What is Perpendicular: Definition and 506 Discussions
In elementary geometry, the property of being perpendicular (perpendicularity) is the relationship between two lines which meet at a right angle (90 degrees). The property extends to other related geometric objects.
A line is said to be perpendicular to another line if the two lines intersect at a right angle. Explicitly, a first line is perpendicular to a second line if (1) the two lines meet; and (2) at the point of intersection the straight angle on one side of the first line is cut by the second line into two congruent angles. Perpendicularity can be shown to be symmetric, meaning if a first line is perpendicular to a second line, then the second line is also perpendicular to the first. For this reason, we may speak of two lines as being perpendicular (to each other) without specifying an order.
Perpendicularity easily extends to segments and rays. For example, a line segment
A
B
¯
{\displaystyle {\overline {AB}}}
is perpendicular to a line segment
C
D
¯
{\displaystyle {\overline {CD}}}
if, when each is extended in both directions to form an infinite line, these two resulting lines are perpendicular in the sense above. In symbols,
A
B
¯
⊥
C
D
¯
{\displaystyle {\overline {AB}}\perp {\overline {CD}}}
means line segment AB is perpendicular to line segment CD. For information regarding the perpendicular symbol see Up tack.
A line is said to be perpendicular to a plane if it is perpendicular to every line in the plane that it intersects. This definition depends on the definition of perpendicularity between lines.
Two planes in space are said to be perpendicular if the dihedral angle at which they meet is a right angle (90 degrees).
Perpendicularity is one particular instance of the more general mathematical concept of orthogonality; perpendicularity is the orthogonality of classical geometric objects. Thus, in advanced mathematics, the word "perpendicular" is sometimes used to describe much more complicated geometric orthogonality conditions, such as that between a surface and its normal.
Problem statement : I copy and paste the (slightly different) problem statement as it appeared in the text to the right.
Attempt : By inspection, we find that the vector ##\vec B'## perpendicular to ##\vec B = 3\hat i+4\hat j## is ##\boldsymbol{\vec B' = 4\hat i -3\hat j}##, remembering that...
There are a couple of problems with the same setup. On plugging in the values and solving for the integral, I am not getting the expected values of the force. Is there something wrong in the solution attached?
For this problem,
How do we calculate the moment of inertia of (2) and (3)?
For (3) I have tried,
##I_z = \int r^2 \, dm ##
## ds = r ## ##d\theta ##
##\lambda = \frac {dm}{ds}##
##\lambda ## ##ds = dm ##
## \lambda r ## ##d\theta = dm ##
##I_z = \lambda \int r^3 d\theta ##
##I_z = \lambda...
I first tried to use a method based on Gram Schmidt orthogonalization
method:
$$
v_{\parallel}=\left(v\ldotp\frac{u}{\left\Vert u\right\Vert }\right)\frac{u}{\left\Vert u\right\Vert }+\left(v\ldotp\frac{w}{\left\Vert w\right\Vert }\right)\frac{w}{\left\Vert w\right\Vert },
$$
and
$$...
I know that each material is made up of tiny magnets due to electrons orbiting the nucleus and also from electron spinning about its own axis. In ferromagnetic or paramagnetic rod these tiny magnets align with the applied field causing the net field in the rod to increase. But for diamagnetic...
This question appeared in a university entrance exam.Basically, if magnetic flux passing through a surface of a loop changes over time ,only then e.m.f will be induced to that loop.But here only a straight line is used and there's no chance of forming any area.So by definition there's no chance...
That was my approach:
$$P_f - P_i = [(m-dm)(v + dv) + dm(u+v+dv)] - [m(v)]$$
$$= mdv - dmv + dmu + dmv = mdv + dmu = 0$$
Since the variation of the rocket's velocity is perpendicular to itself, $$ dv = v d \theta => m v d \theta + dm u = 0$$
So we have $$\frac{dm}{m} = \frac{-v d \theta}{u}$$...
