MHT CET · Maths · Vector Algebra
Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) be vectors of magnitude 2,3 and 4 respectively. If \(\bar{a}\) is perpendicular to \((\bar{b}+\bar{c}), \bar{b}\) is perpendicular to \((\bar{c}+\bar{a})\) and \(\vec{c}\) is perpendicular to \((\bar{a}+\bar{b})\), then the magnitude of \(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is equal to
- A 29
- B \(\sqrt{29}\)
- C 26
- D \(\sqrt{26}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{29}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \overline{\mathrm{a}} \perp(\overline{\mathrm{b}}+\overline{\mathrm{c}}), \quad \overline{\mathrm{b}} \perp(\overline{\mathrm{c}}+\overline{\mathrm{a}}) \text { and } \overline{\mathrm{c}} \perp(\overline{\mathrm{a}}+\overline{\mathrm{b}}) \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}=0, \overline{\mathrm{~b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{a}}=0, \overline{\mathrm{c}} \cdot \overline{\mathrm{a}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{b}}=0 \\ & \Rightarrow \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}=0\end{aligned}\)
\(\begin{aligned}|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2 & =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & =2^2+3^2+4^2+2(0) \\ & =4+9+16\end{aligned}\)
\(\begin{array}{ll}\therefore \quad & |\bar{a}+\bar{b}+\bar{c}|^2=29 \\ & \Rightarrow|\bar{a}+\bar{b}+\bar{c}|=\sqrt{29}\end{array}\)
\(\begin{aligned}|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|^2 & =|\overline{\mathrm{a}}|^2+|\overline{\mathrm{b}}|^2+|\overline{\mathrm{c}}|^2+2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \\ & =2^2+3^2+4^2+2(0) \\ & =4+9+16\end{aligned}\)
\(\begin{array}{ll}\therefore \quad & |\bar{a}+\bar{b}+\bar{c}|^2=29 \\ & \Rightarrow|\bar{a}+\bar{b}+\bar{c}|=\sqrt{29}\end{array}\)
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