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Can someone help me with this questions?.But please dont give me the answer.I just dont understand this lesson

1)Find the cross product and show it is orthogonal to both u and v.
u=<-3,2,3>
v=<0,1,0>
 
2)Find the cross product and show it is orthogonal to both u and v.
     u=6k
     v=-i+3j+k
 
 
3)Find the area of the parrallelogram that has the two vectors as adjacent sides
 
u=2i+4j+6k
v=i+k
 
4)Find the cross product of the unit vectors
                   J×k
                 i j k
                0 1 0
                0 0 1
 
 
THE LESSON IS CROSS PRODUCT OF TWO VECTORS
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1 Answer

The vector cross product is defined as
 
u x v= (uyvz-uzvy)+ (uzvx-uzvz)j + (uxvy-uyvx)k where ux is the x component of U
 
u=uxi+uyj+uz and likewise for v
 
1. you were given the components of u and v using the above definition of the cross product calculate the vector a=uxb. The next step is to recall that if two vectors are orthogonal then one is perpendicular to the other and the dot product is zero. So you have to show that a• u =0 and a•v =0
 
Does this help get you started?
Jim

Comments

u cross v  is the definition of the vector cross product you calculate the components by multiplying the components of u and v together as I've shown above.
Would you like me to work the whole problem for you completely?
jim
Daisy,
              you were given u an v as components u=<-3,2,3>
v=<0,1,0> 
 
so u=-3i+2j+3k and v=j
 
and u x v= (uyvz-uzvy)i + (uzvx-uxvz)j + (uxvy-uyvx)k =(2*0-3*1)i+(3*0-(-3)*(0))j + (-3*1-2*0)K=-3i+0j-3K=u cross v and this vector is perpendicular to v because (-3i-3k)•j=0 because i and k are perpendicular to j so i•j=0 and k•j=0.
Hope this helps a little
Jim
 
 
 
 
Daisy,
          Try this link for additional help with the cross product.
Jim
Can you tell my if my answers are wron? 
 
For number 4. I found the cross product to be 1i-0j+0k
Yes you are correct jxk=i You got it now nice work
Jim

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