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 $b \vec{s}=-\vec{r}$
$\therefore |b \vec{s}|^{2}+|\vec{r}+b \vec{s}|^{2}=|-\vec{r}|^{2}+|\vec{r}-\vec{r}|^{2}=|\vec{r}|^{2}$