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I don't know what you mean by "zero component along a line."
If you look at the representation of a vector on paper using a Cartesian coordinate system -- in other words, one using x and y axes -- the orthogonal components of the vector are the projections of the vector on the x and y axes. If the vector is parallel to one of the axes, its projection on the other axis will be zero. But the vector will still have a non-zero magnitude. Its entire magnitude will project on only one axis.
But a vector must have magnitude AND direction. And if it has zero magnitude, its direction cannot be determined.
Still trying to make heads or tails out of your question.......
If you draw a random vector on a Cartesian grid, it will have an x component and a y component, which are both projections of the original vector upon the axes. However, it could also be represented by projecting it onto a new set of orthogonal axes -- call them x' and y' -- where the x' axis is oriented to be parallel to the original vector and the y' vector is perpendicular to it. In that case, the x' component will have a magnitude equal to the magnitude of the original vector -- in other words, a non-zero value along a line parallel to the x' axis -- and a zero magnitude in the y' direction.
What else can I help you with?
Can a vector have zero component along a line and still have non-zero magnitude?
Yes, if it has a non-zero component along some other line - usually, but not necessarily orthogonal.
Can a vector have a component equal to zero and still have a nonzero magnitude?
Yes. For instance, the 2-dimensional vector (1,0) has length sqrt(1+0) = 1 A vector only has zero magnitude when all its components are 0.
Can the component of a non-zero vector be zero?
Yes, the component of a non-zero vector can be zero. A non-zero vector can have one or more components equal to zero while still having a non-zero magnitude overall. For example, in a two-dimensional space, the vector (0, 5) has a zero component in the x-direction but is still a non-zero vector since its y-component is non-zero.
Is their any physical quantity having both magnitude and direction still not a vector?
I don't think so - is something has a magnitude and a direction, by definition it is a vector.
How do you find the y-component of a vector if you are not given the magnitude of the vector?
It depends on what other information you are given. If you are given the x-component and the angle, then you can still use trig by doing the following. Tan(angle)=y/x If you do not have the x component, then you will need to use other kinematic equations to find the answer. The different ways to find this information are too numerous to list out, but I will give an example of a common one. If you are given the horizontal distance and time the object flies, use d=vt to find the x component of velocity, then use the first strategy.
Can a vector have zero component along a line and still have non-zero magnitude?
Yes, if it has a non-zero component along some other line - usually, but not necessarily orthogonal.
Can a vector have a component equal to zero and still have a nonzero magnitude?
Yes. For instance, the 2-dimensional vector (1,0) has length sqrt(1+0) = 1 A vector only has zero magnitude when all its components are 0.
Is their any physical quantity having both magnitude and direction still not a vector?
I don't think so - is something has a magnitude and a direction, by definition it is a vector.
How can a null vector be a vector if there is no direction?
A null vector does not have a direction but still satisfies the properties of a vector, namely having magnitude and following vector addition rules. It is often used to represent the absence of displacement or a zero result in a vector operation.
Can unit vector have negative component?
Yes, a unit vector can have negative component since a unit vector has same magnitude and direction as a negative unit vector. Here is the general work out of the problem: Let |v| be the norm of (v1, v2). Then, the unit vector is (v1/|v|, v2/|v|). Determine the "modulus" or the norm |(v1/|v|, v2/|v|)| to get 1, which is the new norm. If we determine the norm of |(-v1/|v|, -v2/|v|)|, we still have the same norm 1.
What is it when two vectors' dot product is one?
That fact alone doesn't tell you much about the original two vectors. It only says that (magnitude of vector-#1) times (magnitude of vector-#2) times (cosine of the angle between them) = 1. You still don't know the magnitude of either vector, or the angle between them.
How do you find the y-component of a vector if you are not given the magnitude of the vector?
It depends on what other information you are given. If you are given the x-component and the angle, then you can still use trig by doing the following. Tan(angle)=y/x If you do not have the x component, then you will need to use other kinematic equations to find the answer. The different ways to find this information are too numerous to list out, but I will give an example of a common one. If you are given the horizontal distance and time the object flies, use d=vt to find the x component of velocity, then use the first strategy.
Can you find a vector quantity that has a magnitude of zero but components that are different from zero?
No, that's not possible - at least, not with vectors over real numbers. The magnitude of a vector of components a, b, c, d, for example, is the square root of (a2 + b2 + c2 + d2), and as soon as any of those numbers is different from zero, its square, the sum, and the square root of the sum will all be positive. It is not possible (in the real numbers) to compensate this with a negative number, since the square of a real number can only be zero or positive. Another answer: In special relativity we use a metric for vectors different from the Euclidean one mentioned above. If (t, x, y, z) is a 4-vector in Minkowski space the squared "length" is defined as t2 - x2 - y2 - z2. As you can see this can be negative (for spacelike vectors), positive (for timelike vectors) or zero (for null, or lightlike vectors). See related link for more information
Is velocity a vector or a scalar?
no its a vector quantity,not a scalar quantity,bcz still it z a velocity bt NT a speed On a typical journey the average velocity is the straight-line distance between the start and finish, divided by the time taken, and it also has a direction. The average speed is the actual distance run, divided by the speed. The average speed might not be equal to the magnitude of the average velocity. For example on a round trip the average speed might be 40 mph, while the average velocity is zero.
Are two vectors having equal magnitude?
It depends. Magnitude is technically the length of the vector represented by v. our equation of the magnitude is given by: v= SQRT( x^2 + y^2) You can have 2 similar vectors pointing at different directions and still get the same magnitude.
Is the average speed equal to the magnitude of the average velocity justify your answer?
Yes usually and no rarely, velocity is defined as a vector, having both a direction and a magnitude (which is speed in the case of velocity). For instance 100 mph (speed) east (0o) (direction). In this form it is easy to see that the magnitude is 100 mph but mathematically to determine the magnitude of a vector you would divide the vector by its direction. 100 mph 0o / 0o = 100 mph Average speed and average velocity share the same relationship as instantaneous speed and instantaneous velocity so divide out the average direction from your average velocity to determine your average speed. If this is over a time period and you know the beginning and ending places in space your averages will simply be the difference from the starting to the ending places. So yes so long as you define speed to actually be the magnitude of the vector. However, if speed is taken without direction over time it may become something different. If an object travels along a vector with a negative magnitude its speed will not be negative but its vector magnitude will. Ex: A car travelling in reverse still has a positive speed but a compass will show it to be heading in the opposite direction of travel, a negative vector value...
Is force a vector quantity if so why?
Yes, force is a vector quantity, because it is exerted in a specific direction (even in the case of a symmetrical explosion, in which force is exerted in all directions, that is still a type of vector).
