Vector Multiplication in a Triangle on the Cartesian Plane

AI Thread Summary
The discussion focuses on calculating the vector multiplication of three vectors in a triangle on the Cartesian plane, specifically A, B, and C with magnitudes of 16, 12, and 20 respectively. The magnitude of A x B is confirmed to be 192, with the direction along the -x axis. The use of determinants for vector multiplication is emphasized, highlighting that the cross product is not commutative and results in a new vector. A mnemonic for remembering the cross product relationships between unit vectors is provided, aiding in understanding the calculations. Ultimately, all answers for the magnitudes were found to be 192, with some complexity in determining the directions.
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Homework Statement



For the vectors in a Triangle, with a = 16, b = 12, and c = 20 what are

(a) the magnitude and (b) the direction of A x B

(c) the magnitude and (d) the direction of A x C

(e) the magnitude and (f) the direction B x C

this is Vector Multiplication.

Homework Equations


ABsin(ø)
pythagorean theorem
tan^-1 (y/x)
3. The Attempt at a Solution

16i x 12j = 192 (magnitude of A x B it is correct) and the direction is along the -x axis (believed)

since this is vector multiplication, I realized that the best way to do it was using the determinant. which is a value associated with a square matrix. We can get the above relations following matrix algebra procedures to calculate the determinant of a 3×3 matrix, which in favor ends up giving us the vector product. if I am not mistaken the vector product can be expressed by multiplying the vectors, now here is where the problem arises. multiplying vector is not commutative and when multiplying vectors you get a new one (on the Z direction?)
 
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You should learn the different results of the cross-products between the unit vectors.

A handy mnemonic (a memory device) is

i j k | i j

which translated means (reading from left to right):

i x j = k

j x k = i

k x i = j

if you read the mnemonic in reverse, from right to left, you get:

j x i = -k

i x k = -j

k x j = -i

Any unit vector crossed with itself is zero:

i x i = j x j = k x k = 0

This can be confirmed also by writing out the cross products using determinants.
 
ah Indeed thank you very much. surprise to find out that after all the attempts all of the answers were 192. and the directions were a little bit tricky but I got the right answer.

thanks for the guidance.
 
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