somewhat understand what you are saying, so are you saying I need to isolate the unknowns? I still don't fully understand this.
 
Yes, solve the linear system of equations.
 
Ok, trying this with the perpendicular qustion it'd be "(2i+mj-10k) . (i-3j+nk)=0". I would have 2 unknowns, how would I solve this. I still don't know how to go about the parallel question.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top