MHT CET · Maths · Vector Algebra
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
- A \(\mathrm{a}, \mathrm{c}, \mathrm{b}\) are in A.P.
- B \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in G.P.
- C \(\mathrm{a}, \mathrm{c}, \mathrm{b}\) are in G.P.
- D \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) are in A.P.
Answer & Solution
Correct Answer
(C) \(\mathrm{a}, \mathrm{c}, \mathrm{b}\) are in G.P.
Step-by-step Solution
Detailed explanation
Given vectors are coplanar.
\(
\begin{array}{l}
\therefore\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0\end{array}\)
\(\therefore \mathrm{a}(0-\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{c}(\mathrm{c}-0)=0 \Rightarrow-\mathrm{ac}-\mathrm{ab}~+\) \(\mathrm{ac}+\mathrm{c}^{2}=0\)
\(\therefore \mathrm{c}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}, \mathrm{c}, \mathrm{b} \text { are in G.P.}\)
\(
\begin{array}{l}
\therefore\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0\end{array}\)
\(\therefore \mathrm{a}(0-\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{c}(\mathrm{c}-0)=0 \Rightarrow-\mathrm{ac}-\mathrm{ab}~+\) \(\mathrm{ac}+\mathrm{c}^{2}=0\)
\(\therefore \mathrm{c}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}, \mathrm{c}, \mathrm{b} \text { are in G.P.}\)
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