u=2i+4j+6k

v=i+k

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Eric Y. | SAT PrepSAT Prep

The magnitude of the cross product is the area of the parallelogram with two sides A and B.

First find the cross product; the answer should be a vector (x,y,z).

Then change that vector into a single magnitude.

u x v = (2i+4j+6k) x (i+k)

= 2(ixi)+2(ixk)+4(jxi)+4(jxk)+6(kxi)+6(kxk)

Use Right Hand Rule:

u x v = 2(0)+2(-j)+4(-k)+4(i)+6(j)+6(0)

= 4i+4j-4k

|u x v| = √(4^2+4^2+(–4)^2) = √(3(4^2)) = 4√(3)

area of the parallelogram = 4√(3)

Thank you but why did i×i became 0 and i×k became -j? i didnt get that part.sorry i am terrible with cross products

a x b = |a| |b| sin(θ) n, where θ is angle between a and b and n is a unit vector perpendicular to both in direction determined by right hand rule. So if the two vectors are the same, then the angle is 0 and the sine is 0. E.g., i × i = 1*1*sin(0) = 1*0 = 0. Instead of right hand rule you can use signed angles. CCW in x-y plane is positive, CCW in y-z plane is positive, and CCW in z-x plane is positive. i × k = 1*1*sin(–90°) j = 1*1*(–1) j = – j.

I thought that the are was 2 because.

|u×v|=√4+4-4=√4=2

I mean area

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