Find the vector, not with determinants, but by using properties of cross products.


Find the vector, not with determinants, but by usi

Answer

We recall the cross product property that

i times i =0,  j times j =0,  k times k =0,

i times j =k,  j times k =i,  k times i =j,

j times i =-k,  k times j =-i,  i times k =-j

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Answer: 1.        We need to find k times (i -7j)

   k times (i -7j) = ktimes i - 7(ktimes j)

       

= {color{Blue} 7i-1j+ 0k}

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Answer 2.   Now we find (i+j)times (i-j)

(i+j)times (i-j) = itimes (i-j) +jtimes (i-j)

(i+j)times (i-j) = i times i- i times j+j times i-j times j

(i+j)times (i-j) = 0- k+(-k)-0

(i+j)times (i-j) = {color{Blue} - 2k}

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Answer 4. Given a = left langle 2, -1,  2 right rangle and b = left langle 5,  2,  1 right rangle

atimes b =left langle 2,  -1,  2 right rangle times left langle 5,  2,  1 right rangle

Using distributive property and the properties of cross product

atimes b =4 k -2j +5k -1i +10j -4i  

atimes b ={color{Blue} -5i +8j +9k}

Now we find btimes a

btimes a=left langle 5,  2,  1 right rangle times left langle 2,  -1,  2 right rangle

btimes a = -5k -10j -4k +4i +2j +1i

btimes a ={color{Blue} 5i -8j -9k}

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Note: The problem #3 was cut-off in the image, so I skipped that.


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