Question

If $\vec{r}$ and $\vec{s}$ are non-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

• A. $2|\vec{r}{|}^2$
• B. $\frac{|\vec{r}{|}^2}{2}$
• C. $3|\vec{r}{|}^2$
• D. $|\vec{r}{|}^2$

Answer 1

Answer: (D)

For minimum value $|\vec{r}+b\vec{s}|$.
Let $\vec{r}$ and $\vec{s}$ are anti-parallel so $b\vec{s}=-\vec{r}$
$\therefore |b\vec{s}{|}^2+|\vec{r}+b\vec{s}{|}^2=|-\vec{r}{|}^2+|\vec{r}-\vec{r}{|}^2=|\vec{r}{|}^2$