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MHT CET · Maths · Vector Algebra

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

  1. A \(81\)
  2. B \(9\)
  3. C \(5\)
  4. D \(4\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(9\)

Step-by-step Solution

Detailed explanation

\(
\begin{aligned}
& |\bar{a}|=1,|\bar{b}|=8,|\bar{c}|=4, \text { and } \\
& \bar{a} \cdot(\bar{b}+\bar{c})+\bar{b} \cdot(\bar{c}+\bar{a})+\bar{c} \cdot(\bar{a}+\bar{b})=0 \\
& \Rightarrow 2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=0
\end{aligned}
\)
Now,
\(
\begin{aligned}
& |\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=1+64+16+0 \quad \ldots[\text { From (i) }] \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2=81 \\
& \Rightarrow|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|=9
\end{aligned}
\)