Question: Given u = ( 2 , 4 , − 3 ) , v = ( 3 , − 1 , 7 ) , and i,j,k being the standard basis vector, find: | v − u + 3 k | My work: = | ( 3 , − 1 , 7 ) − ( 2 , 4 , − 3 ) + 3 k | = | ( ( 3 − 2 ) , ( − 1 − 4 ) , 7 − ( − 3 ) ) + 3 k | = | ( 1 , − 5 , 10 ) + 3 k | I know that the last line is not the full answer but I don't know what to do to continue to get the final result. I was thinking on making the coordinates (1,−5,10) into i−5j+10k and then adding the 10k with the 3k to get 13k but I am still confused on what to do with the absolute expression afterwards. If anyone can help me out, I'd appreciate it.

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Question: Given   u  =  (  2  ,  4  ,  −  3  )  ,  v  =  (  3  ,  −  1  ,  7  )  , and i,j,k being the standard basis vector, find:      |    v  −  u  +  3  k      |  My work:  =      |    (  3  ,  −  1  ,  7  )  −  (  2  ,  4  ,  −  3  )  +  3  k      |    =      |    (  (  3  −  2  )  ,  (  −  1  −  4  )  ,  7  −  (  −  3  )  )  +  3  k      |    =      |    (  1  ,  −  5  ,  10  )  +  3  k      |  I know that the last line is not the full answer but I don&#039;t know what to do to continue to get the final result. I was thinking on making the coordinates (1,−5,10) into i−5j+10k and then adding the 10k with the 3k to get 13k but I am still confused on what to do with the absolute expression afterwards. If anyone can help me out, I&#039;d appreciate it.

nuramaaji2000fh · Accepted Answer

You should convert 3k to the vector (0,0,3).      |    (  1  ,  −  5  ,  10  )  +  (  0  ,  0  ,  3  )      |    =      |    (  1  ,  −  5  ,  13  )      |  Now, in case you weren&#039;t aware, the magnitude of a vector       |              v      →            |   is the distance between its two ends. This means that if we have a vector             v      →        =  (      v    1    ,      v    2    ,      v    3    ), then by Pythagoras&#039;s theorem we have             v      →        =            v      1      2        +          v      2      2        +          v      3      2      Using this definition we find the magnitude of   (  1  ,  −  5  ,  13  ) is            1      2        +    (    −    5          )      2        +          13      2        =      1    +    25    +    169    =      195

Bruno Thompson · Accepted Answer

You are right! Finish using this:      |    a  i  +  b  j  +  c  k      |    =            a      2        +          b      2        +          c      2        .

Question: Given u=(2,4,−3),v=(3,-1,7), and i,j,k being the standard basis vector, find: |v-u+3k|

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