If $\left[\begin{array}{llll}\bar{a} \times \bar{b} & \bar{b} \times \bar{c} & \bar{c} \times…

If $\left[\begin{array}{llll}\bar{a} \times \bar{b} & \bar{b} \times \bar{c} & \bar{c} \times \bar{a}\end{array}\right]=\lambda\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]^2$, then $\lambda$ is equal to
  1. 3
  2. 0
  3. 1
  4. 2

Solution

$\begin{aligned} & {[\overline{\mathrm{a}} \times \overline{\mathrm{b}} \overline{\mathrm{b}} \times \overline{\mathrm{c}} \overline{\mathrm{c}} \times \overline{\mathrm{a}}] } \\ & =(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \times(\overline{\mathrm{c}} \times \overline{\mathrm{a}})\} \\ & =(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{((\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}-((\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{c}}) \overline{\mathrm{a}}\} \\ & =(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot\{((\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}) \overline{\mathrm{c}}\} \ldots[\because(\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{c}}=0] \\ = & \{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \cdot \overline{\mathrm{c}}\}((\overline{\mathrm{b}} \times \overline{\mathrm{c}}) \cdot \overline{\mathrm{a}}) \\ = & {[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}][\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}] \ldots .[\because[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]=[\overline{\mathrm{b}} \overline{\mathrm{c}} \overline{\mathrm{a}}]] } \\ & =[\overline{\mathrm{a}} \overline{\mathrm{b}} \overline{\mathrm{c}}]^2\end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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