I looked at this question and i wanted to ask if we could also use; ##C## =## c_2 ##(##-\dfrac {3}{2}i## +## j - 3k)## ... cheers
This problem can also be solved by using the approach of cross product ##A×B##...
Initially, I calculate the moment of inertia of of a square lamina (x-z plane). Thr this square is rotated an angle $\theta$ about a vertex and I need to calculate the new moment of inertia about that vertex.
Can I split the rotated square to two squares in the x-z plane and y-z plane to find...
Hey,
I wonder how I can calculate the force of a plate that is submerged in water.
I thought maybe the drag equation would be suitable for this with the drag coefficient set to 1.28 (https://www.grc.nasa.gov/www/k-12/airplane/shaped.html). But is implies the flow is according to the image...
a. I solved a but I don't fully understand how it works.
$$z = f_x'(1, -1)(x -1) + f_y'(1, -1)(y+1) = 2(x-1) + 3(y+1)$$
Eitherway it's b that's my issue.
I can find the gradient of both plane and surface, but trying to do "dot-product of both normals = 1" will give an equation involving two...
I've a disc which can rotate freely about two perpendicular axis (fixed to the body)
If I simultaneous try to rotate it about the two axis, what will happen?
For easy calculation, I will use a for m_1 and b for m_2 and then back substitute for a and b.
We have (0, 0) and (1, a).
d_1 = sqrt{(1 - 0)^2 + (a - 0)}
d_1 = sqrt{(1)^2 + (a)^2}
d_1 sqrt{1 + a^2}
For d_2, we are going to need (0, 0) and (1, b).
I say d_2 = sqrt{1 + b^2}...
Determine whether the lines are parallel,
perpendicular, or neither.
Line 1: y = (x/4) − 1
Line 2: y = 4x + 7
Looking at the slopes of the lines, I say neither. Is (1/4) the negative reciprocal of 4?
I say no.
My answer is neither.
Here is my attempt to draw a diagram for this problem:
I'm confused about the "the perpendicular bisector of ##BC## cuts ##BA##, ##CA## produced at ##P, \ Q##" part of the problem.
How does perpendicular bisector of ##BC## cut the side ##CA##?
A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity of the particle toward the north-west relative to the observer?
I ask because while the particles total...
I copied and pasted some code from another thread on this forum, and it has worked, however I am unsure how to change the length of the line. On the attached image, the line is the one labelled H(X,T). I want the line to be perpendicular to the plane- to pass through the blue node labelled (dfd)...
In ##\mathbb{R}^2##, there are two lines passing through the origin that are perpendicular to each other. The orientation of one of the lines with respect to ##x##-axis is ##\psi \in [0, \pi]##, where ##\psi## is uniformly distributed in ##[0, \pi]##. Also, there are two vectors in...
I have tried this question but can't get my head around to doing part b)
This is the sketch I have drawn and I'm guessing 'x' is the distance that I have to fine that's perpendicular to force Fb, though I'm not sure if I have set it up correctly?
Any help would be appreciated! Thanks
Hi I'm very stuck on what to do for these 2 questions I got wrong
Can someone please help me on what triangle I need to sketch out in order to find the 2 components of forces for F1 and F2. I'm assuming you have to make use of the sine or cosine rule here
I'd be grateful for any help given...
$$\tau = I\alpha$$
$$FL/2 = I\omega^2L/2$$
$$T = 1/\theta \sqrt{F/I}$$
would this be correct?
I came up with this more basic question to solve a slightly harder question so I do not know the answer to the above-stated problem.
This is my sketch:
I think the only possible way to get emergent beam perpendicular to AB is when the incoming ray is refracted towards side AC then total internal reflection occurs as shown in my sketch.
Angle QAR = 60o and angle ARQ = 90o so angle AQR = 30o
It means angle CQP is also 30o...
It is said that lift on a wing always acts perpendicular to the relative wind.
