Vector question: i, j, k


by DB
Tags: vector
DB
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#1
Nov4-05, 07:47 PM
P: 501
i've been reading on the internet about vectors and i keep seeing this stuff about "i, j, k". im learning vectors in physics but my teacher hasnt mentioned any of this stuff. ive read that if "i, j, k" are the components of the x, y and z axis? and that when u multiply two different letters u get 0 and to like letters u get 1. basically what im asking is can u help me understand wat these letter means n wat we use them for?

thanks
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z-component
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#2
Nov4-05, 07:59 PM
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That's correct. This form of vector expression is called unit vector notation. Vectors can be broken into i j and k, representing the x y and z axes, respectively. Basically it's a more standard way of expressing vectors without any relative angles.

For example, I can express "50 N at an angle of 30 degrees relative to the horizontal" in unit vector notation by finding the x and y components like usual using cos 30 and sin 30 times the magnitude, 50. With unit vector notation, you can just say 43.3i + 25j N (50 cos 30 = 43.3 & 50 sin 30 = 25).

Makes sense?
DB
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#3
Nov4-05, 08:53 PM
P: 501
ya thanks z.c, so basically its like saying 43.3 units along the x axis, 25 units along the y?

wat about the fact that i*j=0 and i*i=1?

ahrkron
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#4
Nov4-05, 09:29 PM
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Vector question: i, j, k


Quote Quote by DB
wat about the fact that i*j=0 and i*i=1?
The "*" there stands for the "dot product", or "inner product" of the vectors. It is basically an orthogonal proyection; it returns the length of the "shadow" that one vector projects over the other one (where this "shadow" falls perpendicularly to the vector into which you are projecting).

Two vectors with similar direction will have large shadows, almost as long as they are. Two vectors with the same exact direction will have shadows exactly as long as they are (hence i*i=1); finally, vectors perpendicular to each other will not project any shadow on each other (i*j=0).

On intermediate directions, you need a cosine to saw this all together.
z-component
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#5
Nov4-05, 11:07 PM
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Quote Quote by DB
ya thanks z.c, so basically its like saying 43.3 units along the x axis, 25 units along the y?
wat about the fact that i*j=0 and i*i=1?
That's right, and it's convenient to have this form of a combined resultant vector that shows the resultant x, y and z components all together. If we had a k term, we'd have the location along the z axis as well.

Knowing that [tex]v = \frac{dx}{dt}[/tex] (velocity is equal to the derivative of position x with respect to time t), if we have a vector of position in terms of time, we can find velocity by deriving the unit vector.
For example, given [tex]x = 75t^{2}\hat{i} + 1.25t\hat{j}[/tex],
Find velocity by: [tex]v = \frac{d(75t^{2}\hat{i} + 1.25t\hat{j})}{dt}[/tex]


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