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If $a, b, c$ are distinct positive numbers and vectors $a \hat{\imath}+a \hat{\jmath}+c \hat{k}, \hat{\imath}+\hat{k}$
and $c \hat{\imath}+c \hat{\jmath}+b \hat{k}$ lie in a plane, then
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and $c \hat{\imath}+c \hat{\jmath}+b \hat{k}$ lie in a plane, then
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Verified Answer
The correct answer is:
$c$ is G.M. of a and b
(B)
since, three vectors are coplanar
$\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$
$\left|\begin{array}{lll}
0 & a & c \\
1 & 0 & 1 \\
0 & c & b
\end{array}\right|=0$
Expanding along $C_{1}$, we get
$-1\left(a b-c^{2}\right)=0 \Rightarrow a b=c^{2} \Rightarrow c$ is G.M. of a and b
since, three vectors are coplanar
$\left|\begin{array}{lll}
\mathrm{a} & \mathrm{a} & \mathrm{c} \\
1 & 0 & 1 \\
\mathrm{c} & \mathrm{c} & \mathrm{b}
\end{array}\right|=0$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$
$\left|\begin{array}{lll}
0 & a & c \\
1 & 0 & 1 \\
0 & c & b
\end{array}\right|=0$
Expanding along $C_{1}$, we get
$-1\left(a b-c^{2}\right)=0 \Rightarrow a b=c^{2} \Rightarrow c$ is G.M. of a and b
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