Vectors 1 Question 30
31. Let $\overrightarrow{\mathbf{A}}(t)=f _1(t) \hat{\mathbf{i}}+f _2(t) \hat{\mathbf{j}}$ and $\overrightarrow{\mathbf{B}}(t)=g(t) \hat{\mathbf{i}}+g _2(t) \hat{\mathbf{j}}$, $t \in[0,1], f _1, f _2, g _1 g _2$ are continuous functions. If $\overrightarrow{\mathbf{A}}(t)$ $\overrightarrow{\mathbf{A}}(t)$ and $\overrightarrow{\mathbf{B}}(t)$ are non-zero vectors for all $t$ and $\overrightarrow{\mathbf{A}}(0)=2 \hat{\mathbf{i}}, \quad \overrightarrow{\mathbf{A}}(1)=6 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{B}}(0)=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}} \quad$ and $\overrightarrow{\mathbf{B}}(1)=2 \hat{\mathbf{j}}+6 \hat{\mathbf{j}}$. Then, show that $\overrightarrow{\mathbf{A}}(t)$ and $\overrightarrow{\mathbf{B}}(t)$ are parallel for some $t$.
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Solution:
- $\overrightarrow{\mathbf{A}}(t)$ is parallel to $\overrightarrow{\mathbf{B}}(t)$ for some $t \in[0,1]$, if and only if $\frac{f _1(t)}{g _1(t)}=\frac{f _2(t)}{g _2(t)}$ for some $t \in[0,1]$
or $f _1(t) \cdot g _2(t)=f _2(t) \cdot g _1(t)$ for some $t \in[0,1]$
Let $h(t)=f _1(t) \cdot g _2(t)-f _2(t) g _1(t)$
$\therefore h(0)=2 \times 2-3 \times 3=-5<0$
and $h(1)=f _1(1) \cdot g _2(1)=f _2(1) \cdot g _1(1)$
$=6 \times 6-2 \times 2=32>0$
Since, $h$ is a continuous function and $h(0) \cdot h(1)<0$, Therefore, here is some $t \in[0,1]$ for which $h(t)=0$, i.e. $\overrightarrow{\mathbf{A}}(t)$ and $\overrightarrow{\mathbf{B}}(t)$ are parallel vectors for this $t$.