Why is this so? Is it because we arbitrarily choose to analyze its components with one purposely chosen to be perpendicular or is it a matter of physics that the lift component is naturally perpendicular. If so...
I think i could deal with this problem interpreting this force like a central force, what seems pretty nice to me, since in a circular orbit the force will always pass through the center, if it is perpendicular to the velocity.
I thought, since the force is central and in this case, spherically...
There's a constant magnetic field B. If a particle is acted on by a force qv*B (* cross) only, and the initial velocity v0 is normal to B, is the motion certainly a circular one (for any m, q, v0)?
mv''=qv*B
If one solves this equation (vector), it doesn't seem obvious.
If route perpendicullar forces supose to have no affect, why isn't this the case when somebody tries to cross a river? the motion is perpendicullar when water speeds aprox. at the same direction.. what am i missing?
Let's say we have two people that travel with respect to each, other in some inertial frame. So if we took the relative velocity of them will they be the same in as v^12+v2^2
Hey, not sure how to translate this from my native language, I hope you understand what I mean.
Write down for the line
y = 2x + 3 perpendicular and parallel lines passing through the point
(1; 1) equations.
Hi everyone!
So I got a question shown below here and I’m struggling in what the question is trying to say here and what it wants me to give?
Here is what I done
For when resolving the individual x and y component of forces, my angle has always been measured counter clockwise from the...
μThe magnetic field is supposed to be B=μ0i/(2πr). I think that the force would be aimed upwards.
Now I tried to divide the wire BC to infinite smaller wires where B is roughly constant in each one of them.
What I get is:
However this is not correct when I plug the relevant numbers in it...
Say that these pictures are accurate.
Personally, I think I can solve this problem, but the issue is, that I had a debate on it with my Physics teacher.
This is how my teacher would solve it; my teacher says that each Fk on each side is constructed from half of F and Fx. So, F = 2 * Fk * sin...
Homework Statement
I have a conceptual question regarding why the x and y-axis must be perpendicular. Chapter 2 in section 2-1 of Giancoli's Physics states that " The x and y axes are always perpendicular". This chapter is one-dimensional motion and the section is Reference frames and...
I have a question for my E&M assignment (I'm not putting it in the homework thread since I just need more of a concept check) that involves a straight wire with current through it. If you have a wire with a current through it, and you have a rectangular wire also with current, to the right of...
Hi, I am really stuck! I need to find the equation of the plane through the line x=2y=3z perpendicular to the plan 5x+4y-3z=8. Can anyone give me any pointers of where to start with this? Not expecting a full solution, just an idea of where to start.
THanks!
since it is known that ##\vec{A_\perp} = -{mG \over R^2}## why did the professor write it as ##\vec{A_\perp} = {- R G \rho \over 3}## for perfect sphere with perfect mass distribution ? Shouldn't it be ##\vec{A_\perp} = -{4 \over 3} \pi R G \rho##? I need help thanks.
Homework Statement
A 15-cm-diameter CD has a mass of 21 g .
What is the CD's moment of inertia for rotation about a perpendicular axis through its center?Homework Equations
I = (1/2)MR^2
3. The Attempt at a Solution
I = (1/2)(0.21g).((15/2)^2)
= 5.9*10^-4 kg*m^2
cant see what I am...
Hi, i get the math that is involved but if I have only the x,y coordinates for 2 points to connect and if i want to know what will be the perpendicular line to the line connecting these two points, how can I do that?
I have a Force Vector = 100N, making an Angle = 45 degrees with x-axis.
When I find their Components trigonometrically, I get 70N each; as
Fx = 100xCos(45) = 70N
Fy = 100xSin(45) = 70N
Verifying the result, by Head-to-Tail method, I get 70N + 70N = 140N.
Why is there discrepancy or where am I...
Let's say that we have two guys, each weighs 50 kg. They are pulling a piece of against each other as hard as they can. The tension of the rope will therefore be the force of friction acting on each of them (the same magnitude).
Now, let's say that a third person joins the party and grabs the...