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If $\vec{r}$ and $\vec{s}$ arenon-zero constant vectors and the scalar $b$ is chosen such that $|\vec{r}+b \vec{s}|$ is minimum, then the value of $|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}$ is equal to
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The correct answer is:
$|\vec{r}|^{2}$
For minimum value $|\vec{r}+b \vec{s}|=0$. Let $\vec{r}$ and $\vec{s}$ are anti-parallel so $b \vec{s}=-\vec{r}$
$\therefore|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$
$\therefore|b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$
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