Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
MathematicsVector AlgebraBITSATBITSAT 2020
Options:
  • A $2|\vec{r}|^{2}$
  • B $|\vec{r}|^{2} / 2$
  • C $3|\vec{r}|^{2}$
  • D $|\vec{r}|^{2}$
Solution:
2561 Upvotes Verified Answer
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 $b \vec{s}=-\vec{r}$

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

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.