If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-negative distinct numbers and $\mathrm{a}…

If $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are non-negative distinct numbers and $\mathrm{a} \hat{\imath}+\mathrm{a} \hat{\jmath}+\mathrm{c} \hat{k}, \hat{\imath}+\hat{k}$ and $\mathrm{c} \hat{\imath}+\mathrm{c} \hat{\jmath}+\mathrm{b} \hat{k}$ are coplanar vectors, then
  1. $\mathrm{a}, \mathrm{c}, \mathrm{b}$ are in A.P.
  2. $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in G.P.
  3. $\mathrm{a}, \mathrm{c}, \mathrm{b}$ are in G.P.
  4. $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are in A.P.

Solution

Given vectors are coplanar. $\begin{array}{l} \therefore\left|\begin{array}{lll} \mathrm{a} & \mathrm{a} & \mathrm{c} \\ 1 & 0 & 1 \\ \mathrm{c} & \mathrm{c} & \mathrm{b} \end{array}\right|=0 \\ \therefore \mathrm{a}(0-\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{c}(\mathrm{c}-0)=0 \Rightarrow-\mathrm{ac}-\mathrm{ab}+\mathrm{ac}+\mathrm{c}^{2}=0 \\ \therefore \mathrm{c}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}, \mathrm{c}, \mathrm{b} \text { are in G.P. } \end{array}$

Asked in: MHT CET 2020 (15 Oct Shift 1)

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