Find the direction cosines of the a vector in the plane of b → =&lt; 2 , 1 , 3 &gt; and c → =&lt; − 1 , 2 , 1 &gt; and perpendicular to c → Let the vector a → = ⟨ x , y , z ⟩Since it is perpendicular to c → − x + 2 y + z = 0And | x y z 2 1 3 − 1 2 1 | ⟹ − 5 x − 7 y + 5 z = 0since a, b, c are coplanarSo y=0Then x=ySo the answer should be ⟨ 1 2 , 0 , 1 2 ⟩ , but that is not correct.What is wrong with my solution?

Question

Find the direction cosines of the a vector in the plane of             b      →        =&amp;lt;  2  ,  1  ,  3  &amp;gt; and             c      →        =&amp;lt;  −  1  ,  2  ,  1  &amp;gt; and perpendicular to             c      →      Let the vector             a      →        =  ⟨  x  ,  y  ,  z  ⟩Since it is perpendicular to             c      →        −  x  +  2  y  +  z  =  0And      |                            x                          y                          z                                      2                          1                          3                                      −          1                          2                          1                      |      ⟹    −  5  x  −  7  y  +  5  z  =  0since a, b, c are coplanarSo y=0Then x=ySo the answer should be       ⟨          1              2              ,    0    ,          1              2              ⟩   , but that is not correct.What is wrong with my solution?

Piper Pruitt · Accepted Answer

Let             r      →        =  p            b      →        +  q            c      →      , take dot product of this with             c      →      , and use             r      →        .            c      →        =  0, then            r      →        .            c      →                  =      →        p            b      →        .            c      →        +  q            c      →        .            c      →          ⟹    p  =  −  q            c      2                                b          →                    .                        c          →                      =  −  q      6    3    =  −  2  q  .So the required vector             r      →       is:            r      →        =  q  [  −  2            b      →        +            c      →        ]  =  q  (  −  5            i      →        −  5            k      →        )  .q is arbitrary real number. Next             r      ^        =                              i          →                    +                        k          →                            2      It also means that            r      →        =  α               c      →        ×  (            b      →        ×            c      →        )  ,namely, the vector triple product.

Aidyn Crosby · Accepted Answer

Note that the projection of             b      →       onto the direction perpendicular to             c      →       is            b      →        −  (            b      →        ⋅            c      →        )                    c        →                            |            c                        |                2              =&amp;lt;  2  ,  1  ,  3  &amp;gt;  −      1    2    &amp;lt;  −  1  ,  2  ,  1  &amp;gt;=&amp;lt;      5    2    ,  0  ,      5    2    &amp;gt;Normalize to get             a      →        =&amp;lt;      1          2        ,  0  ,      1          2        &amp;gt;

Find the direction cosines of the a vector in the plane of vec(b) =<2,1,3> and vec(c)=<−1,2,1> and perpendicular to vec(c)

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