Solving for R2: Calculating Vector Cross Product

AI Thread Summary
R2 is expressed as a vector, R2 = 1.043j - 1.143k, which cannot equal a scalar value like 1.547. Instead, the magnitude of the vector R2 is approximately 1.547, derived from the Pythagorean theorem. The discussion clarifies that the magnitude is calculated as the square root of the sum of the squares of its components. It emphasizes that the cross product is unrelated to this calculation, as it involves two vectors. Understanding the distinction between a vector and its magnitude is crucial for grasping the concepts discussed.
hatchelhoff
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I am trying to figure out the following
If R2 = 1.043j -1.143k
Then how can
R2 = 1.547
 
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hatchelhoff said:
I am trying to figure out the following
If R2 = 1.043j -1.143k
Then how can
R2 = 1.547
A vector can't be equal to a number. However, \vec R_2=1.043\vec j-1.143\vec k implies that |\vec R_2|\approx 1.547. This follows immediately from the definition of |\vec x| for arbitrary \vec x, or if you prefer, from the Pythagorean theorem. It doesn't have anything to do with the cross product.
 
Not that there is anything wrong with Fredrik's post above, but my guess is that if you (the OP) don't know what's going on in your own post, then you won't understand the reply.

Fredrik used correct notation to show when talking about a vector and when talking about the magnitude of a vector. Your first mention of R2 is close to being in unit vector notation (without the overhead arrows or crowns or whatever convention). The second time you are simply stating the magnitude of the vector R2.

The magnitude of the vector is the square root of the squares of the magnitudes of the components. Nothing you have posted has anything to do with a cross product, which is an operation on two vectors.
 
Thanks lads, I have have a bit to learn about vectors.
 
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