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Author Topic: Vector and Scalar Resolutes  (Read 43205 times)  Share 

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madoscar65

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Vector and Scalar Resolutes
« on: December 17, 2010, 09:34:57 pm »
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Hi guys,

For Vector and Scalar resolutes, I don't really understand the concepts behind it, like what it means and why it is used. Can anyone explain to me what scalar resolutes are and what it is used for??
Thanks :)
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luffy

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Re: Vector and Scalar Resolutes
« Reply #1 on: December 18, 2010, 06:19:31 pm »
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I am not 100% sure on this, so others can correct me if I'm wrong (because I'm doing Spesh 3/4 next year).

It would be much easier to understand if you simply drew a diagram. However, I will try my best:
Pretty much when you are completing a resolute, you are finding the projection of two different vectors in the direction of one of them.

Let there be two vectors 'a' and 'b'. Throughout these explanations, we will be carrying out a resolute of 'a' and 'b' in the direction of 'b'.

So, the scalar resolute of a and b is defined as the projection of vector 'a' times the projection of 'b'. In order to find the scalar resolute, we use the dot product. In this case, it would be a.b. I'm assuming you know how to carry out a dot product - its pretty simple anyway.

Now, when you find the dot product, you realise that its a scalar quantity (i.e. a magnitude with no direction), whereas vectors have a direction as well. Therefore, when you actually want to find the vector itself, you must find the magnitude, which is the scalar resolute and multiply it by the unit vector.
i.e. The vector resolute of 'a' and 'b' in the direction of 'b' is found by the scalar resolute multiplied by the unit vector of 'b'. Therefore, it is (a.b) x b(hat).

Notes:
- The scalar resolute is in fact, the "magnitude" of the vector resolute.
- Often, they will ask you to find the vector resolute of 'a' and 'b' parallel to the direction of 'b'. If you draw a right-angled triangle, you will quickly see, it is simple the vector 'a' subtracted by the vector resolute (a.b) x b(hat)

Yeah, I dont think I explained it well enough - tell me if I didn't. Nonetheless, a simple diagram will make it much more easier to understand.

bblovee

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Re: Vector and Scalar Resolutes
« Reply #2 on: December 18, 2010, 08:22:17 pm »
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It's too hard to explain resolutes using words, as i noticed while reading the above hahaha

but if you want a definition for scalar resolute, it is basically the magnitude of the resolute of a vector. since scalar quantities only have magnitude and no direction.

you should refer to the essential spesh book 2.2 and 2.4

otherwise you could wait until school starts and let your teacher explain it, trying to understand it yourself without any prior knowledge is very distracting and difficult
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