Find two unit vectors that are parallel to the xy plane

AI Thread Summary
To find two unit vectors parallel to the xy plane and perpendicular to the vector [1, -2, 2], the z component must be zero, resulting in vectors of the form [a, b, 0]. The condition for perpendicularity requires the dot product with [1, -2, 2] to equal zero, leading to the equation a - 2b = 0. This equation has infinite solutions, but to find unit vectors, the magnitude must equal one. For example, the vectors [2, 1, 0] and [4, 2, 0] can be scaled to unit length, yielding the final answers. Ultimately, the requirement for unit vectors limits the solutions to specific normalized forms.
PiRsq
Messages
112
Reaction score
0
I got the answer to this question but I don't quite understand why its like that...The question is:

Find two unit vectors that are parallel to the xy plane and perpendicular to the vector [1,-2,2]
 
Last edited by a moderator:
Physics news on Phys.org
Find two unit vectors that are parallel to the xy plane and perpendicular to the vector [1,-2,2].

It would be better to show us what you had done and exactly what it is that you "dont quite understand".

However, since this is very simple: Saying that a vector is "parallel to the xy plane" means that its z component i 0- you are looking for a vector of the form [a, b, 0].

Saying that the vector is "perpendicular to the vector [1, -2, 2] means that its dot product with that vector is 0: [a, b, 0].[1, -2, 2]= a- 2b= 0.

That is one equation in 2 unknowns so it have an infinite number of solutions. In particular, if you take b= 1, then a-2= 0 so a= 2.
[2, 1, 0] is a vector "parallel to the xy plane and perpendicular to the vector [1, -2, 2]". Taking b= 2, a- 4= 0 so [4, 2, 0] is another. In fact, it is just [1, -2, 0] multiplied by 2. Obviously, if one vector is "parallel to the xy plane and perpendicular to the vector [1, -2, 2]", then any multiple of it is also because it is in the same direction.
 
Plus, it says 'unit vectors'. So there's only 2 answers to this.
 
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