Let `A(t)=f_1(t)veci+f_2(t)vecj and vecB(t)=g_1(t)veci+g_2(t)vecj, tepsilon[0,1] where f_1,f_2,g_1,g_2` are continuous functions. If `vecA (t) and vecB(t)` are non zero for all `tepsilon [0,1] and vecA(0)=2veci+3vecj, vecA(1)=6veci=2vecj, vecB(0)=3veci+2vecj and vecB(1)=2veci+6vecj` prove that `vecA(t) and vecB(t)` are parallel for some `tepsilon(0,1)`

Explanation

`vecA(t)` is parallel to `vecB(t)` for some `t in [0, 1]` if and only if `(f_1(t))/(g_1(t))= (f_2(t))/(g_2(t))` for some `t in [0, 1]`
or `f_1(t)*g_2(t) = f_2(t)g_1(t)` for some `t in [0, 1]`
Let `h(t) = f_1(t)*g_2(t) -f_2(t)*g_1(t)`
`" " h(0) = f_1(0)*g_2(0)- f_2(0)*g_1(0)`
`" " = 2xx 2-3 xx 3=-5 lt 0`
`" " h(1) = f_1(1)*g_2(1)-f_2(1)*g_1(1)`
`" " = 6xx 6-2 xx 2 = 32 gt 0`
Since h is a continuous function, and `h(0)*h(1) lt 0`, there are some `t in [0,1]` for which `h(t) =0`, i.e.,
`vecA(t) and vecB(t)` are parallel vectors for this `t`.