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 $\overline{\mathrm{a}}, \overline{\mathrm{b}}, \overline{\mathrm{c}}$ are three vectors such that $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}+\overline{\mathrm{c}})+\overline{\mathrm{b}} \cdot(\overline{\mathrm{c}}+\overline{\mathrm{a}})+\overline{\mathrm{c}} \cdot(\overline{\mathrm{a}}+\overline{\mathrm{b}})=0$ and $|\overline{\mathrm{a}}|=1$, $|\overline{\mathrm{b}}|=8$ and $|\overline{\mathrm{c}}|=4$, then $|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|$ has the value
MathematicsVector AlgebraMHT CETMHT CET 2023 (11 May Shift 1)
Options:
  • A $81$
  • B $9$
  • C $5$
  • D $4$
Solution:
2791 Upvotes Verified Answer
The correct answer is: $9$
$$
\begin{aligned}
& |\bar{a}|=1,|\bar{b}|=8,|\bar{c}|=4, \text { and } \\
& \bar{a} \cdot(\bar{b}+\bar{c})+\bar{b} \cdot(\bar{c}+\bar{a})+\bar{c} \cdot(\bar{a}+\bar{b})=0 \\
& \Rightarrow 2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=0
\end{aligned}
$$

Now,
$$
\begin{aligned}
& |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1+64+16+0 \quad \ldots[\text { From (i) }] \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=81 \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=9
\end{aligned}
$$

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.