Vectors in different planes add up to give a zero resultant?

AI Thread Summary
Three or four vectors in different planes can indeed add up to give a zero resultant, provided they are arranged correctly in three-dimensional space. The discussion highlights that while vectors can be resolved into components along the i, j, and k axes, they must not all lie in the same defined plane to satisfy the problem's conditions. For example, two vectors can define a plane, and a third vector must not be in that same plane to achieve a zero resultant. The concept of a "crooked" plane is introduced, suggesting that vectors can still be in a single plane even if they don't align with the standard xy, xz, or yz planes. Overall, the arrangement and orientation of the vectors are crucial for achieving a zero resultant.
xphloem
Messages
10
Reaction score
0

Homework Statement



Can
1. three
2. four
vectors in different planes add up to give a zero resultant?


Homework Equations





The Attempt at a Solution



1. Yes.
2. Yes.

1. suppose that we resolve the 3 vectors in i,j,k components. Putting each one of them zero in the respective three vectors to make them in 3 separate plane. Cant they be zero?
2. similar as above. the only thing is that the fourth vector has none of its components zero.



Am I right?
 
Physics news on Phys.org
pl reply me
 
Hi xphloem,

For #1, were you thinking of something like this:

<br /> \begin{align}<br /> \vec A &amp;= (1,1,0)\nonumber\\<br /> \vec B &amp;= (-1,0,1)\nonumber\\<br /> \vec C &amp;= (0,-1,-1)\nonumber<br /> \end{align}<br />

They definitely add to zero, but they are in a single plane, so they don't satisfy the requirements of the problem.

For this problem: if you start with any two vectors, they define a plane. Then you want a third vector that is not in that same plane. Can those add to zero?
 
alphysicist said:
Hi xphloem,

For #1, were you thinking of something like this:

<br /> \begin{align}<br /> \vec A &amp;= (1,1,0)\nonumber\\<br /> \vec B &amp;= (-1,0,1)\nonumber\\<br /> \vec C &amp;= (0,-1,-1)\nonumber<br /> \end{align}<br />

They definitely add to zero, but they are in a single plane, so they don't satisfy the requirements of the problem.

For this problem: if you start with any two vectors, they define a plane. Then you want a third vector that is not in that same plane. Can those add to zero?

How can all of them be in one plane
the first on is in xy plane
second one in xz plane
and the third in yz plane
am I wrong?
 
The three (essentially only 2) vectors A, B and C together form a 'new' plane, not like the xy, xz or yz plane, but you can also have a 'crooked' plane for example. Then they are still in the same plane, even though the plane is not the xy, yz or zx plane..
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top