How to find a vector with two 3d points?A:(2,3,1); B:(8,-2,4); C:(3,0,5).find a so A → B → is orthogonal to A → B → + a . A → C → I think the answer is ( A → B → ) . ( A → B → + a . A → C → ) = 0 and then find a.(if this isn't the answer please tell me.)but how can I write or find A → B → . I think with matrices because that's what we are studying now.

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How to find a vector with two 3d points?A:(2,3,1); B:(8,-2,4); C:(3,0,5).find a so             A      →                  B      →       is orthogonal to             A      →                  B      →        +  a  .            A      →                  C      →      I think the answer is  (            A      →                  B      →        )  .  (            A      →                  B      →        +  a  .            A      →                  C      →        )  =  0  and then find a.(if this isn&#039;t the answer please tell me.)but how can I write or find             A      →                  B      →      . I think with matrices because that&#039;s what we are studying now.

Cynthia Lester · Accepted Answer

First, note that                                                        A              B                        →                                              =                              B            →                          −                              A            →                                                            =        (        8        −        2        ,        −        2        −        3        ,        4        −        1        )                                          =        (        6        ,        −        5        ,        3        )                                                                    A              C                        →                                              =                              C            →                          −                              A            →                                                            =        (        3        −        2        ,        0        −        3        ,        5        −        1        )                                          =        (        1        ,        −        3        ,        4        )                                                                    A              B                        →                          +        a                                            A              C                        →                                              =        (        6        ,        −        5        ,        3        )        +        a        (        1        ,        −        3        ,        4        )                                          =        (        6        +        a        ,        −        5        −        3        a        ,        3        +        4        a        )            Then, as you have mentioned, we need to have                     A        B            →        ⋅      (                            A          B                →              +    a                            A          C                →              )    =  0  :  (  6  ,  −  5  ,  3  )  ⋅  (  6  +  a  ,  −  5  −  3  a  ,  3  +  4  a  )  =  0    36  +  6  a  +  25  +  15  a  +  9  +  12  a  =  0    33  a  +  70  =  0    ⟹    a  =  −      70    33

iroroPagbublh · Accepted Answer

Here&#039;s the answer to your question:Let   A  =      (                            2                                      3                                      1                      )   ,   B  =      (                            8                                      −          2                                      4                      )   and   C  =      (                            3                                      0                                      5                      )  .  You can compute                     A        B            →        =  B  −  A  =      (                            6                                      −          5                                      3                      )   and                     A        C            →        =  C  −  A  =      (                            1                                      −          3                                      4                      )  Note that the dot product between two vectors             u      →        ,            v      →       is given by             u      →        ⋅            v      →        =      |    u      |        |    v      |    c  o  s  (  θ  ), so             u      →        ⋅            v      →        =  0    ⟹    cos  ⁡  θ  =  0    ⟹    θ  =  π      /    2      (      90    ∘    )This way you can find a as you describe in the question:  (                    A        B            →        )  .  (                    A        B            →        +  a                    A        C            →        )  =        (    4        a    +    3    )    +  6        (    a    +    6    )    −  5        (    −    3        a    −    5    )    =  0 . This means   a  =  −      70    33

How to find a vector with two 3d points? A:(2,3,1); B:(8,-2,4); C:(3,0,5). find a so vec(A) vec(B) is orthogonal to vec(A) vec(B) +a.vec(A) vec(C).

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