Q219 / 270 · NDA 2018 (Phase 1) · Mathematics

If $\vec{a}+2 \vec{b}+3 \vec{c}=\overrightarrow{0}$ and $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}=\lambda(\vec{b} \times \vec{c})$, then what is the value of $\lambda$ ?

  • A
    2
  • B
    3
  • C
    4
  • Correct answer
    D
    6

Solution

$\begin{aligned} & \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}}+3 \overrightarrow{\mathrm{c}}=\overrightarrow{0}...(1) \\ & \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})...(2) \\ & (1) \times \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow-\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+0+3 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=3 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}...(3) \\ & (1) \times \overrightarrow{\mathrm{c}} \Rightarrow \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{c}}=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}+0=0 \\ & \Rightarrow \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=2 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}....(4) \\ & \therefore(2) \Rightarrow \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \\ & \Rightarrow 3 \overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+2(\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}})=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) from (3)and(4) \\ & \Rightarrow 6(\overrightarrow{\mathrm{~b}} \times \overrightarrow{\mathrm{c}})=\lambda(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}) \\ & \therefore \lambda=6 \end{aligned}$

Hence, the correct answer is option: 'D'.

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