Question 31236
if vector B is perpendicular to vector A and A is solely in the x-y plane then vector B has to be solely in the z-plane, so its vector is B=ck...some value of k... we do not know where in the z-plane it lies.


So, C = 3i+4j+ck


I think this is what you require.


jon.
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OK, after your reply, B is also in the xy plane: my mistake, sorry.


OK, in the xy plane, a.b = 0 for 2 orthogonal vectors.

we have {{{(matrix(2,1, 3,4))}}}.{{{ (matrix(2,1, x,y)) }}} = 0


so, 3x+4y=0


Now, if both terms were 12 and one negative, then we would have zero. So, how about x=4 and y=-3... that would equal zero.


So, {{{ (matrix(2,1, 3,4)) }}}.{{{ (matrix(2,1,4,-3)) }}} = 0 holds true


So, C = {{{ (matrix(2,1, 3,4)) }}}+{{{ (matrix(2,1,4,-3)) }}}
{{{ (matrix(2,1,(3+4),(4-3))) }}}
{{{ (matrix(2,1,7,1) )}}}


this is one version of the answer, since the scalar product could also have been: {{{ (matrix(2,1, 3,4)) }}}.{{{ (matrix(2,1,-4,3)) }}} = 0


in which case, B would be {{{ (matrix(2,1,-4,3)) }}}. And hence C would be:
{{{ (matrix(2,1,(3-4),(4+3))) }}}
{{{ (matrix(2,1,-1,7)) }}}


either is correct.
Jon