Let a → = i + 2 j + 3 k and b → = 2 i + 5 k. For which value of t when − 2 ≤ t ≤ 2 holds the length of the vector c t → = t a → + ( 1 − t ) b → is as small as possible?How should one approach this? The minimum would be when c t → is perpendicular to some vector d → = b → − a → or is there something else Im not seeing?

Question

Let             a      →        =  i  +  2  j  +  3  k and             b      →        =  2  i  +  5  k. For which value of t when   −  2  ≤  t  ≤  2 holds the length of the vector                     c        t            →        =  t            a      →        +  (  1  −  t  )            b      →        is as small as possible?How should one approach this? The minimum would be when                     c        t            →       is perpendicular to some vector             d      →        =            b      →        −            a      →       or is there something else Im not seeing?

Jazlene Dickson · Accepted Answer

Hint:Write out the magnitude as a function of   t and then differentiate to find the maximum.

Pierre Holmes · Accepted Answer

Write                     c        t            →        =  t  (            a      →        −            b      →        )  +            b      →      . Then,                              |                                                    c              →                                t                                      |                    2                                    =                  |                                      a            →                          −                              b            →                                                |                    2                          t          2                +        2        (                              a            →                          −                              b            →                          )        ⋅                              b            →                                  t        +                  |                                      b            →                                                |                    2                                                  =                  |                                      a            →                          −                              b            →                                                |                    2                                      (            t            +                                          (                                                      a                    →                                                  −                                                      b                    →                                                  )                ⋅                                                      b                    →                                                                                                |                                                                      a                    →                                                  −                                                      b                    →                                                                                        |                                    2                                                      )                    2                −                              (            (                                          a                →                                      −                                          b                →                                      )            ⋅                                          b                →                                                    )              2                                                          |                                                      a                →                                      −                                          b                →                                                                    |                            2                                      +                  |                                      b            →                                                |                    2                                                  ≥                              (                          |                                                      a                →                                      −                                          b                →                                                    |                                      |                                                      b                →                                                    |                                      )              2                        −            (            (                                          a                →                                      −                                          b                →                                      )            ⋅                                          b                →                                                    )              2                                                          |                                                      a                →                                      −                                          b                →                                                                    |                            2                                      =                  |                                                    c              →                                t                                      |                                m            i            n                    2

Let vec(a) =i+2j+3k and vec(b) =2i+5k. For which value of t when −2 <= t <= 2 holds the length of the vector vec(c_t)=t vec(a) +(1−t)vec(b) is as small as possible?

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