Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
are coplanar vectors, then
Solution:
1675 Upvotes
Verified Answer
The correct answer is:
$\mathrm{a}, \mathrm{c}, \mathrm{b}$ are in G.P.
Given vectors are coplanar.
$$
\begin{array}{l}
\therefore\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0 \\
\therefore \mathrm{a}(0-\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{c}(\mathrm{c}-0)=0 \Rightarrow-\mathrm{ac}-\mathrm{ab}+\mathrm{ac}+\mathrm{c}^{2}=0 \\
\therefore \mathrm{c}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}, \mathrm{c}, \mathrm{b} \text { are in G.P. }
\end{array}
$$
$$
\begin{array}{l}
\therefore\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0 \\
\therefore \mathrm{a}(0-\mathrm{c})-\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{c}(\mathrm{c}-0)=0 \Rightarrow-\mathrm{ac}-\mathrm{ab}+\mathrm{ac}+\mathrm{c}^{2}=0 \\
\therefore \mathrm{c}^{2}=\mathrm{ab} \Rightarrow \mathrm{a}, \mathrm{c}, \mathrm{b} \text { are in G.P. }
\end{array}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.