What Is the Result of -4.00 C·(4.00 A×B) in Vector Calculations?

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SUMMARY

The discussion focuses on calculating the result of the expression -4.00 C·(4.00 A×B) in vector calculations. Participants emphasize the importance of first computing the cross product A×B using the formula a x b = (a2b3 - a3b2) i + (a3b1 - a1b3) j + (a1b2 - a2b1) k, before applying the scalar multiplication. A suggestion is made to avoid multiplying the vectors by 4 initially to prevent complications with large numbers. The final result is derived by first finding C·(A×B) and then multiplying by -16.

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Homework Statement


For the following three vectors, what is -4.00 C·(4.00 A×B)?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c03/eq03_86.gif"

Homework Equations


a x b = (a2b3 - a3b2) i + (a3b1 - a1b3) j + (a1b2 - a2b1) k

The Attempt at a Solution


I've multiplied 4 to each of the numbers in A, then plugged in numbers of A and B to the equation, next I multiplied -4 x C and used those to multiply the answer of A and B to get 224 + 1024 + 68k. I really am lost here.

Homework Statement


Use the definition of scalar product, vectors a ·b = ab cos θ, and the fact that vectors a · b= axbx + ayby + azbz to calculate the angle between the two vectors given by = 8.0i + 8.0j + 8.0k and = 7.0i + 5.0j + 6.0k.

The Attempt at a Solution


I found the magnitude of A = 11.3 and B = 8.6 using pythagorean theorem. Then I did 8x7 + 8x4 + 8x6 = 144. Then cos-1(144/97.2), but I don't get an angle...
 
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Hi Gattz! :smile:

i] don't multiply the vectors by 4 at the start …

it makes the numbers too large :rolleyes: …

find C.(AxB) first, then multiply it by -16. :wink:
Gattz said:

Homework Equations


a x b = (a2b3 - a3b2) i + (a3b1 - a1b3) j + (a1b2 - a2b1) k

ii] yes, this is correct … use it! (for AxB)

(and don't bother with the (a1b2 - a2b1)k … you won't need it when you dot-product with C, will you? :smile:)
 

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