Question

Let $\vec A(t) = {f_1}(t)\hat i + {f_2}(t)\hat j\,\,and\,\,\vec B(t) = {g_1}(t)\hat i + {g_2}(t)\hat j,$$t \in
[0,1],\,\,where\,\,{f_1},{f_2},{g_1},\,{g_2}$ are continuous functions. If $\vec A(t)\,\,and\,\,\vec B(t)$are non-zero
vectors for all t and \[\vec A(0) = 2\hat i + 3\hat j,\,\,\vec A(1) = 6\hat i + 2\hat j,\,\,\vec B(0) = 2\hat i + 3\hat
j,\,\,\vec B(1) = 6\hat i + 2\hat j\]. Then show that $\vec A(t)\,and\,\,\vec B(t)\,$are parallel for some t.

Answer 1

Hint: Firstly, put value of $\hat i\,\,and\,\,\hat
j\,\,then\,\,divide\,\,{f_1}(t),\,{g_1}(t)\,\,and\,\,{f_2}(t)\,\,to\,\,{y_2}(t)$ because they are continuous functions.
A continuous function is a function that does not have any abrupt changes in value known as discontinuities. Two vectors are parallel if they are scalar multiples of one another. If u and v are two non-zero vectors and u=cv, then u and v are parallel.
Now, Let $\vec A(t) = {f_1}(t)\hat i + {F_2}(t)\hat j$ and
$\vec B(t) = {g_1}(t)\hat i + {g_2}(t)\hat j,\,\,t \in [0,1]$, where f 1 , f 2 , g 1 , g 2 are continuous functions.
If $\vec A(t)\,\,and\,\,\vec B(t)$ are non-zero vectors for all t and $\vec A(0) = 2\hat i + 3\hat j,\,\,\vec A(1) = 6\hat i +
2\hat j$, $\vec B(0) = 2\hat i + 3\hat j,\,\,\vec B(1) = 6\hat i + 2\hat j$.
To show:- $\vec A(t)\,\,and\,\,\vec B(t)$ are parallel for some t.

Complete step-by-step answer:
$\vec A(t)$is parallel to $\vec B(t)$for some $t \in [0,1]$if and only if
$\dfrac{{{f_1}(t)}}{{{g_1}(t)}} = \dfrac{{{f_2}(t)}}{{{g_2}(t)}}$for some t [0,1].
Now, cross-multiply the numbers, then
\[{f_1}(t) \times {g_2}(t) = {f_2}(t) \times \,{g_1}(t)\] for some t [0,1]
f 1 (t) g 2 (t) – f 2 (t) g 1 (t) = 0
Let h(t) = \[{f_1}(t) \times {g_2}(t) - {f_2}(t) \times {g_1}(t)\] …..(1)
Now, put the value of t =0 in equation (1), then we will get,
h(0) = ${f_1}(0) \times {g_2}(0) - {f_2}(0) \times {g_1}(0)$
$
\therefore h(0) = 2 \times 2 - 3 \times 3 \\
h(0) = 4 - 9 \\
h(0) = - 5 < 0 \\
$
Now, put the value of t =1 in equation (1), then we will get,
$
h(1) = {f_1}(1) \times {g_2}(1) - {f_2}(1) \times {g_1}(1) \\
= 6 \times 6 - 2 \times 2 \\
= 36 - 4 \\
h(1) = 32 > 0 \\
$
Since, A is continuous function and h(0) h(1) < 0
There is some t [0,1] for which h(t) = 0.
i.e. $\vec A(t)\,\,and\,\,\vec B(t)$ are parallel for this vector t.

Note: First put the t = 0 and 1 then we will see that function is less than 0 or greater than 0 for proving two vectors are
parallel as the same variable.